7 Geometry

8 Geometriy 7

1. IntroductionIn this chapter we will look at the fundamental concepts we need in order to begin our study

of geometry.

A. POINT, LINE, AND PLANE

Definition

The word geometry comes from two Greek words, ‘geo’ and ‘metric’, which together mean ‘tomeasure the earth.’ Geometry is now the branch of mathematics that studies space, shape,area, and volume.

geometry

Nature displays an infinite array of geometric shapes,from the smallest atom to the biggest galaxy. Snowflakes,the honeycomb of a bees’ nest, the spirals of seashells, spiders’ webs, and the basic shapes of many flowers arejust a few of nature’s geometric masterpieces.

The Egyptians and Babylonians studied the area and volume of shapes and established general formulas.However, the first real book about geometry was written bya Greek mathematician, Euclid. Euclid’s book, TheElements, was published in about 300 BC. It defined themost basic concepts in geometry and proved some of theirproperties.

Geometry as a science has played a great role in the development of civilization. Throughout history, geometryhas been used in many different areas such as architecture, art, house design, and agriculture.

After studying this section you will be able to:

1. Understand the fundamental geometric concepts of point, line, and plane.

2. Describe the concepts of line segment, ray, and half line.

3. Understand the concepts of plane and space.

4. Describe the relation between two lines.

5. Describe the relation between a line and a plane.

6. Describe the relation between two planes.

Objectives

9Geometric Concepts

The three most basic concepts of geometry are point, line,and plane. Early mathematicians tried to define theseterms. In fact, it is not really possible to define them usingany other concepts, because there are no simpler concepts for us to build on. Therefore, we need to understand these concepts without a precise definition.Let us look instead at their general meaning.

2. PointWhen you look at the night sky, you see billions of stars, each represented as a small dot of

light in the sky. Each dot of light suggests a point, which is the basic unit of geometry.

3. Line

Nature’s Great Book iswritten in mathemati-cal symbols.

Galileo Galilei

All geometric figures consist of collections of points, and many terms in geometry are defined

using points.

We use a dot to represent a point. We name a point with a capital letter such as A, B, C, etc.

Concept

A point is a position. It has no size, length, width, or thickness, and it is infinitely small.

point

Concept

A line is a straight arrangement of points. It is the second fundamental concept of geometry.

There are infinitely many points in a line. A line has no width or thickness, and extends

without end in both directions.

line

10 Geometriy 7

Let none unversed ingeometry enter here.

Plato

The arrows at each end of a line show that the line extends to infinity in both directions.

If any point C is on a line AB or a line d, we write C AB, or C d.

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Property

There exists exactly one line passing through any two distinct points.

By this property, a line is determined by two distinct points. However, remember that a line

consists of more than just two points. There are infinitely many points on a line.

4. Plane

A plane is suggested by a flat surface such as a table top, a wall, a floor, or the

surface of a lake. We represent a plane with a four-sided figure, like a piece of

paper drawn in perspective. Of course, all of these things are only parts of

planes, since a plane extends forever in length and in width.

We use a capital letter (A, B, C, ...) to name a plane. We write plane P, or (P),

to refer to a plane with name P.

Concept

A plane is the third fundamental concept of geometry. A plane has length and width but no

thickness. It is a flat surface that extends without end in all directions.

plane

A line is usually named by any two of its points, or by a lower-case letter.

Look at the diagram. The line that passes through points A and B is written AB. We say it isline AB. The line on the right is simply called line .

11Geometric Concepts

For example, in the diagram above, points A, B, and C lie on the same line d. Therefore A, B,

and C are collinear points. However, point P is not on line so M, P, and N are not collinear

points. We say that, M, P, and N are noncollinear ppoints.

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5. Collinear PointsDefinition

Points that lie on the same line are called collinear ppoints.

collinear ppoints

EXAMPLE 1 Look at the given figure.

a. Name the lines.

b. Write all the collinear points.

c. Give two examples of noncollinear points.

Solution a. There are three lines, AC, CN, and SR.

b. The points A, B, C, the points S, T, R, andthe points C, M, N are on the same line, sothey are collinear.

c. The points A, N, C and the points M, T, Nare not on the same line. They are exam-ples of noncollinear points.

Now consider the three noncollinear points inthe figure on the right. Since we know that twodistinct points determine a straight line, wecan draw the lines AB, AC and BC passingthrough A, B, and C. Therefore, there are threelines that pass through three noncollinear points.

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We can now understand the meaning of the terms point, line, and plane without a formal

definition. We can use these undefined terms to define many new geometric figures and

terms.

12 Geometriy 7

When we say, n triwise noncollinear points,

we mean that any three of n points are

noncollinear.

For example, the diagram opposite shows five

triwise noncollinear points. Any set of three

points in the diagram is noncollinear.

Definition

If three points are noncollinear then they are also called triwise points.

triwise ppoints

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Theorem

different lines pass through n triwise points. ( – 1)

2n n

EXAMPLE 2 How many different lines pass through each number of triwise noncollinear points?

a. 4 b. 5 c. 9 d. 22

Solutiona. b.

c. d. = = 22 (22 – 1) 22 21

2312 2

= = 9 (9 – 1) 9 8

36 lines2 2

= = 5 (5 – 1) 5 4

10 lines2 2

= = 4 (4 – 1) 4 3

6 lines2 2

Check Yourself 11. Describe the three undefined terms in geometry.

2. Name the collinear points in the figure.

3. Look at the figure.

a. Name the lines.

b. Name all the collinear points.

c. Give two examples of noncollinear points.

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4. How many different lines can pass through each number of triwise noncollinear points?

a. 8 b. 14 c. 64 d. 120

Answers1. Point: A point is a position. It has no size, length, width, or thickness, and it is infinitely

small. Line: A line a straight arrangement of points. There are infinitely many points in a

line. A line has no width or thickness, and extends without end in both directions. Plane:

A plane has length and width but no thickness. It is is a flat

surface that extends without end in all directions.

2. The points A, B, C and the points D, B, E are collinear points.

3. a. The lines: AC, AB, DG b. The points A, E, B, the points D, B, G are collinear points c.

The points A, F, G, and the points D, B, C are non collinear points.

4. a. 28 b. 91 c. 2016 d. 7140

1. Line Segment

B. LINE SEGMENT, RAY, AND HALF LINE

Definition

The line ssegment AB is the set of points

consisting of point A, point B, and all the

points between A and B. A and B are called theendpoints of the segment. We write [AB] to

refer to the line segment AB.

line ssegment

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This definition describes one type of line segment: a closed line segment. There

are three types of line segment.

a. Closed Line SegmentA line segment whose endpoints are included in the

segment is called a closed lline ssegment.

[AB] in the diagram is a closed line segment.

b. Open Line SegmentA line segment whose endpoints are excluded from the

segment is called an open lline ssegment.

The line segment AB in the diagram is an open line segment and denoted by ]AB[.

We use an empty dot ( ) to show that a point is not included in a line segment.

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Physical model of a linesegment: a piece of string.

14 Geometriy 7

EXAMPLE 3 Name the closed, open and half-open line seg-

ments in the figure on the right.

Solution Closed line segment: [AB]

Half-open line segments: [AC[, [BC[, and [BD[

Open line segment: ]CD[

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Property

If C is a point between A and B, then

[AC] + [CB] = [AB].

Using this property, we can conclude that if

three points are collinear, then one of them is

between the other points.

Point B is between the points A and C.

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c. Half-Open Line SegmentA line segment that includes only one of its endpoints is called a half-oopen lline ssegment.

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2. Ray

In the diagrams, each ray begins at a point and extends to infinity in one direction. A is the

endpoint of [AB, and C is the endpoint of [CD.

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Definition

The ray AB is the part of the line AB that contains point A and all the points on the line

segment that stretches from point A through point B to infinity. The ray AB is denoted by [AB.

ray


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1. PlaneWe can think of the floor and ceiling of a room as parts of horizontal planes. The walls of a

room are parts of vertical planes.

A point can be an element of a plane.

In the diagram, point A is an element of plane P. We can write A (P). Similarly, B (P), C (P), D (P), and E (P).

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C. PLANE AND SPACE

a. Coplanar Points

In the figure, points A, B, and C are all in the plane P. They

are coplanar points. Points K, L, and M are also coplanar

points. A, K, and M are not coplanar points, because they

do not lie in the same plane.

Definition

Points that are in the same plane are called coplanar ppoints.

coplanar ppoints

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16 Geometriy 7

b. Coplanar Lines

For example, in the figure, the lines m and n are both in

the plane P. They are coplanar lines.

Definition

Lines that are in the same plane are called coplanar llines.

coplanar llines

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2. Space

We have seen that lines and planes are defined by sets of points.

According to the definition of space, all lines and planes can be considered as subsets of space.

Definition

Space is the set of all points.

space

In the figure, the plane P is determined by the

noncollinear points A, B, and C.�

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Theorem

For any three points, there is at least one plane that contains them. For any three non-

collinear points, there is exactly one plane that contains them.

1. Intersecting LinesTwo lines that intersect each other in a plane are calledintersecting llines, or concurrent llines.

In the figure on the left, line d and line l intersect each

other at point A. They are intersecting lines.

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D. RELATION BETWEEN TWO LINES


2. Parallel LinesTwo lines are parallel if they are in the same plane and

do not have a common point.

In the figure on the left, line d and line l are parallel

lines. We write d l to show that lines d and l are par-

allel.

3. Coincident LinesTwo lines are coincident if each one contains all the

points of the other.

In the figure on the left, line d and line l are coincident

lines. We write d = l to show that lines d and l are

coincident.

4. Skew LinesTwo lines are skew if they are non-coplanar and they do

not intersect.

In the figure on the left, E and F are two non-parallel

planes. Hence, lines d and l are in different planes, and

since they do not intersect, they are skew lines.

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EXAMPLE 4 In the figure there are three intersecting lines. Decide whether each statement is true or

false.

a. point A is the intersection of l and d

b. point C is the intersection of d and l

c. point B is the intersection of l and m

Solution a. True, since point A is the common point of l and d.

b. False, since point C is not a common point of d and l.

c. True, since point B is the common point of l and m.

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18 Geometriy 7

We have seen the different possibilities for the relation between two lines. Let us look at the

possible relations between a line and a plane.

1. The Intersection of a Line and a PlaneA line can intersect a plane at one point.

In the diagram on the left, the line d intersects the plane

E at point A.

2. Parallelism of a Line and a PlaneA line can be parallel to a plane.

In the diagram on the left, there is no common point

between line d and plane E. They are parallel.

3. A Line Lies in a PlaneIf at least two points of a line lie in a plane, then the line

lies in tthe pplane. We write d (E) to show that line d lies

in plane E.

In the diagram, points A and B are in plane E, so the line

AB lies in the plane E.

E. RELATION BETWEEN A LINE AND A PLANE

F. RELATION BETWEEN TWO PLANES

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1. Parallel PlanesIf two planes have no common point, they are called parallel pplanes. We write (A) (B) to show that two

planes are parallel. The opposite walls of a room are an

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2. Intersecting PlanesIf two planes have only one common line, they are called

intersecting planes.

3. Coincident Planes If two planes have three noncollinear points in common,

they are called coincident planes. (P) and (Q) in the

figure are coincident planes. We write (P) = (Q) to show

that planes P and Q are coincident.

4. Half Planes A line in a plane separates the plane into two disjoint

regions that are called half planes. (E1) and (E2) in the

figure are half planes of (E).

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EXERCISES 1.1

20 Geometriy 7

11.. Explain why the concepts of point, line, and plane

cannot be defined in geometry.

22.. Draw five points on a piece of paper, and make

sure that no three are of them collinear. Draw all

the lines passing through these points. How many

lines can you draw?

33.. Explain the difference between a ray and a half

line.

44.. At least how many points determine a line?

55.. At least how many noncollinear points determine

a plane? Why?

66.. Give examples from daily life to illustrate the

concepts of point, line, and plane.

77.. Write words to complete the sentences.

a. A point has no __________ and no __________.

b. Two points determine a ___________ .

c. Three noncollinear points determine a _____ .

d. Two lines that lie in different planes and do

not intersect are called ___________ lines.

88..

Determine whether the statements are true or

false for the given figure.

a. A, B, and C are collinear points

b. points D and E are not in the line l

c. B l

d. E l

e. C, D, and E are noncollinear

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1133.. Write the coplanar points

in the given figure.

1144.. Draw a diagram to show that the intersection of

two planes can be a line.

1155.. Draw a diagram to show that the intersection of

three planes can be a point.

1122.. Describe the intersection of the line and the

plane in each figure.

1166.. Look at the figure.

a. How many planes are

there?

b. Write the intersection

of the planes.

c. How many lines pass through each point?

1111.. Write the meaning of the following.

a. [CD] b. [PQ[ c. ]AB[ d. [KL e. ]MN f. EF

1100.. Name all the lines, rays, line segments, and half

lines in the given figure.

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99.. How many different lines can pass through each

number of triwise noncollinear points?

a. 5 b. 7 c. 21 d. 101

a. b. c.

21Chapter Review Test 1A

CHAPTER REVIEW TEST 1A

11.. Which concept is precisely defined in geometry?

A) point B) line C) plane

D) space E) ____

22.. A plane has no

A) thickness. B) length. C) width.

D) surface. E) ____

33.. A ray with an open endpoint is called

A) a line. B) a half line.

C) a line segment. D) an open line segment.

E) ____

44.. According to the figure,

which statement is true?

A) A, B and E are collinear points

B) l d = {B}

C) C l

D) D, B, and E are noncollinear points

E) _

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which statement is

false?

A) l d = {C} B) l m = {A}

C) l d m = {A, B, C} D) m d = {B}

E) ____

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88.. Space is

A) the intersection of two planes.

B) the set of all points.

C) a subset of a plane.

D) a very large plane.

E) ?

77.. How many lines do five points determine if no

three of the points are collinear?

A) 15 B) 12 C) 10 D) 9 E) ?

66.. ABCD is a rectangle in a plane P. E is a point such

that E (P). How many planes are there that

include point E, with one or more of points A, B,

C, and D?

A) 7 B) 8 C) 9 D) 10 E) ?

22 Geometriy 7

1122.. Let [AB] + [BC] = [AC], and [MN] + [NK] = [MK].

Which points are between two other points?

A) A and M B) B and M

C) C and K D) B and N

E) ____


which statement is

false?

A) C (E)

B) l (E) = {A}

C) l d = {B}

D) l and d are skew lines

E) ?

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which statement is false?

A) (P) l = l

B) (P) m = m

C) (P) n = n

D) l m n = {A}

E) ?

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1100.. Which figure shows ]AB?

A) B)

C) D)

E) ____

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24 Geometriy 7

1. AngleOne of the basic figures in geometry is the angle.

A television antenna is a physical model of an angle.

Changing the length of the antenna does not change the

angle. However, moving the two antennae closer

together or further apart changes the angle.

A. REGIONS OF AN ANGLE

Definition

An angle is the union of two rays that have a common endpoint. The rays are called the sides

of the angle. The common endpoint is called the vertex of the angle.

angle

Look at the diagram. [BA and [BC are the sides of the

angle. The vertex is the common endpoint B.

The symbol for an angle is . We name the angle in the

diagram ABC, or CBA, and say ‘angle ABC’, or ‘angle

CBA’. We can also name angles with numbers or

lower-case letters, or just by their vertex.

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NoteIn three-letter angle names the letter in the middle must always be the vertex.


1. Define the concept of angle and the regions an angle forms.

2. Measure angles.

3. Classify angles with respect to their measures.

4. Classify angles with respect to their positions.

5. Classify angles with respect to the sum of their measures.

Objectives

25Angles

EXAMPLE 1 Name the angles in the diagrams.

Solution a. AOB or BOA b. A

c. 1 d. a

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a. b. c. d.

EXAMPLE 2 Answer the questions for the angle ABC on the right.

a. Which points are in the interior region of the angle?

b. Which points lie on the angle?

c. Which points are in the exterior region of the angle?

Solution a. The points D and E are in the interior region of the angle.

b. The points A, B, C, and H lie on the angle.

c. The points G and F are in the exterior region of the angle.

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Definition

The region that lies between the sides (arms) of an angle is called the interior rregion of the angle.

The region that lies outside an angle is called the exterior rregion of the angle.

interior aand eexterior rregion oof aan aangle

B. MEASURING ANGLESAngles are measured by an amount of rotation. We measure this rotation in unitscalled degrees. One full circle of rotation is360 degrees. We write it as 360°.

We can show the size of an angle on a diagramusing a curved line between the two rays at thevertex, with a number. When we write the sizeof an angle, we write a lowercase m in front ofthe angle symbol.

For example, mAOB = 45° means that angle AOB measures 45° degrees.

Look at some more examples of angle measures in the diagrams.

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Babylonian astronomers chosethe number 360 to represent onefull rotation of a ray back on toitself.

Why this number was chosen?

It is because 360 is close to thenumber of days in a year and it isdivisible by 2, 3, 4, 5, 6, 8, 9, 10,12, and many other numbers.

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26 Geometriy 7

Notice that the symbol for a 90° angle is a small square at the vertex. A 90° angleis also called a right aangle in geometry.

It is important to read angles carefully in geometry problems. For example, anangle in a problem might look like a right angle (90°). However, if it is not labelledas a right angle, it may be a different size. We can only use the given informationin a problem. We calculate other information using the theorems in geometry.

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To measure angles with a protractor, follow the three steps below.

1. Place the central hole (dot) of the protractor on the vertex of the angle.

2. Place the zero measure on the protractor along one side of the angle.

3. Read the measure of the angle where the other side of the angle crosses the protractor’s scale.

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Notice that there are two semicircular scales of numbers on the protractor. If the angle measure is smaller than 90° then we read the angle using the scale with the smaller number.If the angle measure is greater than 90° then we use the scale with the larger number.

Definition

The geometric tool we use to measure angles on paper is called a protractor.

A protractor has a semi-circular shape and a scale with units from 0 to 180.

protractor

EXAMPLE 3 Read the protractor to find the measure of

each angle.

a. mAOB b. mAOC c. mAOD

d. mAOE e. mAOF f. mBOC

g. mCOF h. mDOE

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27Angles

Solution a. mAOB = 22° b. mAOC = 68°

c. mAOD = 90° d. mAOE = 140°

e. mAOF = 175° f. mBOC = mAOC – mAOB = 68° – 22° = 46°

g. mCOF = mAOF – mAOC = 175° – 68° = 107°

h. mDOE = mAOE – mAOD = 140° – 90° = 50°

For example, let us use a protractor to draw an angle of 56°.

1. Draw a ray.

2. Place the centre point of the protractor on the endpoint (A) of the ray.

Align the ray with the base line of the protractor.

4. Remove the protractor and draw [AC.

3. Locate 56° on the protractor scale. Make a dot at that point and label it as C.

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After learning to how use a protractor we can easily draw and measure angles.

Check Yourself 11. Name the following angles.

2. Find the following sets of points in the figure.

a. L {X}

b. int L {X}

c. ext L L

d. int L ext L

e. int L {S}

f. L {T, S, }

g. int L {Z, Y}

h. ext L {Z, Y}

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a. b. c. d.

28 Geometriy 7

3. Measure each angle using a protractor.

4. Draw the angles.

a. 45° b. 83° c. 174° d. 180° e. 225°

Answers1. a. AOB b. A c. 3 d. b

2. a. b. {x} c. d. e. f. {S, K} g. h. {Z, Y}

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We can classify angles according to their measures.

1. Acute AngleAn angle that measures less than 90° is called an acute

angle.

The angles on the left are all examples of acute angles

because they measure less than 90°.

2. Right AngleAn angle that measures exactly 90° is called a right aangle.

The angles on the left are all examples of right angles

because they measure exactly 90°. We use a special

square symbol at the vertex to show a right angle.

3. Obtuse AngleAn angle that measures between 90° and 180° is called

an obtuse aangle.

The angles on the left are all obtuse angles.

4. Straight AngleAn angle that measures exactly 180° is called a straight

angle. In the diagram, A is a straight angle.

C. TYPES OF ANGLE WITH RESPECT TO THEIR MEASURES

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29Angles

5. Complete AngleAn angle that measures exactly 360° is called a complete

angle. In the diagram, E is a complete angle. ��

EXAMPLE 4 Classify the angles according to their

measure.

Solution a. 180° is a straight angle.

b. 360° is a complete angle.

c. 125° is between 90° and 180°, so

it is an obtuse angle.

d. 90° is a right angle.

e. 35° is less than 90°, so it is an acute angle.

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d. e.

D. TYPES OF ANGLE WITH RESPECT TO THEIR POSITION

In the diagram, the angles AOC and BOC have a

common vertex and a common side ([OC) with

non-intersecting interior regions.

Therefore, AOC and BOC are adjacent angles.

EXAMPLE 5 Determine whether the pairs of angles are vertical or not,

using the figure.

a. a, b b. a, c c. d, a d. b, d

Solution The lines l and k intersect at one point. Therefore,

a. a and b are not vertical angles,b. a and c are vertical angles, because they are in opposite directions,c. d and a are not vertical angles, andd. b and d are vertical angles.

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1. Adjacent Angles

Definition

Adjacent aangles are two angles in the same plane that have a common vertex and a common

side, but do not have any interior points in common.

adjacent aangles

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TheoremIf two angles are vertical then they are also congruent, i.e. they have equal measures.

30 Geometriy 7

E. TYPES OF ANGLE WITH RESPECT TO THE SUM OFTHEIR MEASURES

Each angle is called the complement of the other angle.

For example, in the diagram opposite, ANB and CMD

are complementary angles, because the sum of their

measures is 90°:

mANB + mCMD = 30° + 60° = 90°.

1. Complementary AnglesDefinition

If the sum of the measures of two angles is 90°, then the

angles are called complementary aangles.

complementary aangles

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In the diagram, XYZ and MNO are supplementary

angles because the sum of their measures is 180°:

mXYZ + mMNO = 40° + 140° = 180°.

2. Supplementary Angles

Definition

If the sum of the measures of two angles is 180°, then the angles are called supplementary

angles. Each angle is called the supplement of the other angle.

supplementary aangles

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EXAMPLE 6 Find x if the given angles are

complementary.

Solution a. If x and 2x are

complementary, then

x + 2x = 90°.

Therefore, x = 30°.

b. 2x + 10° + 3x + 30° = 90°

2x + 3x + 10° + 30° = 90°

5x = 50°

x = 10°

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a. b.

31Angles

EXAMPLE 7 Find x if the given angles are

supplementary.

Solution a. If 2x and 4x are supplemen-

tary, then

2x + 4x = 180°.

Therefore, x = 30°.

b 2x + 60° + 3x + 50° = 180°

2x + 3x + 60° + 50° = 180°

5x + 110° = 180°

5x = 70°

x = 14°

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a. b.

Check Yourself 21. Find x if the given angles are complementary.

2. Find x if the given angles are supplementary.

Answers1. a. 18° b. 30° c. 9°

2. a. 20° b. 25° c. 30°

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32 Geometriy 7

A. CORRESPONDING ANGLES AND ALTERNATE ANGLESDefinition

Let m and n be two lines in a plane. A third line l that intersects each of m and n at different

points is called a transversal of m and n.

In the diagram, line AB is a transversal of m and n.

supplementary aangles

Definition

In a figure of two parallel lines with a transversal, the

angles in the same position at each intersection are

called corresponding aangles.

corresponding aangles

In the diagram, 1 and 5 are corresponding angles.

Also, the angle pairs 2 and 6, 3 and 7, and 4 and

8 are corresponding angles.

1. Corresponding Angles

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Let us look at the types of angle formed in the figure of two parallel lines with a transversal.

Remember the notation for parallel lines: m n means that m is parallel to n.

Property

Corresponding angles are congruent.

Therefore, in the diagram, m1 = m5,

m2 = m6,

m3 = m7, and

m4 = m8.

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1. Identify corresponding angles, alternate interior angles, and alternate exterior angles.

2. Identify interior angles on the same side of a transversal.

3. Describe the properties of angles with parallel sides.

4. Define an angle bisector.

Objectives

33Angles

In the diagram, the angles 4 and 6 are alternate

interior angles. Also, 3 and 5 are alternate interior

angles.

2. Alternate Interior Angles

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Property

Alternate exterior angles are congruent.

Therefore, in the diagram,

m1 = m7, and m2 = m8.

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Definition

In a figure of two parallel lines with a transversal, the interior angles between the parallellines on opposite sides of the transversal are called alternate iinterior aangles.

alternate iinterior aangles

In the diagram, the angles 1 and 7 are alternate

exterior angles. Also, 2 and 8 are alternate exterior

angles.

3. Alternate Exterior Angles

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Definition

In a figure of two parallel lines with a transversal, the angles outside the parallel lines on

opposite sides of the transversal called alternate eexterior aangles.

alternate eexterior aangles

Property

Alternate interior angles are congruent.

Therefore, in the diagram,

m4 = m6, and m3 = m5.

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34 Geometriy 7

4. Interior Angles on the Same Side of a TransversalDefinition

In a figure of two parallel lines intersected by a

transversal, interior angles on the same side of the trans-

versal are supplementary.

Therefore, in the diagram, mx + my = 180°.

alternate iinterior aangles

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If [BA [CF then ABC

and x are supplementary.

mABC + mx = 180°

100° + mx = 180°

mx = 80°

EDC and y are

alternate interior

angles.

mEDC = my

my = 30°

mBCD = mx + my

= 80° + 30°

= 110°

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EXAMPLE 8 In the digaram, [BA DE.

Find mBCD.

Solution

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5. Angles with Parallel SidesTheorem

The measures of two angles with parallel sides in the same direction are equal.

Proof Consider the diagram on the right.

1. AOB and LTB are corresponding angles.

AOB LTB

2. LTB KLM (corresponding angles)

AOB KLM

mAOB = mKLM�

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[OA [LK

[OB [LM

35Angles

Proof�

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mCAB = mFDE

2x + 40° = 6x – 20°

40° + 20° = 6x – 2x

60° = 4x

15° = x

So mCAB = 70°.

EXAMPLE 9 In the figure, [AC [DF, [AB [DE, mCAB = 2x + 40°,

and mFDE = 6x – 20°. Find mCAB.

Solution

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mAOB = mOAE + mOBF

mAOB = 40° + 30°

= 70°

EXAMPLE 10 In the figure, [AE [BF, mA = 40°, and mB = 30°.

Find mAOB.

Solution

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TheoremThe measures of two angles with parallel sides in opposite directions are equal.

1. KLM BPL (corresponding angles)

2. AOB BPL AOB KLM

Property

In the figure, if d k and B is the intersection of [BA and

[BC, then

mb = ma + mc.

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36 Geometriy 7

Property

In the figure, if d k and B is the intersection of [BA and

[BC, then

ma + mb + mc = 3360°.

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Property

In a figure such as the figure opposite, the sum of the

measures of the angles in one direction is equal to the

sum of the measures of the angles in the other direction.

mx + my + mz = ma + mb + mc + md

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mFHC + mFHD = 180° (supplementary angles)

mFHC = 180° – 2x

mAEF + mEFH + mFHC = 360°

5x + 3x + 180° – 2x = 360°

6x + 180° = 360°

6x = 180°

x = 30°

EXAMPLE 11 In the figure, AB CD, mAEF = 5x, mEFH = 3x, and

mFHD = 2x. Find x.

Solution

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35° + 25° = 4x + x

60° = 5x

x = 12°

Therefore, mABC = 48°.

EXAMPLE 12 In the figure, [AE [DF, mEAB = 35°, mBCD = 25°,

and mABC = 4 mCDF. Find mABC.

Solution

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AB CD

37Angles

Property

In the diagram, if [OA [LK and [OB [LB then

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EXAMPLE 13 In the figure, [BA [FD, [BC [FE, [BA [FG,

mABC = 60°, and mGFE = x. Find x.

Solution

Let us draw a line [BF parallel to [ED.

mDEB + mEBF = 180°

118° + mEBF = 180°

mEBF = 62°

mBAC + mABF = 180°

mBAC + 85° = 180°

mBAC = 95°

EXAMPLE 14 In the figure, [AC [ED, mABE = 23°,

and mBED = 118°, Find mBAC.

Solution ��

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Two lines are called perpendicular lines if they intersect

at right angles. We write AB CD to show that two lines

AB and CD are perpendicular.

(interior angles on the same

side of a transversal)

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mGFD = 90°

mGFD + mGFE + mDFE = 360°

90° + mGFE + 120° = 360°

90° + x + 120° = 360°

x + 210° = 360°

x = 150°

mABC + mDFE = 180°

60° + mDFE = 180°

mDFE = 120°"#$

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38 Geometriy 7

In the figure, [OB is the bisector of AOC:

mAOB = mBOC

= AOC.12

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Let us draw a line [BK parallel to [AD.

mDAB+mABK =180°

112° + mABK =180°

mABK =68°

mBCE =mABC + mABK

120° =mABC + 68° mABC =52°

EXAMPLE 15 In the figure, [AD [CE, mDAB = 112°,

and mBCE = 120°.

Find mABC.

Solution

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6. Bisector of an AngleDefinition

A ray that divides an angle into two congruent angles is called the bisector of the angle.

angle bbisector

mABD + mCBD = 180°

mEBF = 90°m ABD m CBD+ =

2 2

EXAMPLE 16 In the figure, [BE and [BF are the bisectors of ABD and

CBD respectively. Find mEBF.

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Definition

The bisectors of two adjacent supplementary angles are perpendicular to each other.

39Angles

Let [OH AG. mHOC = mGCF = cmHOB = mCBE = bmHOD = mBAD = aLet mCOB = mBOA = x.c + x = b x = b – c b + x = ab + b – c = aTherefore, a + c = 2 b.

EXAMPLE 17 In the figure, [OE is the bisector of FOD.

mBAD = a, mEBC = b, and mFCG = c.

Show that a + c = 2 b.

Solution

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EXERCISES 2

40 Geometriy 7

11.. Using the given fig-

ure, find each set of

points.

a. O {P}

b. O {N} c. O {K, O, M}

d. int O {P} e. int O {N}

f. int O {K, O, M} g. ext O {N}

h. ext O {P} i. O int O

j. O ext O k. int O ext O O

44.. Draw the angles.

a. 20° b. 35° c. 75° d. 120°

e. 175° f. 210° g. 240° h. 330°�

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22.. Find each set of

points for the given

figure.

a. ABC ACD

b. int ABC CAD

c. ABC int CAD

d. ext ABC CAD

e. ABC ext CAD

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33.. Measure the angles using a protractor.

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55.. Classify the types of angle.

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66.. Find x in each figure if the angles are

complementary.

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77.. Find x in each figure if the angles are

supplementary.

88.. In the figure, m n,

l is a transversal and

m7 = 115°.

Find the measures.

a. m1 b. m2

c. m3 d. m4

e. m5 f. m6

g. m8

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41Angles

99.. Given [BA [DE, find

mx.

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1100.. In the figure,

[BA [ED,

[BC [EF,

mABC = 3x – 30°, and

mDEF = 4x – 70°.

Find x.�

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[BA [DE,

mBCD = 40°, and

mCDE = 120°.

Find mABC.

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[BA [FG,

mEFG = 120°, and

mABC = 130°.

Find mx.

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[BA [ED] and

[CD] [EF.

Find the relation

between x, y, and z.

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[BA [EF,

mBCD = 100°,

mCDE = 25°, and

mFED = 105°.

Find mABC.

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1177.. In the figure, d l.

Find x.

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[BA [DE,

[BC [DF and

[BC [BD], and

mGDE = 40°.

Find mABC.��

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[BC [DF,

[BA [DG, and

[ED] is the angle

bisector of mGDF.

Find mABC.

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[AB [CD.

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AB CD.

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d l.

Find mx.

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d l.

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CHAPTER REVIEW TEST 2

42 Geometriy 7


[BC is the angle

bisector of ABC.

Find mx.

A) 65° B) 55° C) 50° D) 45°

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11.. The complement of an angle x is 10° more than

three times mx. Find the measure of the bigger

angle.

A) 50° B) 60° C) 70° D) 80°

22.. The sum of the measures of the supplementary

and complementary angles of an angle x is 250°.

Find mx.

A) 10° B) 20° C) 30° D) 40°

33.. What is the measure of the angle between the

bisectors of two adjacent supplementary angles?

A) 45° B) 60° C) 75° D) 90°

44.. In the figure, [OA [OB,

mAOC = a,

mCOB = b, and

. Find b.

A) 30° B) 36° C) 54° D) 60°

23

a =b

66.. In the figure, l k.

Find mx.

A) 110° B) 100° C) 90° D) 80°

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55.. The ratio of two complementary angles is . Find

the measure of the supplementary angle

of the smaller angle.

A) 170° B) 160° C) 150° D) 110°

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77.. In the figure,

m n,

mKAB = 130°, and

mLCD = 40°.

Find mABC.

A) 100° B) 90° C) 80° D) 70°

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88.. In the figure,

d l. Find x.

A) 40° B) 30° C) 20° D) 10°

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99.. In the figure,

[AB] [BE.

Find mx.

A) 30° B) 40° C) 50° D) 60°

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[BC [DE and

[BA [DF.

What is the relation

between mx and

my?

A) mx + my = 90° B) mx + my = 180°

C) mx = my D) mx – my = 30°

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44 Geometriy 7

A. THE TRIANGLE AND ITS ELEMENTSThe roofs of many buildings have a triangular cross-section. A triangle makes a simple musi-

cal instrument, and many traffic signs have a triangular shape. These are just some exam-

ples of how triangles are used in the world around us.

In this section we will consider the main features of triangles and how we can use them to

solve numerical problems.

1. DefinitionThe word triangle means ‘three angles’. Every triangle has three angles and three sides.

Definition triangle, vvertex, sside

A triangle is a plane figure which is formed by three line segments joining three noncollinear

points. Each of the three points is called a vertex of the triangle. The segments are called thesides of the triangle.

The plural of vertex isvertices.

We name a triangle with the symbol followed by three capital letters, each

corresponding to a vertex of the triangle. We

can give the letters in any order, moving

clockwise or counterclockwise around the

triangle.

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Challenge!Without lifting your pencil,join the following four

points with three segmentsto form a closed figure.

Activity

Make a poster to show how triangles are used in everyday life. You can take photographs,

make drawings or collect pictures from magazines or newspapers to show buildings,

designs, signs and artwork which use triangles.

MMaakkiinngg aa PPoosstteerr - TTrriiaanngglleess


1. Define a triangle.

2. Name the elements of a triangle.

3. Describe the types of triangle accordin to sides.

4. Describe the types of triangle according to angles.

Objectives

45Triangles and Construction

Definition interior aand eexterior aangles oof aa ttriangle

In a triangle ABC, the angles BAC, ABC and

ACB are called the interior aangles of the

triangle. They are written as A, B and C,

respectively. The adjacent supplementary

angles of these interior angles are called theexterior aangles of the triangle. They are

written as A, B and C, respectively.

We can refer to the sides of a triangle ABC by

using the line segments AB, BC and AC, or by

using the lower-case form of the vertex

opposite each side.

For instance, in ABC at the right:

a is the side opposite vertex A,

b is the side opposite vertex B, and

c is the side opposite vertex C.

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B Ca

bc

EXAMPLE 1 Look at the figure.

a. Name all the triangles in the figure.

b. Name all the interior angles of MNE.

c. Name all the vertices of NEP.

d. Name all the sides of MNP.

e. Name all the exterior angles of ENM.

Solution a. MNE, NEP and MNP

b. M (or NME), MNE and MEN.

c. points N, E and P

d. segment MP, segment PN and segment

NM

e. E, N and M

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Notice that a triangle isdefined as the union ofthree line segments. Sincean angle lies between tworays (not two line segments),a triangle has no angles bythis definition. However,we can talk about theangles of a triangle byassuming the existenceof rays: for example, therays AB and AC formangle A of a triangle ABC.

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For example, we can refer to the triangle shown at the right as ABC. We can also call it BCA,

CAB, ACB, BAC or CBA. The vertices of ABC are the points A, B and C. The sides of ABC

are the segments AB, BC and CA.

46 Geometriy 7

EXAMPLE 2 In the figure, P(ABC) = P(DEF). Find x.

Solution P(ABC) = P(DEF)

x + 2 + x +10 = 16 + 14 + x (given)

2x + 12 = x + 30

x = 18

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Check Yourself 11. Three distinct points K, M and N lie on a line m, and a fourth point T is not on the line

m. Point T is joined to each of the other points. Find how many triangles are formed and

name each one. 2. Find and name all the triangles in the figure at the right.

3. Polygon ABCDE is a regular polygon and its diagonals are

shown in the figure. Name

a. all the triangles whose three vertices lie on the polygon.

b. all the triangles which have exactly one vertex on the

polygon.

c. all the triangles which have two sides on the polygon.

d. the sides of all the triangles which do not have a side on the polygon.

4. The side AC of a triangle ABC measures 12.6 cm, and the two non-congruent sides AB and

BC are each 1 cm longer or shorter than AC. Find P(ABC).

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A regular ppolygon is apolygon in which all sideshave the same length andall angles are equal.

Triangles in the worldaround us

For instance, the perimeter of the triangle

ABC in the figure is

P(ABC) = BC + CA + AB = a + b + c.

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Challenge!Move exactly threetoothpicks in the followingarrangement to make fivetriangles.

Definition perimeter oof aa ttriangle

The sum of the lengths of the three sides of a triangle is called the perimeter of the triangle.

We write P(ABC) to mean the perimeter of a triangle ABC.


5. Point X is on the side KN of a triangle KMN. Find the length

of MX if the perimeters of the triangles KXM, XMN and KMN

are 24, 18, and 30, respectively.

Answers1. Three triangles are formed: KMT, MNT and TKN.

2. AEL, LEB, LBC, AKL, AGK, ALB, ABC, AFC,

ADF, AGL, ADC

3. a. ABC, BCD, CDE, DEA, EAB

b. BKL, CLM, DMN, ENP, APK

c. ABC, BCD, CDE, DEA, EAB

d. sides of BKL: BK, KL, BL; sides of CLM: CL, ML, CM; sides of DMN: DM, MN, DN;

sides of ENP: EN, NP, EP; sides of APK: AP, KP, AK

4. 37.8 cm 5. 6

2. Regions of a TriangleAny given triangle ABC separates the plane which contains it into three distinct regions:

1. The points on the sides of the triangle form the ttriangle iitself.

2. The set of points which lie inside the triangle form the interior oof tthe ttriangle, denotedint ABC.

3. The set of points which lie outside the

triangle form the exterior oof tthe ttriangle,

denoted ext ABC.

The union of a triangle with its interior and

exterior region forms a plane. In the figure

opposite, the plane is called E. We can write

E = int ABC ABC ext ABC.

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EXAMPLE 3 Write whether each statement is true or falseaccording to the figure opposite.

a. Point T is in the interior of DFE.

b. M ext BDE

c. ADF BED =

d. ext FDE int FCE = FCE

e. Points T and K are in the exterior of DFE.

Solution a. false b. true c. false d. false e. true

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The picture shows the ‘food triangle’ of different types of food.Can you see what the differentregions mean?

48 Geometriy 7

We usually use the capital letter V to indicate

the length of a median. Accordingly, the

lengths of the medians from the vertices of a

triangle ABC to each side a, b and c are

written as Va, Vb and Vc, respectively. As we

can see, every triangle has three medians.

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3. Auxiliary Elements of a TriangleThree special line segments in a triangle can often help us to solve triangle problems. These

segments are the median, the altitude and the bisector of a triangle.

Definition median

In a triangle, a line segment whose endpoints are a vertex and the midpoint of the side

opposite the vertex is called a median of the triangle.

In the figure, the median to side BC is the

line segment AD. It includes the vertex A and

the midpoint of BC.

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a. Median

Auxiliary eelements areextra or additionalelements.

Check Yourself 2Answer according to the figure.

a. Name five points which are on the triangle.

b. Name three points which are not on the triangle.

c. Name two points which are in the exterior of the triangle.

d. What is the intersection of the line ST and the triangle ABC?

e. What is the intersection of the segment NS and the exterior of the triangle ABC?

Answersa. points A, B, C, T and S b. points J, L and N c. points J and L d. points S and T

e.

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A physical model of atriangle with its interiorregion


EXAMPLE 4 Name the median indicated in each triangle and indicate its length.

Solution a. median MD, length Vm

b. median TE, length Vt

c. median PF, length Vp

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a. b. c.

Activity

1. Follow the steps to construct the median of a triangle by paper folding.

2. Cut out three different triangles. Fold the triangles carefully to construct the three

medians of each triangle. Do you notice anything about how the medians of a triangle

intersect each other?

Take a triangular piece of paperand fold one vertex to anothervertex. This locates themidpoint of a side.

Fold the paper again from themidpoint to the opposite vertex.

DM is the median of EF.

PPaappeerr FFoollddiinngg - MMeeddiiaannss

Definition centroid oof aa ttriangle

The medians of a triangle are concurrent. Their common point is called the centroid of the

triangle.

50 Geometriy 7

In the figure, AN is the angle bisector which

divides BAC into two congruent parts.

We call this the bisector of angle A

because it extends from the vertex A.

Since AN is an angle bisector, we can write

m(BAN) = m(NAC).

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We usually use the letter n to indicate the

length of an angle bisector in a triangle.

Hence the lengths of the angle bisectors of a

triangle ABC from vertices A, B and C are

written nA, nB and nC, respectively. As we can

see, every triangle has three angle bisectors.

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The centroid of a triangle is the center of gravity of the triangle. In other words, a triangular

model of uniform thickness and density will balance on a support placed at the centroid of

the triangle. The two figures below show a triangular model which balances on the tip of a

pencil placed at its centroid.

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Definition triangle aangle bbisector

An angle bbisector of a triangle is a line segment which bisects an angle of the triangle and

which has an endpoint on the side opposite the angle.

b. Angle bisector

Concurrent llines are lineswhich all pass through acommon point.


Definition incenter oof aa ttriangle

The angle bisectors in a triangle are concurrent

and their intersection point is called theincenter of the triangle. The incenter of a

triangle is the center of the inscribed circle of

the triangle.

As an exercise, try drawing a circle centered at the incenter of each of your triangles fromthe previous activity. Are your circles inscribed circles?

We have seen that nA, nB and nC are the bisectors of the interior angles of a triangle ABC. Wecan call these bisectors interior aangle bbisectors. Additionally, the lengths of the bisectors ofthe exterior angles A, B and C arewritten as nA, nB and nCrespectively. Thesebisectors are called the exterior aanglebisectors of the triangle.

In the figure at the right, segment KN is theexterior angle bisector of the angle K inKMT and its length is nK.

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The inscribed ccircle of atriangle is a circle whichis tangent to all sides ofthe triangle.

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Activity

Follow the steps to explore angle bisectors in a triangle.

1. Cut out three different triangles.

2. Fold the three angle bisectors of each triangle as shown below.

3. What can you say about the intersection of the angle bisectors in a triangle?

PPaappeerr FFoollddiinngg - AAnnggllee BBiisseeccttoorrss

Folding the angle bisector of A. AN is the angle bisector of A. BM is the angle bisector of B.

52 Geometriy 7

EXAMPLE 5 Find all the excenters of KMN in the figure

by construction.

Solution To find the excenters, we first construct the

bisector of each exterior angle using the

method we learned in Chapter 1. Then we

use a straightedge to extend the bisectors

until they intersect each other.

The intersection points E1, E2 and E3 are the

excenters of KMN.

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An escribed ccircle of atriangle is a circle which istangent to one side of thetriangle and the extensionsof the other two sides.

Definition excenter oof aa ttriangle

The bisectors of any two exterior angles of a

triangle are concurrent. Their intersection is

called an excenter of the triangle.

In the figure, ABC is a triangle and the bisectors

of the exterior angles A and C intersect at

the point O. So O is an excenter of ABC. In

addition, O is the center of a circle which is

tangent to side AC of the triangle and the

extensions of sides AB and BC of the triangle.

This circle is called an escribed ccircle of

ABC.

As we can see, a triangle has three excenters and three corresponding escribed circles.

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Definition altitude oof aa ttriangle

An altitude of a triangle is a perpendicular line segment from a vertex of the triangle to the

line containing the opposite side of the triangle.

In the figure, AH is the altitude to side

BC because AH is perpendicular to BC.

In a triangle, the length of an altitude is called a height of the triangle.

The heights from sides a, b and c of a triangle

ABC are usually written as ha, hb and hc,

respectively. As we can see, every triangle has

three altitudes.

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c. Altitude

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EXAMPLE 6 Name all the drawn altitudes of all the

triangles in the figure.

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Solution There are eight triangles in the figure. Let us look at them one by one and name the drawn

altitudes in each.

54 Geometriy 7

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Activity

To fold an altitude, we fold a triangle so that a side matches up with itself and the fold

contains the vertex opposite the side.

Cut out three different triangles. Fold them carefully to construct the three altitudes ofeach triangle. What can you say about how the altitudes intersect?

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Definition orthocenter oof aa ttriangle

The altitudes of a triangle are concurrent. Their common point is called orthocenter of the

triangle.

Since the position of the altitudes of atriangle depends on the type of triangle, theposition of the orthocenter relative to thetriangle changes. In the figure opposite, theorthocenter K is in the interior region of thetriangle. Later in this chapter we will lookat two other possible positions for theorthocenter.

Once we know how to draw an altitude of a triangle, we can use it to find the area of thetriangle.

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Definition area oof aa ttriangle

The area of a triangle is half the product of the length of a side (called the base of the

triangle) and the height of the altitude drawn to that base. We write A(ABC) to mean the

area of ABC.

For example, the area of ABC in the figure

is Area is usually

expressed in terms of a square unit.

( )= = .2 2

BC AH a hA ABC

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EXAMPLE 7 Find the area of each triangle.

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Solution a. (Definition of the area of a triangle)

(Substitute)

= 20 cm2 (Simplify)

10 4=

2

( )=2

BC AHA ABC

56 Geometriy 7

Definition circumcenter oof aa ttriangle

The intersection point of the perpendicular bisectors of a triangle is called the circumcenter

of the triangle. The circumcenter of a triangle is the center of the circumscribed circle of the

triangle.

The circumscribed ccircleof a triangle is a circlewhich passes through allthe vertices of the triangle.

EXAMPLE 8 Find the circumcenter of each triangle by construction.

a. b. c.

Definition perpendicular bbisector oof aa ttriangle

In a triangle, a line that is perpendicular to a side of the triangle at its midpoint is called aperpendicular bbisector of the triangle.

In the figure, HN, DN and EN are the

perpendicular bisectors of triangle ABC.

Perpendicular bisectors in a triangle are

always concurrent.

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b. (Definition of the area of a triangle)

(Substitute)

= 35 cm2 (Simplify)

c. (Definition of the area of a triangle)

(Substitute)

= 24 cm2 (Simplify)

6 8=

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5 14=

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FT DEA DEF

The picture below hangsstraight when the hooklies on the perpendicularbisector of the picture’stop edge.


As an exercise, draw three more triangles on a piece of paper and construct their

circumcenters. Check that each circumcenter is the center of the inscribed circle.

Solution First we construct the perpendicular bisector of each side of the triangle. Their intersection

point is the circumcenter of the triangle.

a. b. c.

Check Yourself 31. Name the auxiliary element shown in each triangle using a letter (n, h or V) and a vertex

or side.

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Activity

There are three main faculties ona university campus. The universitywants to build a library on thecampus so that it is the samedistance from each faculty.

1. Make a geometric model of the

problem.2. Find the location of the library in

the picture opposite.

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58 Geometriy 7

2. In a triangle MNP, the altitude NT of side MP and the median MK of side NP intersect at

the point R.

a. Name all the triangles in the figure formed. b. Name two altitudes of MTN.

3. In a triangle DEF, EM is the median of side DF. If DE = 11.4, MF = 4.6 and the perimeter

of DEF is 27, find the length of side EF.

4. In a triangle KLM, LN is the altitude of the side KM. We draw the angle bisectors LE and

LF of angles KLN and MLN respectively. If the angles between the angle bisectors and the

altitude are 22° and 16° respectively, find m(KLM).

5. In the figure, A(ABH) = A(AHC). Find x.

6. Write one word or letter in each gap to

make true statements about the figures.

a. Point O is a(n) __________ .

b. Segment ________ is a median.

c. Point _______ is an excenter.

d. Segment ________ is an altitude.

e. Point B is a(n) _____________.

f. Segment ER is a(n) __________

___________.

g. Point _________ is a circumcenter.

h. Line TM is a(n) __________

__________.

i. Point ________ is a centroid.

Answers1. a. nB b. hp c. Vx d. Vl e. hn f. nL

2. a. MNK, MKP, MNT, NTP, MRT, MNR, RNK, MNP b. NT, TM

3. 6.4 4. 76° 5. 5

6. a. incenter b. ET c. K d. AB (or BC) e. orthocenter (or vertex) f. angle bisector

g. M h. perpendicular bisector i. G

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B. TYPES OF TRIANGLESome triangles are given special names according to the lengths of their sides or themeasures of their angles.

1. Types of Triangle According to SidesA triangle can be called scalene, isosceles or equilateral, depending on the lengths of its sides.

Definition scalene ttriangle

A triangle is called scalene if all of its sides

have different lengths. In other words, a

scalene triangle has no congruent sides.

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Definition isosceles ttriangle

A triangle is called isosceles if it has at least

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Dynamic geometrysoftware is a powerful toolfor studying geometricconcepts. Geometryprograms allow us tochange and manipulatefigures, so that we canexplore and experimentwith geometricalconcepts instead of justmemorizing them.

Activity

The Euler lline of a triangle is the line which passes through

the orthocenter, circumcenter and centroid of the triangle.

Draw a scalene triangle and find its Euler line using

a. a ruler and set square. b. a compass and straightedge. c. dynamic geometry software.

Which method was easier?

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60 Geometriy 7

EXAMPLE 9 Segment EM is a median of an isosceles

triangle DEF with base DF. Find the length of

EM if the perimeter of EMF is 65 and the

perimeter of DEF is 100.

Solution Let us draw an appropriate figure.

In the figure opposite,

in DEM, a + b + x = 65, (1)

in DEF, 2(a + b) = 100. So a + b = 50. (2)

Substituting (2) into (1) gives us 50 + x = 65; x = 15. So EM = 15.

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In an isosceles triangle, the congruent sidesare called the legs of the triangle. The thirdside is called the base of the triangle.

The two angles between the base and thelegs of the triangle are congruent. They arecalled the base aangles of the triangle.

The angle opposite the base is called thevertex aangle.

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Activity GGoollddeenn TTrriiaanngglleess

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A golden ttriangle is a triangle in which the

ratio of the length of the legs to the length

of the base is the golden ratio. The angle

between the two legs of a golden triangle is

always 36°.

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To construct a golden triangle, first draw a

square ABCD and mark the midpoint E of

AB. Find the point F on the extension of AB

by making EF = EC. Then find G by making

AF = AG and BA = BG. Finally, draw FG

and AG. Then BGF is a golden triangle.

1. Construct a golden triangle using a straightedge and compass.

2. Repeat the construction using dynamic geometry software.

3. In both constructions, check the measures of the interior angles.

The sides of the Great Pyramidof Giza are golden triangles.

The head of this kneehammer forms an isoscelestriangle.

Line segments AB andBC are in the goldenratio if

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EXAMPLE 10 In KMN, K . Given that KN is 4 cm

less than MN and MK is 2 cm more than

three times KN, find the perimeter of KMN.

Solution We begin by drawing the figure opposite.

If MK = x then KN = x – 4. Also, MK = MN

because K N.

Also, we are given MN = 3KN + 2

x = 3(x – 4) + 2

10 = 2x

5 = x.

Since P(KMN) = 3x – 4, P(KMN) = (3 5) – 4 = 11 cm.

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EXAMPLE 11 In ABC opposite, O is the intersection point

of the bisectors of the interior angles of the

triangle. Given that OE BC, OD AB,

AD = 4 cm, DE = 5 cm and EC = 6 cm, find

P(EOD).

Solution Let us join points A and C to O. We know

from the question that OA and OC are the

bisectors of A and C, respectively.

Since OD AB,

m(OAB) = m(AOD). (Alternate Interior

Angles Theorem)

So ODA is an isosceles triangle and

AD = OD = 4 cm. (1)

Similarly, since OE BC,

m(EOC) = m(OCB). (Alternate Interior Angles Theorem)

So EOC is also an isosceles triangle and OE = EC = 6 cm. (2)

By (1) and (2), P(EOD) = OE + OD + DE = 6 + 4 + 5 = 15 cm.

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62 Geometriy 7

In an equilateral triangle, all of the interior

angles are congruent and measure 60°.

Notice that an equilateral triangle is also an

isosceles triangle, but an isosceles triangle is

not always equilateral.

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EXAMPLE 12 The three sides of a triangle measure 5n + 8, n+12 and 3n+10 with n N. Which value of

n makes this triangle equilateral?

Solution If the triangle is equilateral, all the sides must be congruent.

So 5n + 8 = n + 12 = 3n + 10. Let us solve the first equality to find n:

5n + 8 = n + 12

4n = 4

n = 1.

If we substitute 1 for n, the side lengths become (5 1) + 8 = 13, 1 + 12 = 13 and

(3 1) + 10 = 13. So the triangle is equilateral when n = 1.

Definition equilateral ttriangle

A triangle is called equilateral if it has three

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How many equilateraltriangles can you see inthe figure below?

You will probably ‘see’ twotriangles, one on top of theother. This is actually anoptical illusion, though,as the white triangle isnot actually drawn.

Activity

Find six toothpicks and try to do each thing below. Some things may not be possible.

Can you explain why?

1. Make one equilateral triangle with six toothpicks.

2. Make two equilateral triangles with six toothpicks.

3. Make three equilateral triangles with six toothpicks.

4. Make four equilateral triangles with six toothpicks.

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Check Yourself 4

1. In ABC opposite, DE BC and point O is the incenter of the

triangle. If BD = 6 and EC = 4, find DE.

2. The perimeter of an isosceles triangle is 18.4 and its base measures 4 units more than the

length of one leg. Find the length of a leg of this triangle.

3. The sides of an isosceles triangle have lengths in the ratio 4 : 5 : 5. Find the length of the

base of the triangle if its perimeter is 28.

4. The perimeter of an isosceles triangle is 22.8. An equilateral triangle is drawn such that one

side is congruent to the base of the isosceles triangle. If the perimeter of the equilateral

triangle is 24.6, find the length of one leg of the isosceles triangle.

5. In an isosceles triangle NTM, MN = NT, MN = 35, TN = 4x +15 and MT = 40 – x2. Find

MT.

6. In the figure, all triangles are equilateral,

AG = 24.12 cm and AC = 3CE = 2EG. Find

the perimeter of each triangle.

7. The three sides of a triangle measure 3a,

a+10 and 6a – 15. Which value of a makes the triangle equilateral?

8. Construct an isosceles and an equilateral triangle.

Answers1. 10 2. 4.8 3. 8 4. 7.3 5. 15 6. 12.6 cm, 6.3 cm, 4.2 cm 7. 5

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2. Types of Triangle According to AnglesA triangle can be called acute, right or obtuse, depending on the measures of its angles.

Definition acute ttriangle, rright ttriangle, oobtuse ttriangle

A triangle is called an acute ttriangle if all its angles are acute.

A triangle is called a right ttriangle if it has a right angle.

A triangle is called an obtuse ttriangle if it has an obtuse angle.

The picture shows a puzzlecalled the Three CompanionsPuzzle. Get your own andtry to free one of the trianglesfrom the string. Can you doit?

64 Geometriy 7

In a right triangle, the sides adjacent to the

right angle are called the legs of the triangle.

The side opposite the right angle is called thehypotenuse of the triangle.

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NoteNotice that a triangle can be only one of obtuse, acute or right.

EXAMPLE 14 Classify each triangle according to its side lengths and angle measures.

Solution a. isosceles right triangle b. scalene acute triangle c. isosceles acute triangle

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EXAMPLE 13 Name all the right triangles in the figure.

Solution There are four smaller right triangles (ABK,

BKC, CKD and DKA) and four larger

triangles (ABC, BCD, CDA and DAB).

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Challenge!Try to change theequilateral triangle in thefigure so that it pointsupwards by moving onlythree balls.

Then try to make thetriangle in this figurepoint downwards by usingthe least number of ballspossible.


EXAMPLE 15 Draw a right triangle and divide it using

a. two parallel lines which are perpendicular to one of the legs.

b. two parallel lines which are not perpendicular to legs.

c. two perpendicular lines to create two more right triangles.

d. two intersecting lines which are not perpendicular to each other to create two more righttriangles.

Activity ‘Tangram’ is a fun puzzle and a good way to exercise your brain. The name comes fromtan, which means ‘Chinese’, and gram, which means ‘diagram’ or ‘arrangement’. Thepuzzle first appeared in China thousands of years ago, and it is now known all over theworld. There are seven pieces in a tangram set: five triangles, one square and oneparallelogram. The challenge of the puzzle is to use the seven pieces together to makedifferent shapes. You must use all the pieces, and they must all touch but not overlap.

All seven tangram pieces are made up of right triangles with this shape:

The first tangram challenge is to make a square with all seven pieces. The solution isshown below.

Find a tangram set, or copy the figure above to make your own.

1. Make one right triangle using all of the pieces.

2. Can you make an obtuse triangle by using all of the pieces?

3. Can you make an acute triangle by using all of the pieces?

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66 Geometriy 7

Solution a. b. c. d.

EXAMPLE 16 Draw each triangle and use a set square to find its orthocenter. Write the orthocenter as an

intersection of lines or line segments.

a. acute scalene b. right scalene c. obtuse scalene

Solution Remember that the orthocenter of a triangle is the intersection point of its altitudes. We draw

the altitudes in each triangle by using a set square.

a. orthocenter K,

K = AF BD EC

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b. orthocenter A,

A = AB CA AD

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T = AT BT TC

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Ealier in this chapter we said that the position of the orthocenter of a triangle depends on

the type of triangle. One position is in the interior of the triangle. Can you see what the other

two possible positions are, after studying the example above? How do they correspond to the

types of triangle shown?

a 30°-60° set square

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Check Yourself 51. Classify each triangle according to its angle measures.

2. Name all the right triangles in the figure.

3. At most how many of each type of angle can one triangle have?

a. acute angle b. right angle c. obtuse angle

(Hint: Try to draw a suitable figure for each case using a protractor.)

4. Draw a right triangle and divide it using

a. two intersecting lines which are perpendicular to each other.

b. two intersecting lines which are not perpendicular to each other, to make three more

right triangles.

5. Construct a right isosceles triangle.

Answers1. a. right triangle b. acute triangle c. acute triangle

d. obtuse triangle e. obtuse triangle

2. DKB, KAB, KBC, KCD, KDA

3. a. three b. one c. one

4. a. b.

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a. b. c. d. e.

This shark’s fin forms aright triangle with the

water.

How many triangles?

68 Geometriy 7

22.. How many triangles can be formed by joining anythree points D, E, F and G if no three of the givenpoints are collinear? Name each triangle.

33.. In a triangle ABC, AB is of AC, AB = BC and

AC = 15 cm. Find P(ABC).

65

A. The Triangle and Its Elements

11.. Find and name all the

triangles in the figure.

EXERCISES 3.1

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44.. In a triangle KMN, KM = cm, MN is 75% of

KM and KN is 0.1 cm more than KM. Find

P(KMN).

185

99.. In an isosceles triangle DEF, DF is the base

and FT is a median. Given that P(DEF) = 23 cm

and P(EFT) is 1 cm more than the perimeter of

triangle DTF, find DF.

1100.. Draw three triangles ABC

as in the figure and

construct each element

separately, using a compass

and straightedge.

a. ha b. Va c. nA

66.. Answer according to

the figure.

a. Name fourcollinear points onABC.

b. Name a point which is in the interior ofADC.

c. What is the intersection of ABC and ADC?

d. What is the intersection of ABC and int MAC?

77.. Draw four figures to show how two triangles canintersect to form a four-sided, five-sided,six-sided and three-sided polygon.

88.. In a triangle ABC, two points different to A and B

on the side AB are joined to the vertex C by line

segments. Similarly, three points different to B

and C on side BC are joined to the vertex A by line

segments. How many regions inside the triangle

are formed by the intersection of these segments?

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55.. The side AC of a triangle ABC measures 12.8 cm,which is 2.6 cm less than the sum of the lengthsof the other sides. Find the perimeter of thistriangle.

1111.. Draw four triangles KMN as inthe figure and find eachpoint separately usingonly a compass andstraightedge.

a. centroid b. incenter

c. orthocenter d. circumcenter

1133.. Find the excenters of the triangle in question 11

by using a protractor and ruler.

1122.. Repeat question 11 with a protractor and ruler.

1144.. The sides AB, BC and AC of a triangle ABC measure

13, 14 and 15 units respectively. Given that the

length of the altitude to side BC is 12, find the

lengths of the remaining altitudes.

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2211.. Write always, sometimes or never to make truestatements.

a. If a triangle is isosceles then it is ______________equilateral.

b. If a triangle is equilateral then it is ___________isosceles.

c. If a triangle is scalene then it is ______________isosceles.

d. If a triangle is obtuse then it is _______________isosceles.

e. An obtuse triangle is __________________ a righttriangle.

f. In a triangle DEF, if DE EF then DF is________________ perpendicular to EF.

g. A scalene triangle ________________________ hasan acute angle.

h. If a triangle has two complementary anglesthen it is ____________________ a right triangle.

2200.. Complete the table showing the location (in theinterior, on the triangle or in the exterior) of theintersection of the segments or lines for each typeof triangle.

Perpendicularbisectors

Anglebisectors

MediansLine ccontaining

the aaltitudes

Acutetriangle a. b. c. d. Right

triangle e. f. g. h. Obtusetriangle i. j. k. l.

1188.. The sides of a triangle measure 2x + 8, 3x – 6,and 12 + x.

a. Find the value(s) of x that make(s) the triangleisosceles.

b. Which value(s) of x make(s) the triangleequilateral?

1199.. The sum of the lengths of the legs of anisosceles right triangle is 22 cm. Find the area ofthis triangle.

2222.. In each case, draw a triangle with the given

property.

a. All three angle bisectors are medians.

b. No altitude is a median.

c. Only one angle bisector is the perpendicular

bisector of a side.

d. Only one altitude is in the interior region of

the triangle.

e. The medians, altitudes and angle bisectors

coincide.

f. Exactly one of the three altitudes is also a

median.

2233.. Divide any right triangle using two lines so that

the figure contains a total of

a. five right triangles. b. six right triangles.

B. Types of Triangle

1155.. Look at the figure and

name

a. an isosceles triangle.

b. three right triangles.

c. an obtuse isosceles triangle.

d. an acute triangle.

e. an equilateral triangle.

1177.. State whether each type of triangle is possible ornot.

a. an isosceles acute triangle

b. a right equilateral triangle

c. a scalene acute triangle

d. an obtuse isosceles triangle

e. an obtuse equilateral triangle

1166.. Find the circumcenter of a right triangle using a

ruler and protractor.

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70 Geometriy 7

A. THE CONCEPT OF CONGRUENCEIn the previous section we studied triangles and their features and properties. In this sectionwe will look at possible relations between two or more triangles.

If we are given two triangles, how can we compare them? We might notice that they are thesame size and shape. This important relation in geometry is called congruence. Let us startour study of congruence with a general definition of congruence in figures and polygons.

1. Congruent FiguresThe world around us is full of objects of various shapes and sizes.If we tried to compare some of these objects we could put themin three groups:

objects which have a different shape and size,

objects which are the same shape but a different size, and

objects which are the same shape and size.

The tools in the picture at the right have different shape and size.

The pictures below show tools which have the same shape but different size. In geometry, figures like this are called similar ffigures. We will study similar figures in Chapter 3.

Congruence is a basicgeometric relationship.


1. Identify congruent triangles

2. Construct a circle

3. Construct congruent segments

4. Find the midpoint of a segment

5. Construct perpendicular lines and parallel lines

6. Construct congruent angles and an angle bisector

7. Construct atriangle tram given information

8. Desctibe and use the properties of isosceles, equilateral and right triangles.

9. Describe and use the triangle Angle Bisector Theorem.

Objectives


The pictures below show objects which are the same size and shape. In this section, we willstudy figures which have this property.

Factories often need toproduce many partswith exactly the samesize and shape.

Definition congruent ffigures

Figures that have the same size and shape are called congruent ffigures. We say ‘A is congruent

to B’ (or ‘B is congruent to A’) if A and B are congruent figures.

The pictures at the bottom of the previous page show some examples of congruent objects.

The pictures below show two more examples. In these two examples there is only one piece

left to fit in the puzzle. Therefore, without checking anything, we can say that each piece and

its corresponding place are congruent.

Activity

Make a poster to show congruent figures in everyday life. You can take photos, draw

pictures or collect pictures from magazines or newspapers that show buildings, designs,

signs and artwork with congruent parts.

MMaakkiinngg aa PPoosstteerr - CCoonnggrruueenntt FFiigguurreess

Congruence in nature:the petals of this flower

are congruent.

EXAMPLE 17 Which piece is congruent to the empty space?

Solution If we compare the vertices and sides, we can easily see that only c. fits into the space.

a. b. c. d.

72 Geometriy 7

A car has manycongruent parts.

Activity

When you learned common fractions, you probably learned them by working with

figures divided into congruent parts. Often the figures are circles and rectangles, as

these are the easiest to divide into any number of congruent parts.

Dividing (also called dissecting) a figure into congruent parts can also be a puzzle. As an

example, can you see how to dissect the first figure below into two congruent pieces?

Answer:

Now try the two puzzles below. The answers are at the back of the book.

1. Dissect each figure into four congruent pieces.

2. The polygon below left can be dissected into four congruent polygons, as shown in the

figure below right. There is also a way to divide this polygon into five congruent

polygons. Can you find it?

CCoonnggrruueenntt DDiisssseeccttiioonnss


Challenge!Remove five toothpicksto make five congruenttriangles.

In the figure below, ABC and DEF are congruent because their corresponding parts arecongruent. We can write this as follows:

A D AB DE

B E and BC EF

C F AC DF.

We can show this symbolically in a figure as follows:

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Definition corresponding eelements oor pparts

The points, lines and angles which match perfectly when two congruent figures are placed

one on top of the other are called corresponding eelements or corresponding pparts of the

congruent figures.

We can see that by definition, corresponding pparts oof ccongruent ffigures are congruent. We

can write this in a shorter way as CPCFC.

You are already familiar with congruent segments (segments that have equal lengths) and

congruent angles (angles that have equal measures). In the rest of this section we will look

at congruent figures which are made up of segments and angles. These figures are polygons

and especially triangles.

Sometimes we need tomove or modify a figureto see that it is congruentto another figure. Thebasic changes that wecan make to a figure arereflection (flipping),rotation (turning) andtranslation (sliding). Wewill study these inChapter 3.

Definition congruent ttriangles

Two triangles are congruent if and only if their corresponding sides and angles are congruent.

We write ABC DEF to mean that ABC and DEF are congruent.

2. Congruent Triangles

We can think of congruent figures as figures that are exact copies of each other. In other

words, we can put congruent figures one on top of the other so that each side, angle and

vertex coincides (i.e. matches perfectly).

74 Geometriy 7

EXAMPLE 18 Given that MNP STK, state the congruent angles and sides in the two triangles without

drawing them.

Solution The figure at the right shows how the

vertices of each triangle correspond to each

other. Because MNP STK and CPCTC

(corresponding parts of congruent triangles

are congruent), we can write

M S MN ST

N T and NP TK

P K PM KS.

As we can see, the order of the vertices in congruent triangles is important when we are

considering corresponding elements. Any mistake in the ordering affects the correspondence

between the triangles.

If two triangles are congruent then we can write this congruence in six different ways. For

instance, if ABC is congruent to DEF, the following statements are all true:

ABC DEF

ACB DFE

BAC EDF

BCA EFD

CAB FDE

CBA FED.

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A short history of the symbol:

Gottfried WWilhelmLeibniz

(1640-1716)introduced for congruencein an unpublished manuscript in

1679.

In 1777,Johann FFriedrich

Häseler(1372-1797)

used (with the tilde reversed).

In 1824,Carl BBrandan

Mollweide(1774-1825)

used the modern symbol forcongruence in Euclid’s Elements.

EXAMPLE 19 Complete each statement, given that PRS KLM.

a. PR _____ b. _____ K c. _____ SP

d. S _____ e. ML _____ f. L _____

Solution a. PR KL b. P K c. MK SP

d. S M e. ML SR f. L R

EXAMPLE 20 Decide whether or not the two triangles in

the figure are congruent and give a reason for

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EXAMPLE 21 ABC EFD is given with AB = 11 cm, BC = 10 cm and EF + ED = 19 cm. Find the

perimeter of EFD.

Solution Since ABC EFD, AB = EF, BC = FD and

AC = ED by the definition of congruence.

So by substituting the given values we get

11 = EF, 10 = FD and AC = ED.

Since we are given that EF + ED = 19 cm,

we have 11 + ED = 19 cm; ED = 8 cm.

So P(EFD) = EF + ED + FD = 11 + 8 + 10 = 29 cm.

Check Yourself 61. KLM XYZ is given. State the corresponding congruent angles and sides of the

triangles.

2. State the congruence JKM SLX in six different ways.

3. Triangles KLM and DEF are congruent. P(KLM) = 46 cm, the shortest side of KLM

measures 14 cm, and the longest side of the DEF measures 17 cm. Find the lengths of

all the sides of one of the triangles.

4. Triangles DEF and KLM are congruent. If DE = 12.5 cm, EF = 14.4 cm and the perimeter

of the triangle KLM is 34.6 cm, find the length of the side DF.

Solution Let us calculate the missing angles:

m(C) = 60° (Triangle Angle-Sum Theorem in ABC)

m(M) = 30° (Triangle Angle-Sum Theorem in KMN)

Now we can write the congruence of corresponding parts:

AB KM (Given)

BC KN (BC = KN = 4)

AC MN (AC = MN = 8)

A M (m(A) = m(M) = 30°)

B K (m(B) = m(K) = 90°)

C N (m(C) = m(N) = 60°)

Therefore, ABC MKN by the definition of congruent triangles.

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What would happen ifthe blades of this ship’s

propellor or these wheelswere not congruent?

76 Geometriy 7

In this section we will construct geometric figures using only two instruments,

a straightedge and a compass.

1. Basic ConstructionsWe use a straightedge to construct a line,

ray, or segment when two points are given.

A straightedge is like a ruler without num-

bers.

We use a compass to construct an arc or a

circle, given a point O and a length r

(a radius).

Result: [CD] [AB].

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Construction 1

Constructing a congruent segment.Given [AB],

construct [CD] such that [CD] [AB].

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5. Two line segments KL and AB bisect each other at a point T. If AL = 7 and the lengths of

the segments KL and AB are 22 and 18 respectively, find the perimeter of KTB.

Answers1. 2.

3. 14 cm, 15 cm, 17 cm 4. 7.7 cm 5. 27

PKM SLN, KMP LNS, MPK NSL,

PMK SNL, KPM LSN, MKP NLS

K X

L Y

M Z

KL XY

LM YZ

KM XZ


Result: M is the midpoint of [AB].

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Finding the midpoint of a given segment.

Given [AB],

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Construction 3

Constructing a perpendicular to a line at a given point on the line.

Given point M on the line l,

construct [MN] l.

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78 Geometriy 7

Result: [MN] l.

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Construction 4

Constructing a perpendicular to a given line through a point outside the given line.

Given line l and a point N outside the line, construct [MN] l.

Result: AA.

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Constructing a congruent angle.

Given A, construct A such that A A.

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Construction 6

Constructing a parallel to a line through a point outside the line.

Given line l with point N which is not on l, construct a line through N which is parallel to l.

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Constructing an angle bisector.

Given CAB, construct the bisector of CAB.

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80 Geometriy 7

c. Look at construction 6.

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Use a straightedge to draw [AB].Set the compass at thepoints A and B.

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Use a straightedge to draw [AB]. Next, openthe compass to |AB| and draw two arcs, onewith center A and the other with center B.

Label the intersection point C. Draw [AC]and [BC]. All the sides have equal length,so ABC is an equilateral triangle.

Draw another line l.Choose any point online l and label it A.

Use the radius |AB| and setthe compass point at A. Drawan arc intersecting l.Label the point of intersectionB. Now [AB] [AB]

EXAMPLE 22 a. Construct two congruent line segments.

b. Construct an obtuse angle and bisect it.

c. Construct two parallel line segments.

d. Construct an isosceles triangle.

e. Construct an equilateral triangle.

Solution a.

b.

d.

e.

Draw any obtuse angle ABC. Use B as thecenter, and draw an arc AïC. Next, draw two arcs,one with center A and the other with center C.

Label the point D where the twoarcs intersect. Draw [BD.[BD is the angle bisector of ABC.

Draw a line segment [AB]. Use any radius greater

than and draw two arcs with centers A

and B. Name the intersection point C.

1|AB|2

Draw the triangle ABC.

|AC| = |BC|, so the triangle is

isosceles.


Practice Problems 71. Construct a 30° angle. (Hint: construct a 60° angle and bisect it.)

2. Construct a right triangle with legs which are congruent to [AB]

and [CD] in the figure.

3. Construct a right triangle whose legs are in the ratio 2:1.

4. Construct a line segment and divide it into four equal parts.

Answers1. By constructing equilateral triangles:

2.

3.

4.

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Find the midpoint of [AB]. Find the midpoint of [AC]and the midpoint of [CB].

Label the new points D andE.

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Find the midpoint of[AD] (see construction 2)

Draw a perpendicularline to [AB] from A.(see construction 3)

Open your compassto as [AC] then draw[AD] and [DB].

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Open your compassmore than [MA] anddraw two arcs, onewith center M, teother with center N.Draw a line from Ato the point ofintersection.

Open your compassto [CD] and draw anarc with center A.Label the point ofintersection D. Draw[AD] and [DB.

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82 Geometriy 7

2. Constructing TrianglesWe can construct basic geometric figures using only a straightedge and a compass. However,to construct triangles we need a compass, a ruler and a protractor. We use the ruler to measure the sides of triangle, and the protractor to draw the angles.

We have seen that a triangle has six basic elements: three angles and three sides. To construct a triangle, we need to know at least three of these elements, and one of these threeelements must be the length of a side. Let us look at the possible cases.

a. Constructing a Triangle from Three Known SidesLet us construct ABC,

where |AB| = c, |BC| = a, and |AC| = b, given that

a < b < c.

NoteIn any triangle, the sum of any two given angles is less than 180° and the sides satisfy the

triangle inequality.

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Construction 1

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Construction 2

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Open the compass as much as length c and put thesharp point of the compass on A. Then draw an arc.

Name the intersection point B.

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Construction 4

Again open the compass as much as length b and put thesharp point on A. Then draw an arc on the upper side ofd.

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Construction 5

Finally, open the compass as much as length a and putthe sharp point on B. Then draw an arc which intersectsthe other arc drawn before. Name the intersection pointC.

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NoteRemember that in a triangle, side a is opposite A, side b is opposite B, and side c is

opposite C. When we talk about ‘side b’ we mean the side opposite B, or the length of this

side.

Construction 5

After determining the point C, draw [AC] and [BC]. Theresult is the constructed triangle.

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EXAMPLE 23 Construct ABC given |AB| = 10 cm, |BC| = 8 cm, and |AC| = 6 cm.

Solution

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b. Constructing a Triangle from Two Known Angles and a Known SideLet us construct the triangle ABC, where A, B, and the side c are given.

Construction 1

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84 Geometriy 7

Construction 3

Using a protractor, take the point A as a vertex and draw

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EXAMPLE 24 Construct ABC given mB = 40°, mC = 70°, and |BC| = 12 cm.

Solution

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3. Constructing a Triangle from Two Known Sides and a KnownAngle

Finally, let us construct ABC given

|AB| = c, |BC| = a and the known angle B.

Construction 1

Draw a line d and use a compass to locate the points B

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EXAMPLE 25 Construct ABC given |BC| = 5 cm, |AB| = 10 cm, and mB = 70o.

Solution

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86 Geometriy 7

Practice Problems 81. State the things you need to know in order to construct a triangle.

2. Draw an equaliteral triangle with sides 6 cm long.

3. Construct ABC given a = 5 cm, b = 4 cm and c = 2 cm.

4. Construct ABC given a = 7 cm, b = 6 cm and c = 8 cm.

5. Construct DEF given d = 6 cm, e = 8 cm and f = 10 cm.

6. Construct ABC given mA = 40o, mB = 65o and |AB| = 10 cm.

7. Construct KLM given mM = 45o, mL = 70o and |ML| = 7 cm.

8. Construct PQR given mR = 40o, mQ = 60o and |RQ| = 4 cm.

9. Construct MNP given mM = 30o, mN = 65o and |MN| = 15 cm.

10.Construct ABC given mB = 90o, |AB| = 5 cm and |BC| = 12 cm.

11.Construct PQR given mQ = 80o, |PQ| = 7 cm and |QR| = 4 cm.

12.Construct GHK given mH = 50o, |GH| = 6 cm and |HK| = 9 cm.

13.Construct XYZ given mY = 110o, |XY| = 3 cm and |YZ| = 5 cm.

14.Can you draw a triangle from only three given angles?

C. ISOSCELES, EQUILATERAL AND RIGHT TRIANGLES Isosceles, equilateral and right triangles are useful triangles because they have many special

properties. If we can identify one or more of these triangles in a figure then we can often use its

properties to solve a geometric problem. In this section we will look at some fundamental

theorems about isosceles, equilateral and right triangles, and some useful additional properties.

1. Properties of Isosceles and Equilateral Trianglesa. Basic Properties

Theorem

If two sides of a triangle are congruent then the angles opposite these sides are also

congruent.

Proof Let us draw an appropriate figure.

Given: AB = AC

Prove: B C

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Let AN be the bisector of A.

Statements Reasons

1. AB AC 1. Given

2. BAN CAN 2. Definition of an angle bisector

3. AN AN 3. Reflexive property of congruence

4. ABN ACN 4. SAS Congruence Postulate

5. B C 5. CPCTC

Theorem

If two angles in a triangle are congruent then the sides opposite these angles are also

congruent.

Proof Let us draw an appropriate figure.

Given: B C

Prove: AB AC

We begin by drawing the bisector AN, and

continue with a paragraph proof.

Since AN is the angle bisector,

BAN CAN.

It is given that B C.

By the reflexive property of congruence, AN AN.

So ABN ACN by the AAS Congruence Theorem.

Since CPCTC, we have AB AC.

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EXAMPLE 26 In a triangle DEF, T DF such that DT = DE. Given m(EDT) = 40° and m(DEF) = 85°,

find m(TEF).


Since DE = DT, DET is an isosceles triangle.

So by the Isosceles Triangle Theorem,

m(DET) = m(DTE).

So by the Triangle Angle-Sum Theorem in

DET,

m(EDT) + m(DET) + m(DTE) = 180°

40° + 2m(DET) = 180°

m(DET) = 70°.

So m(TEF) = m(DEF) – m(DET) = 85° – 70° = 15°.

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88 Geometriy 7

Corollary

If a triangle ABC is equilateral then it is also equiangular. In other words,

if a = b = c then m(A) = m(B) = m(C).

Corollary oof tthe IIsosceles TTriangle TTheorem

Corollary

If a triangle ABC is equiangular then it is also equilateral, i.e. if m(A) = m(B) = m(C)

then a = b = c.

Corollary oof tthe CConverse oof tthe IIsosceles TTriangle TTheorem

EXAMPLE 27 In the figure, ABC and DEF are equilateral

triangles. If BF = 17 cm and EC = 3 cm, find

AB + AH + DH + DF.

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m(HCE) = 60° and m(HEC) = 60°. (ABC and DEF are equilateral)

So in HEC,

m(H) + m(E) + m(C) = 180° (Triangle Angle-Sum Theorem)

m(H) = 60°.

So HEC is equiangular. (m(C) = 60°, m(E) = 60°, m(H) = 60°)

Therefore HEC is equilateral. (By the previous Corollary)

So HE = HC = EC = 3 cm.

Let a and b be the lengths of the sides of ABC and DEF, respectively.

In ABC, AB = a, BE = a – 3 and AH = a – 3. (EC = 3 cm, given)

In DEF, DF = b, CF = b – 3 and DH = b – 3. (EC = 3 cm, given)

So

AB + AH + DH + DF = a + a – 3 + b – 3 + b

= 2(a + b) – 6. (1)

Moreover, BF = a – 3 + 3 + b – 3 (Segment

Addition Postulate)

17 = a + b – 3

20 = a + b. (2)

Substituting (2) into (1) gives us AB + AH + DH + DF = 34 cm.

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Practice Problems 9

1. In a triangle ABC, the interior angle bisector at the vertex A makes an angle of 92° with

the side opposite A and has the same length as one of the remaining sides. Find all the

angles in ABC.

2. In the figure, CE is the angle bisector of C, HD BC and

HD = 5 cm. Find the length of AC.

3. In the figure, DCE is an equilateral triangle and DC = BC.

is given. Find m(A).

Answers1. 8°, 84° and 88° 2. 10 cm 3. 5°

( ) 1=

( 13m Am B)

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b. Further properties

Properties 61. For any isosceles triangle, the following statements are true.

a. The median to the base is also the angle bisector of the vertex and the altitude to the

base.

b. The altitude to the base is also the angle bisector of the vertex and the median to the

base.

c. The angle bisector of the vertex is also the altitude and the median to the base.

In other words, if AB = AC in any triangle ABC then nA = Va = ha.

2. In an equilateral triangle, the medians, angle bisectors and altitudes from the same

vertex are all the same, i.e., ha = nA = Va, hb = nB = Vb, and hc = nC = Vc.

Moreover, all of these lines are the same length:

ha = hb = hc = nA = nB = nC = Va = Vb = Vc.

In other words, if AB = BC = AC in ABC then ha = hb = hc = nA = nB = nC = Va = Vb = Vc.

3. If AB = AC in any triangle ABC then nB = nC, Vb = Vc and hb = hc.

4. If ha = nA or ha = Va or nA = Va in ABC then ABC is isosceles.

90 Geometriy 7

5. a. If ABC is an isosceles triangle with

AB = AC, P BC, E AC, D AB,

PE AB and PD AC,

then PE + PD = b = c.

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b. If P, E and D are any three points on

the sides of an equilateral triangle ABC

such that PE and PD are parallel to

two distinct sides of ABC, then

PE + PD = AB = BC = AC.

6. a. In any isosceles triangle ABC with

AB = AC, the sum of the lengths of two

lines drawn from any point on the base

perpendicular to the legs is equal to

the height of the triangle from the

vertex B or C. In other words, if

AB = AC, P BC, H AC, D AB,

PH AC and PD AB, then

PH + PD = hc = hb.

b. In any equilateral triangle ABC, the sum of the lengths of two lines drawn from any

point on any side perpendicular to the other sides is equal to the height of the triangle

from any vertex. In other words, if AB = BC = AC, P BC, H AC, D AB, PH AC

and PD AB, then PH + PD = ha = hb = hc.

7. In any equilateral triangle ABC,

if P int ABC and points D, E and F

lie on the sides of ABC such that

PE AB, PD AC and PF BC, then

PE + PF + PD = AB.

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if P int ABC and points D, E and F lie

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PD AC and PF BC,

then PD + PE + PF = AH where

AH BC.

EXAMPLE 28 An isosceles triangle TMS has base TS which measures 8 cm and perimeter 32 cm. The

perpendicular bisector of leg TM intersects the legs TM and MS at the points F and K,

respectively. Find the perimeter of TKS.


Since hk = Vk in TKM, TKM is isoscelesby Property 6.4.

So TK = KM. (1)

By the Segment Addition Postulate,MK + KS = MS. (2)

Since TMS is isosceles andP(TMS) = 32 cm, TM = MS = 12 cm. (3)

So P(TKS) = TK + KS + ST

= KM + KS + ST (By (1))

= MS + ST (By (2))

= 12 + 8 = 20 cm. (By (3))

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EXAMPLE 29 In the figure,

AB = AC,

PD AC and

PE AB.

Given AB + AC = 32 cm, find PD + PE.

Solution Since AB + AC = 32 cm and

AB = AC, we have AB = AC = 16 cm.

So by Property 6.5b, PD + PE = AB = 16 cm.

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92 Geometriy 7

Solution Look at the figure.

Given: AT is a median and AB = AC.

Prove: AT bisects A and is an altitude ofABC.

We will write a two-column proof.

Proof:

EXAMPLE 32 Prove that in any isosceles triangle, the median to the base is also the angle bisector of the

vertex and the altitude to the base.

EXAMPLE 31 In the equilateral triangle ABC shown

opposite, PD BC, PE AB and PF AC.

PE = 6 cm, PF = 5 cm and

P(ABC) = 42 cm are given.

Find the length of PD.

Solution Since ABC is equilateral and its perimeter is 42 cm, AB = =14 cm.

By Property 6.7, PD + PE + PF = AB. So PD + 6 + 5 = 14 and so PD = 3 cm.

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Statements Reasons

1. AC AC 1. Given

2. ABC ACB 2. Isosceles Triangle Theorem

3. BT TC 3. AT is a median.

4. ABT ACT 4. By 1, 2, 3 and the SAS Congruence Postulate

5. m(TAB) = m(TAC) 5. By 4 (CPCTC)

6. AT bisects A 6. Definition of angle bisector

7. m(ATB) = m(ATC) 7. By 4 (CPCTC)

8. m(ATB) = m(ATC) = 90° 8. Linear Pair Postulate

9. AT is also an altitude of ABC 9. Definition of altitude

EXAMPLE 30 In the figure opposite, H, A and B are

collinear points with CH HA,

PE AC, PD AB and AB = AC.

PE = 6 cm and HC = 10 cm are given.

Find the length of PD.

Solution By Property 6.6a, PE + PD = CH.

So 6 + PD = 10 and so PD = 4 cm.

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2. In an equilateral triangle ABC, H BC, N AC, BN is the interior angle bisector of B,

and AH is the altitude to the side BC. Given BN = 9 cm, find AH.

3. In the figure, BH = HC,

m(HAC) = 20° and

m(BCD) = 30°.

Find m(BDC).

4. In the figure, AB = AC = 13 cm,

PE AC, PF AB and PE = 4 cm.

Find the length of PF.

5. In the figure, AB = AC, PE HB,

PF AC and BH HC.

Given CH = 12 cm,

find the value of PE + PF.

6. In the figure, AB = BC,

PD = x + 3, PH = 3x – 1, and

AE = 14. Find the value of x.

Answers

1. 36 2. 9 cm 3. 80° 4. 9 cm 5. 12 cm 6. 3

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Practice Problems 10

1. In the figure, AD and BE are the interior angle

bisectors of A and B, respectively.

AC = BC and BE + AD = 12 cm are given.

Find the value of BE AD.

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94 Geometriy 7

Proof There are many proofs of the Pythagorean

Theorem. The proof we will give here uses

the dissection of a square. It proves the

Pythagorean Theorem for the right triangle

ABC shown opposite.

Imagine that a large square with side length

a + b is dissected into four congruent right

triangles and a smaller square, as shown in

the figure. The legs of the triangles are a and

b, and their hypotenuse is c. So the smaller

square has side length c.

We can now write two expressions for the area S of the larger square:

and S = (a + b)2.

Since these expressions are equal, we can write

This concludes the proof of the Pythagorean Theorem.

2 2

2 2 2

2 2 2

4 + =( + )2

2 + = +2 +

= + .

a bc a b

ab c a ab b

c a b

2= 4 +

2a b

S c

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Theorem

In a right triangle ABC with m(C) = 90°,

the square of the length of the hypotenuse

is equal to the sum of the squares of the

lengths of the legs, i.e.

c2 = b2 + a2.

Pythagorean TTheorem

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2. Properties of Right Triangles

a. The Pythagorean TheoremThe Pythagorean Theorem is one of the most famous theorems in Euclidean geometry, and

almost everyone with a high school education can remember it.


EXAMPLE 33 In the figure, ST SQ. Find x and y.

Solution First we will use the Pythagorean Theorem

in SKT to find x, then we can use it in

SKQ to find y.

SK2 + KT 2 = ST2 (Pythagorean Theorem in SKT)

x2 + 122 = 132 (Substitute)

x2 + 144 = 169

x2 = 25 (Simplify)

x = 5 (Positive length)

SK2 + KQ2 = SQ2 (Pythagorean Theorem in SKQ)

52 + y2 = 62 (Substitute)

y2 =36 – 25 (Simplify)

y = ò11

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x = –5 is not an answerbecause the length of asegment cannot be negative.So the answer is x = 5.From now on we willalways consider onlypositive values for lengths.

Activity

1. Cut out a square with side length c, and cut out four identical right triangles with

hypotenuse c and legs a and b. Place the four triangles over the square, matching the

hypotenuses to the sides and leaving a smaller square uncovered in the centre. Try

to obtain the Pythagorean Theorem by considering the area of the original square

and the areas of the parts that dissect it.

2. Cut out or construct two squares with sides a and b. Try

to dissect these squares and rearrange their pieces to

form a new square. Then use the areas of the original

squares and the new square to write the Pythagorean

Theorem. (Hint: Start by cutting out two right triangles

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Try the following activity to discover two more proofs of the Pythagorean Theorem.

96 Geometriy 7

Theorem

If the square of one side of a triangle equals the sum of the squares of the other two sides,

then the angle opposite this side is a right angle.

Converse oof tthe PPythagorean TTheorem


PT = TS = KS,

PM = 4 cm and KM = 3 cm. Find ST.

Solution Let PT = TS = KS = x.

So SM = KM – KS = 3 – x and TM = PM – PT = 4 – x.

In TMS, TS2 = TM2+ MS2 (Pythagorean Theorem)

x2 = (4 – x)2 + (3 – x)2 (Substitute)

x2 = 16 – 8x + x2 + 9 – 6x + x2 (Simplify)

x2 – 14x + 25 = 0

x1, 2 = (7 ò24) cm (Quadratic formula)

Since 7 + ò24 is greater than 3 and 4 which are the lengths of the sides, the answer isx = |ST| = 7 – ò24 cm.

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Quadratic fformulaThe roots x1 and x2 ofthe quadratic equationax2 + bx + c = 0 are

2

1,2

– – 4= .

2b b ac

xa


m + k = 3 n.

Given A(KMN) = 30 cm2,

find the value of n.

Solution m + k = 3 n (1) (Given)

A(KMN) = 30 cm2 (Given)

(Definition of the area of a triangle)

k m = 60 (2)

In KMN, n2 = k2 + m2 (Pythagorean Theorem)

n2 = (k + m)2 – 2km (Binomial expansion: (k+ m)2 = k2 + 2km + m2)

n2 = (3n)2 – 2 60 (Substitute (1) and (2))

8n2 = 120 (Simplify)

n2 = 15

n = ò15 cm.

= 30

2k m

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Proof We will give a proof by contradiction.

Suppose the triangle is not a right triangle,

and label the vertices A, B and C. Then there

are two possibilities for the measure of angle

C: either it is less than 90° (figure 1), or it is

greater than 90° (figure 2).

Let us draw a segment DC CB such that

DC = AC.

By the Pythagorean Theorem in BCD,

BD2 = a2 + b2 = c2, and so BD = c.

So ACD is isosceles (since DC = AC) and

ABD is also isosceles (AB = BD = c). As a

result, CDA CAD and BDA DAB.

However, in figure 3 we have

m(BDA) < m(CDA) and m(CAD) < m(DAB), which gives m(BDA) < m(DAB) if

CDA and CAD are congruent. This is a contradiction of BDA DAB. Also,

in figure 4 we have m(DAB) < m(CAD) and m(CDA) < m(BDA), which gives

m(DAB) < m(BDA) if CAD and CDA are congruent. This is also a contradiction.

So our original assumption must be wrong, and so ABC is a right triangle.

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EXAMPLE 36 In the triangle ABC opposite, K AC and

AN is the interior angle bisector of A.

AB = 16 cm, AN = 13 cm, AK = 12 cm and

NK = 5 cm are given. Find the area of

ABN.

Solution Let us draw an altitude NH from the vertex

N to the side AB.

To find the area of ABN we need to find NH, because and AB is given

as 16 cm.

132 = 122 + 52, so m(NKA) = 90°. (Converse of the Pythagorean Theorem)

Also, NH = NK = 5 cm. (Angle Bisector Theorem)

So (Substitution) 25 16

( )= = = 40 cm .2 2

NH ABA ABN

( )=2

NH ABA ABN

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98 Geometriy 7

Proof Let us join the point P to the vertices A, B

and C and use the Pythagorean Theorem in

the six right triangles that are created.

In ATP, AT2 + TP2 = AP2. (Pythagorean Theorem)

In ANP, AN2 + NP2 = AP2, (Pythagorean Theorem)

AT2 + TP2 = AN2 + NP2. (1) (Transitive property of equality)

In BKP, BK2 + KP2 = BP2. (Pythagorean Theorem)

In BTP, BT2 + TP2 = BP2, (Pythagorean Theorem)

BK2 + KP2 = BT2 + TP2. (2) (Transitive property of equality)

In CNP, CN2 + NP2 = CP2. (Pythagorean Theorem)

In CKP, CK2 + KP2 = CP2, (Pythagorean Theorem)

CN2 + NP2 = CK2 + KP2. (3) (Transitive property of equality)

From (1), (2) and (3) we get

AT2 +TP2 +BK2 +KP2 +CN2 +NP2 =AN2 +NP2 +BT2 +TP2 +CK2 +KP2 (Addition property

of equality)

AT2 + BK2 + CN2 = AN2 + BT2 + CK2. (Simplify)

Theorem

If a triangle ABC with P int ABC has

perpendiculars PK, PN, and PT drawn to the

sides BC, AC and AB respectively, then

AT2 + BK2 + CN2 = AN2 + BT 2 + CK2.

Carnot’s TTheorem

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We can use the Pythagorean Theorem to prove other theorems. Here is one example.


Practice Problems 111. Find the length x in each figure.

2. In the figure, TN = NK, ST = 12 cm

and SN = ò69 cm. Find the length of TK.

3. In a right triangle ABC, m(A) = 90°, AB = x, AC = x – 7 and BC = x + 1. Find AC.

Answers1. a. 15 b. 25 c. 9 d. 5ñ3 e. 20 f. 10 2. 10 cm 3. 5 cm

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a. b. c.

d. e. f.

Solution AT2 + BK2 + CN2 = AN2 + BT2 + CK2 (Carnot’s Theorem)x2 + 42 + (2x)2 = 22 + 42 + 62 (Substitute)

5x2 = 40 (Simplify)x2 = 8x = 2ñ2

EXAMPLE 37 Find the length x in the figure.�

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100 Geometriy 7

b. Further properties

Properties 71. The length of the median to the hypotenuse of a right triangle is equal to half of the length

of the hypotenuse.

2. a. In any isosceles right triangle, the length of the hypotenuse is ñ2 times the length ofa leg. (This property is also called the 45°-445°-990° TTriangle TTheorem.)

b. In any right triangle, if the hypotenuse is ñ2 times any of the legs then the triangle isa 45°-45°-90° triangle. (This property is also called the Converse oof tthe 45°-445°-990°Triangle TTheorem).

3. In any 30°-60°-90° right triangle,

a. the length of the hypotenuse is twice the length of the leg opposite the 30° angle.

b. the length of the leg opposite the 60° angle is ñ3 times the length of the leg oppositethe 30° angle. (These properties are also called the 30°-660°-990° TTriangle TTheorem.)

4. In any right triangle,

a. if one of the legs is half the length of the hypotenuse then the angle opposite this legis 30°.

b. if one of the legs is ñ3 times the length of the other leg then the angle opposite thisfirst leg is 60°. (These properties are also called the Converse oof tthe 330°-660°-990°Triangle TTheorem.)

5. The center of the circumscribed circle ofany right triangle is the midpoint of thehypotenuse of the triangle. �

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Activity

Fold a corner of a sheet of paper, and cut along the fold to get a right triangle. Label the vertices A, B and C so that m(B) = 90°. Fold each of the other two vertices to match point B. Label the point T on the hypotenuse where the folds intersect.What can you say about the lengths TA, TB and TC? Repeat the steps with a differentright triangle, and see if your conclusions are the same.

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EXAMPLE 39

EXAMPLE 38

In the figure at the right, find m(ADC) if

m(BAC) = 90°,

m(BAD) = 2x,

m(ACB) = 3x and

BD = DC.

Solution Since AD is a median, by Property 7.1 we have

So AD = BD = DC. Hence DCA and BDA are isosceles triangles.

Since DCA is isosceles, m(DAC) = m(ACD) = 3x.

Additionally, m(BAC) = m(BAD) + m(DAC) by the Angle Addition Postulate.

So 2x + 3x = 90° and x = 18°.

By the Triangle Angle-Sum Theorem in DCA, m(ADC) + 3x + 3x = 180°.

So m(ADC) = 180° – (6 18)°, i.e. m(ADC) = 72°.

1= .

2AD BC

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In the figure,

m(BAC) = 90°,

m(C) = 60° and

BD = DC.

Find BC if AD = 2x + 3 and AC = 6x – 1.

Solution Since AD is a median and the length of the median to the hypotenuse of a right triangle

is equal to half the length of the hypotenuse,

By the Triangle Angle-Sum Theorem in ABC, m(B) = 30°.

By the 30°-60°-90° Triangle Theorem, because m(B) = 30° and BC is the

hypotenuse.

So by the transitive property of equality, AC = AD, i.e. 6x – 1 = 2x + 3 and so x = 1.

Finally, BC = 2 AC = 2 AD = 2 (2x + 3) = 10.

AC BC1

=2

1= .

2AD BC

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102 Geometriy 7

EXAMPLE 40 One of the acute angles in a right triangle measures 16°. Find the angle between the

bisector of the right angle and the median drawn from the same vertex.

Solution Let us draw an appropriate figure. We need

to find m(NAT).

According to the figure,

AN is the angle bisector, AT is the

median, and m(BAC) = 90°.

m(ACB) = 16° by Property 5.3.

Since AT is median to hypotenuse, AT = CT = BT.

So ATC is isosceles.

Therefore, by the Isosceles Triangle Theorem, m(TAC) = m(ACT) = 16°.

Since AN is an angle bisector and m(BAC) = 90°, m(NAC) = 45°.

So m(NAT) = m(NAC) – m(TAC)= 45° – 16° = 29°.

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Solution In ABC, since m(C) = 30°,

m(B) = 60°.

In ABH, since m(B) = 60°,

m(BAH) = 30°.

In ABH, by Property 7.3,

AB = 2 BH = 2 2 = 4 cm.

In ABC, again by Property 7.3,

BC = 2 AB = 2 4 = 8 cm.

So HC = BC – BH = 8 – 2 = 6 cm.

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EXAMPLE 41 In the figure, AB AC and AH BC.

Given m(C) = 30° and BH = 2 cm, find the

length of HC.

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This set square is in theform of a 30°-60°-90°triangle.

Property 5.3:

In any triangle ABC, ifm(B) > m(C) orm(B) < m(C) thenha < na < Va.


Solution We will construct a right triangle in which one leg is half of the hypotenuse. Then by the

Converse of the 30°-60°-90° Triangle Theorem, the angle opposite this leg will measure 30°.

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5. 6. 7.

4.

EXAMPLE 43 Construct a 30° angle by using the Converse of the 30°-60°-90° Triangle Theorem.

EXAMPLE 42 Find the value of x in the figure.

Solution Let us draw an altitude from C to AB.

In BHC,

BC = ñ2 BH (45°-45°-90° Triangle

Theorem)

6ñ2 = ñ2 BH (Substitute)

BH = 6. (Simplify)

AB = AH + HB (Segment Addition

Postulate)

10 = AH + 6 (Substitute)

AH = 4. (Simplify)

In AHC,

AC = 2 AH (30°-60°-90° Triangle Theorem)

AC = 2 4 (Substitute)

AC = x = 8. (Simplify)

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104 Geometriy 7

EXAMPLE 44 Construct the circumscribed circle of a given right triangle.

Solution

1. Label ABC with m(A) = 90° and m(B) > m(C).2. Construct the perpendicular bisector of each side and label their point of intersection M. 3. Draw a circle with center M and radius MB. This is the circumscribed circle.

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1. 2. 3.

1. Draw a line m.2. Construct a perpendicular to the line at any point. Name the line n and label the

intersection point of these lines B. 3. Draw an arc with center B and any radius r. Label the intersection point C of this arc and

the line m. 4. Construct the midpoint K of the segment BC and draw the line k perpendicular to m

through this point.5. Draw an arc with center K and radius BC. Label the intersection point A of the arc and line n. 6. Join A and K to make AK = 2 KB. 7. By the Converse of the 30°-60°-90° Triangle Theorem, m(BAK) = 30°.

Activity

The Kobon ttriangle pproblem is an unsolved problem ingeometry which was first stated by Kobon Fujimura. Theproblem asks: What is the largest number of non-overlappingtriangles that can be produced by n straight line segments?This might seem like a simple question, but nobody has yetfound a general solution to the problem. The mathematician Saburo Tamura proved that the largestinteger below n(n – 2) / 3 is an upper bound for the number ofnon-overlapping triangles which can be produced by n lines. Ifthe number of triangles is exactly equal to this upper bound,the solution is called a perfect solution. Perfect solutions areknown for n = 3, 4, 5, 7, 9, 13 and 15, but for other n-valuesperfect Kobon triangle solutions have not been found.

The perfect Kobon solution for five lines creates 5 (5 – 2) / 3 = 5 triangles. Can you find

this solution?

Make Kobon patterns with seven lines. Can you find the perfect solution for seven lines?

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A perfect Kobon solution with15 lines and 65 triangles(T. Suzuki, Oct. 2, 2005)


Check Yourself 121. In an isosceles right triangle, the sum of the lengths of the hypotenuse and the altitude

drawn to the hypotenuse is 27.3. Find the length of the hypotenuse.

2. In the figure, ABC is a right triangle with

m(ABC) = 90° and CF = FE, and CE is the

angle bisector of C. If m(ADB) = 102°, find

the measure of CAB.

3. One of the acute angles in a right triangle measures 48°. Find the angle between the

median and the altitude which are drawn from the vertex at the right angle.

4. In a triangle ABC, m(B) = 135°, AC = 17 cm and BC = 8ñ2 cm. Find the length of AB.

5. In a right triangle, the sum of the lengths of the hypotenuse and the shorter leg is 2.4.

Find the length of the hypotenuse if the biggest acute angle measures 60°.

6. In the figure,m(C) = 60°,HC = 4 cm andDH = 2ñ3 cm. Find the length AC = x.

7. ABC in the figure is an equilateral triangle with

DH BC,

BH = 5 cm and

HC = 3 cm.

Find the length AD = x.

8. The distance from a point to a line k is 10 cm. Two segments non-perpendicular to k are

drawn from this point. Their lengths have the ratio 2 : 3. Find the length of the longer

segment if the shorter segment makes a 30° angle with k.

9. CAB is a right triangle with m(A) = 90° and m(C) = 60°, and D is the midpoint of

hypotenuse. Find the length of the hypotenuse if AD = 3x + 1 and AC = 5x – 3.

10.The hypotenuse of an isosceles right triangle measures 18 cm. Find the distance from the

vertex at the right angle to the hypotenuse.

Answers1. 18.2 2. 22° 3. 6° 4. 7 cm 5. 1.6 6. 5 cm 7. 2 cm 8. 30 cm 9. 14 10. 9 cm

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106 Geometriy 7

D. THE TRIANGLE ANGLE BISECTOR THEOREMTheorem

1. The bisector of an interior angle of a

triangle divides the opposite side in the

same ratio as the sides adjacent to the

angle. In other words, for a triangle ABC

and angle bisector AN,

2. In any triangle ABC, if the bisector AN of

the exterior angle A cuts the extension

of side BC at a point N, then

.AB BNAC CN

= .AB BNAC CN

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Triangle AAngle BBisector TTheorem

Proof of 1 We begin by drawing two perpendiculars NK

and NL from N to the sides AB and AC

respectively, then we draw the altitude

AH BC.

(1) (Definition of the area of a triangle and simplify)

Now let us find the same ratio by using the sides AB and AC and the altitudes NK and NL.

Since N is the point on the angle bisector, by the Angle Bisector Theorem we have NK = NL.

( )=

( )

AHA ABNA ANC

2

BN

AH 2

NC=

BNCN


(By (1), (2) and the transitive property of equality)=AB BNAC CN

( )=

( )

NKA ABNA ANC

2

AB

NL 2

AC=

ABAC

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and NL from N to the extension of the sides

AB and AC respectively, then we draw the

altitude AH BN.


Now let us find the same ratio by using the sides AB and AC, and the altitudes NK and NL.

Since N is the point on the angle bisector, by the Angle Bisector Theorem we have NK = NL.


(By (1), (2) and the transitive property of equality)=AB BNAC CN

( )=

( )

NKA ABNA ACN

2

AB

NL 2

AC=

ABAC

( )=

( )

AHA ABNA ACN

2

BN

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CN=

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Solution (Triangle Angle Bisector

Theorem)

12 6=

8

3 6=

2

= 4

x

x

x

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108 Geometriy 7

1. In any triangle ABC, if AN is an interior

angle bisector then

AN2 = (AB AC) – (BN NC).

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2. In any triangle ABC, if AN is an

exterior angle bisector then

AN2 = (BN CN) – (AB AC).�

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Solution By the Triangle Angle Bisector Theorem,

By Property 8.1,

x2 = 6 4 – 3 2 = 18

x = 3ñ2.

6 3=

2= 4.

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Solution (Triangle Angle Bisector

Theorem)

3 1= ; 3 =12+ ; = 6.

12+x x x

x x

12 4=

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EXAMPLE 48 In the figure, m(CAB) = 2 m(ABC).

Given that AC = 4 cm and AB = 5 cm, find

the length of BC.

Solution Let AD be the bisector of angle A.

Then m(B) = m(DAB) = m(DAC), since m(CAB) = 2 m(ABC).

So DAB is an isosceles triangle. Let AD = DB = x. If BC = a then CD = a – x.

By the Triangle Angle Bisector Theorem in BAC,

5(a – x) = 4x

x = (1)

Now we can use Property 8.1:

x2 = 5 4 – x(a – x)

x2 = 20 – ax + x2

ax = 20. (2)

Substituting (1) into (2) gives

a = 20 ; a2 = 36 ; a = 6 cm.59a

5.

9a

5=

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a x

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Check Yourself 131. In a triangle ABC, N is on side BC and AN is the angle bisector of A. If AB = 8 cm,

AC = 12 cm and BC = 10 cm, find AN.

2. In a triangle KMN, points M, Z, N and T are collinear, KZ is the angle bisector of the

interior angle K, and KT is the angle bisector of the exterior angle K. If MZ = 5 cm and

ZN = 3 cm, find NT.

3. MP is the angle bisector of the interior angle M of a triangle KMN. Find MN if

KP : PN = 3 : 4 and KM = 15 cm.

4. In a triangle ABC, point D is on side BC and AD is the bisector of the angle A. Find BD if

BA : AC = 5 : 3 and BD + DC = 8 cm.

5. ET is an angle bisector in a triangle DEF. Find the length of ET if DE = 14, EF = 21 and

DF = 15.

Answers1. 6ñ2 cm 2. 12 cm 3. 20 cm 4. 5 cm 5. 4ò15

110 Geometriy 7

A. The Concept of Congruence11.. Find two pairs of congruent figures in each

picture. Draw each pair.

a. b.

EXERCISES 3.2

22..

In the figure, polygon ABCDE is congruent to

polygon KLMNP. Find each value, using the

information given.

a. x b. y c. n d. a e. b

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KMN. Find the unknown value in each statement,

using the properties of congruence.

a. AC = 5m – 25, KN = 11 – m

b. m(BCA) = 45° – v, m(MNK) = 2v – 15°,

c. m(B) = 18°, m(M) = u – 4°,

d. BA = 22x – 30, MK = 3 – 2x

55.. Two triangles ABC and CMN are congruent to

each other with AB = 8 cm, CN = 3 cm, and

CM = AC = 6 cm. Find BC and MN.

B. Constructions66.. Construct each angle.

a. 15° b. 105° c. 75° d. 37.5° e. 97.5°

33.. ADE KLN is given. List the congruent

corresponding segments and angles of these

triangles.

77.. a. Construct an isosceles triangle with base 5 cm

and perimeter 19 cm.

b. Construct an equilateral triangle with perimeter

18 cm.

c. Try to construct a triangle with sides of length

2 cm, 3 cm and 6 cm. What do you notice?

Can you explain why this is so?

d. Construct a triangle ABC with side lengths

a = 5 cm and c = 7 cm, and m(B) = 165°.

e. Construct a right triangle ABC in which

m(C) = 90°, b = 5 cm and the hypotenuse

measures 7 cm.

f. Construct ABC with angles m(C) = 120°

and m(B) = 45°, and side b = 5 cm.

g. Construct an isosceles triangle PQR with

vertex angle m(Q) = 40° and side QP = 3 cm.

In each case, construct a triangle using only the

three elements given.

a. a, b, Vc b. a, b, hc

c. a, b, nC d. ha, hb, hc

e. A, a, ha f. Va, Vb, Vc

88..


1122.. In a triangle KLM, m(LKM) = 30°,

m(LMK) = 70° and m(KLM) = 80°.

O int KLM and KO = LO = MO are given.

Find m(OKM), m(OML) and m(OLK).

10.. In the figure,

AB = BC,

AD = DB

and BE = EC.

If DC = 3x + 1 and

AE = 4x – 1,

find the length x.

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11.. In a triangle DEF, m(E) = 65° and m(F) = 15°,

and DK is an angle bisector. Prove that DEK is

isosceles.

C. Isosceles, Equilateral and RightTriangles

9.. KMN is an isosceles triangle with base KN. The

perpendicular bisector of the side MK intersects

the sides MK and MN at the points T and F,

respectively. Find the length of side KN if

P(KFN) = 36 cm and KM = 26 cm.


ABC is an equilateral

triangle.

PE BC, PD AC,

PE = 3 and

PD = 5 are given.

Find the length of one side of the equilateral

triangle.

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triangle, PE AC,

PD BC, and PF AB.

Given P(ABC) = 45,

find the value of

PD + PE + PF.

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BH = HC,

AB = DC and

m(B) = 54°.

Find m(BAC).

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BE = EC and

AD bisects the interior

angle A.

Given AB = 12 and

AC = 7, find the

length of ED.

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O is the center of the

inscribed circle of

ABC, OB = CD,

AB = AC and

m(ECD) = 20°.

Find the measure of

BEC.

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1188.. In a right triangle ETF, the perpendicular

bisector of the leg ET intersects the hypotenuse at

the point M. Find m(MTF) if m(E) = 52°.

1199.. In triangle DEF, DE = EF, m(DEF) = 90° and

the distance from E to DF is 4.8 cm. Find DF.

112 Geometriy 7


m(BAC) = 90°,

m(AHC) = 90°,

m(B) = 30° and

HC = 1 cm.

Find the length of BH.

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m(DBA) = 30°,

m(ABC) = 60° and

AD = 4. Find the

length of BC.

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2266.. In a triangle ABC, BC = 7ñ2, m(BAC) = 30° and

m(BCA) = 45°. Find the length of the side AB.

2277.. The larger acute angle A in an obtuse triangle

ABC measures 45°. The altitude drawn from the

obtuse angle B divides the opposite side into two

segments of length 9 and 12. Find the length of

the side BC.


m(M) = 150°,

LM = 2 and

MN = 3ñ3.

Find the length of KL.

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22.. In the figure,

BC = 12,

m(BAC) = 90°,

m(ADC) = 90° and

m(ABC) = 60°.

If AC is the angle

bisector of C, find the length of AD.

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m(A) = m(B) = 60°,

AD = 7 and

BC = 4.

Find DC.

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AB = AC,

AD = DB and

CD = 8 cm.

Find the length of DB.

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m(A) = 30° and

AB = AC = 16 cm.

Find the value of

PH + PD.

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In the figure, PMN is a right triangle, MH PN

and m(N) = 15°. Find

(Hint: Draw the median to the hypotenuse.)

MHPN

.

2211..

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20.. TF is the hypotenuse of a right triangle TMF, and

the perpendicular bisector of the hypotenuse

intersects the leg MF at the point K. Find

m(KTF) given m(MTF) = 70°.



m(A) = 90°,

m(ADC) = 60°,

m(B) = 30° and

BD = 4 cm.

Find the length of AD.

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3311.. ABC in the figure is

an equilateral triangle.

BH = 5 cm and

HC = 3 cm are given.

Find the length

AD = x.



triangle, BD = 4 cm

and AE = 6 cm. Find

the length of EC.

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m(C) = 90°,

m(BAD) = m(DAC),

BD = DA and

DC = 3 cm. Find the

length of BD.

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3388..

In the figure, NK is the bisector of the interior

angle N and NL = LK. m(NMP) = 90° and

m(P) = 24° are given. Find m(PSM).

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m(BAE) = m(DAE) = 60°,

CB = 6 cm and

CE = 3 cm.

Find the length of CD.

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triangle, PE AC and

PD AB.

PD = 5 cm,

PE = 3 cm and

P(ABC) = 48 cm are

given. Find the length of PH.

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ABC is an isosceles

triangle, AB = AC,

m(BAC) = 30° and

2PE = PD = 4.

Find AC.

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3366.. In a right triangle KLM, m(KLM) = 90° and theperpendicular bisector of the leg LM cuts thehypotenuse at the point T. If m(LMK) = 20°,find m(TLK).


AB = BC

m(ADB) = 30° and

AC = 6.

Find the length of CD.

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114 Geometriy 7


AT = DB = DC and

m(A) = 36°.

Find m(DBT).

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ED = AC,

BD = DC and

m(C) = 63°. Find

m(EDC).

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AB = AC = 18 cm,

PH = 5 cm and

PD = 4 cm.

Find m(HPD).

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triangle, AH = ED,

AE = EC and

CD = 4 cm.

Find the length of

AB.

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AB = AC,

AH = HB,

m(A) = 120° and

PB = 8 cm. Find

the length of CP.

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4422.. A rectangular opening in a wall has dimensions

21 cm by 20 cm. Can an empty circular service

tray with a diameter of 28 cm pass through the

opening?

4433.. Two ships are in the same port. One starts to

travel due west at 40 km/h at 3 p.m. Two hours

later the second ship leaves port, traveling due

south at 60 km/h. How far apart will the ships be

at 7 p.m.?

4444.. The median drawn from the vertex at the right

angle of a right triangle divides the right angle in

the ratio 13 : 5. Find the smallest angle in this

triangle.

4455.. In a triangle DEF, m(DEF) = 120°, DE = 2ñ5

and DF = 8. Find the length of the side EF.


BD = DC and

BC = 12.

Find the length of AD.

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AD = EB,

CD = DB and

m(DEB) = 80°. Find

the measure of ACB.

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4411.. The base and a leg of an isosceles triangle

measure 14 cm and 25 cm respectively. Find the

height of the altitude drawn to the base.


5533.. A line m divides a segment KN with the ratio 2 : 3

at the point M. Find the distances from the points

K and N to m if KN = 40 and the angle between

m and the segment KN is 30°.

5544.. In a triangle KMN, T is the intersection point of

the three angle bisectors and the distance from T

to the side MN is 4. Find the distance from T to

the vertex K if m(K) = 60°.

5555.. Prove that in a right triangle ABC with m(A) = 90°

and acute angles 15° and 75°, the altitude to the

hypotenuse measures of the length of the

hypotenuse.

14

In the figure,


triangle, m(BCE) = 15°,

EF = FC, DF EC and

AD = 2 cm are given.

Find AE.

5588..

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5577.. Prove that if AH is an

altitude and AD is a

median of a triangle

ABC then

|b2 – c2| = 2a HD.

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In the figure,

AC BD,

CE = 2AB and

m(C) = 15°.

Find m(CAF).

5566..

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BE = EC, BD AC,

m(A) = 45°,

m(D) = 30° and

DC = 6. Find the

length x.

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5599.. In the figure,NL = LK andNK is the bisectorof N. Ifm(NMP) = 90°and m(MSP) = 120º, what is m(P)?

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6633.. In a triangle ABC, D lies on the side AC and BD is

the interior angle bisector of B. If AD = 5 cm,

DC = 6 cm and the sum of the lengths of the sides

a and c is 22 cm, find the value of a.

D. The Triangle Angle Bisector Theorem

6611.. In a triangle ABC, D AB and CD is the interior

angle bisector of C. Given AC = 9 cm, BD = 8 cm,

AD = m and BC = n, find the value of m n.

6644.. In a triangle ABC, points C, A and D are collinear

and C, B and E are also collinear. BD is the angle

bisector of EBA. If AC = AD and AB = 6 cm,

find the length of BC.

6622.. In a triangle ABC, D lies on the side BC and AD is

the interior angle bisector of A. If AC = BD,

AB = 9 cm and DC = 4 cm, find the length of BD.

116 Geometriy 7


the interior angle bisector of A. If AB = AD = 12 cm

and AC = 16 cm, find the length of BD.

7722.. In a triangle ABC, points D, B and C are collinear

and AD is the angle bisector of A. If AB = 2m,

AC = 2m + 6, BC = 2m + 2 and DB = 28, find

the value of m.

7700.. In a triangle ABC, points B, C and D are collinear

and AD is the angle bisector of A. If CD = 3 BC

and AB = 12 cm, find the length of AC.

7711.. In a triangle ABC, D AB, E BC and AE and CD

are the angle bisectors of A and C respectively,

intersecting at the point K. If AD = 4, DB = 6 and

AC = 8, find the value of .AKKE

7733.. In the figure, AN and BE

are the angle bisectors of

A and B respectively.

If AB = 4,

AC = 6 and BC = 5,

find the length of BE.

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7744.. In the triangle

ABC at the right,

AD is the angle

bisector of Aand BN is the

angle bisector of

B. DB = m,

BC = 3, AB = 4 and NC = 2 are given. Find the

value of m.

7755.. In a triangle ABC, points D, B and C are collinear

and AD is the angle bisector of A. If AB = 12

and BD = 4 BC, find the length of AC.

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6699.. In the figure, AD and

BE are the angle

bisectors of A and

B, respectively,

and

AC = 12 cm.

Find the length of segment DC.

AEED

3=

2

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6688.. In a triangle DEF, E, F and K are collinear and

DK is the exterior angle bisector of the angle D. If

DE = 4 cm, DF = 3 cm and EF = 2 cm, find DK.


the interior angle bisector of A. If BD = 3 cm,

DC = 4 cm and AB + AC = 14 cm, find the length

of AD.

6655.. In a triangle EFM, points F, N and M are collinear

and EN is the interior angle bisector of E.

If EN = 6, NM = 3 and FN = 4, find the length

of EF.


A. RELATIONS BETWEEN ANGLES

In this section we will look at some basic rules related to the basic and auxiliary elements of

a triangle.

Activity Do the following activities and then try tofind a common conjecture.

1. Cut out three identical triangles andlabel their vertices K, M and N. Draw astraight line and place the trianglesalong the line as in the diagram at thefar right. What can you say about thesum of K, M and N?

2. Cut out an acute triangle, a right triangleand an obtuse triangle. Number theangles of each triangle 1, 2 and 3 andtear them off. Finally, put the threeangles of each triangle adjacent to eachother to form one angle as in the figures at the far right. What can yousay about the sum of the angles in eachtriangle?

3. Cut out a triangle ABC and label itslongest side BC. Fold the triangle so that A lies on the fold line and C lies on BC. Labelthe intersection T of BC and the fold line. Unfold. Now fold the paper so that A, B andC coincide with T. How does this activity support the results of activities 1 and 2 above?

AAnngglleess iinn aa TTrriiaannggllee

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‘It is indeed wonderfulthat so simple a figure

as the triangle is soinexhaustible in

properties.How many as yet

unknown properties ofother figures may there

not be?’August Crelle(1780-1856),

civil engineer andmathematician

After studying this section you will able to1. Describe and use relations between angles2. Describe and use relations between angles

Objectives

118 Geometriy 7

EXAMPLE 49 In the figure, points A, B, F and E, B, C are

respectively collinear. Given that C and F

are right angles, m(E) = 40° and m(A) = ,

find the value of .

Solution In EFB by the Triangle Angle-Sum Theorem,

m(E) + m(F) + m(B) = 180°

40° + 90° + m(B) = 180°

m(B) = 50°.

By the Vertical Angles Theorem,

m(EBF) = m(ABC)

50° = m(ABC).

In ABC by the Triangle Angle-Sum Theorem,

m(A) + m(B) + m(C) = 180°

+ 50° + 90° = 180°

= 40°.

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Theorem

The sum of the measures of the interior angles of a triangle is 180°.

Proof Given: ABC

Prove: m(1) + m(2) + m(3) = 180°

We begin by drawing an auxiliary line DE

through A, parallel to BC. Then we continue

with a two-column proof.

Statements Reasons

1. 1 4; 3 5 1. Alternate Interior Angles Theorem

2. m(DAE)=m(4)+m(2)+m(5)=180° 2. Angle Addition Postulate, by the definition of straight angle

3. m(1) + m(2) + m(3) = 180° 3. Substitute 1 into 2

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Triangle AAngle-SSum TTheorem

An auxiliary line is aline which we add to afigure to help in a proof.


Theorem

The measure of an exterior angle in a triangle is equal to the sum of the measures of its two

nonadjacent interior angles.

Triangle EExterior AAngle TTheorem

Proof Given: ABC

Prove: m(1) = m(3) + m(4)

Statements Reasons

1. m(1) + m(2) = 180°

m(2) = 180° – m(1)1. Linear Pair Postulate

2. m(2) + m(3) + m(4) = 180° 2. Triangle Angle-Sum Theorem

3. 180° – m(1) + m(3) + m(4) = 180° 3. Substitute 1 into 2.

4. m(1) = m(3) + m(4) 4. Simplify.

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Solution m(A) + m(B) + m(C) = 180° (Triangle Angle-Sum Theorem)

3x – 10° + 2x + 20° + 5x = 180°

10x + 10° = 180°

x = 17°

Activity Complete the table for the figure at the right, using the

Triangle Angle-Sum Theorem and the Linear Pair Postulate.

What do you notice about the values in the last two columns of the table?

m(4) m(3) m(2) m(1) m(3) ++ m(4)

75° 55°

63° 135°

77° 46°

39° 85°

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EXAMPLE 50 In a triangle ABC, m(A) = 3x – 10°, m(B) = 2x + 20° and m(C) = 5x.

Find the value of x.

The two interior angleswhich are not adjacentto an exterior angle in atriangle are sometimescalled remote aangles.

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120 Geometriy 7

EXAMPLE 53 In a triangle KMN, D lies on side KM. Decide whether each statement about the figure is

possible or impossible. If it is possible, sketch an example.

a. Triangles KDN and MDN are both acute triangles.

b. Triangles KDN and MDN are both right triangles.

c. Triangles KDN and MDN are both obtuse triangles.

d. Triangle KDN is obtuse and triangle KNM is acute.

Solution a. impossible b. possible c. possible d. possible

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EXAMPLE 52 In the figure, AB = BD, AD = DC

and m(DAC) = 35°. Find m(B).

Solution m(DCA) = m(DAC) (Base angles of an isosceles triangle)

= 35°

m(BDA) = m(DAC) + m(DCA) (Triangle Exterior Angle Theorem)

= 35° + 35°

= 70°

m(BAD) = m(BDA) (Base angles of an isosceles triangle)

= 70°

m(B) + m(BAD) + m(BDA) = 180° (Triangle Angle-Sum Theorem)

m(B) + 70° + 70° = 180°

m(B) = 40°

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EXAMPLE 51 In a triangle ABC, AB AC and m(B) = 136°. Find m(C).

Solution m(B) = m(A) + m(C)

136° = 90° + m(C)

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Theorem

The sum of the measures of the exterior angles of a triangle is equal to 360°.

Proof Given: ABC

Prove: m(A) + m(B) + m(C) = 360°

We will give a flow-chart proof.

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Triangle EExterior AAngle-SSum ttheorem

EXAMPLE 54 In the figure, m(TCA) = 120°,

m(KAB) = 5x and m(PBC) = 7x.

a. Find the value of x.

b. Find m(BAC).

Solution a. m(A) + m(B) + m(C) = 360° (Triangle Exterior Angle-Sum Theorem)

5x + 7x + 120° = 360°

12x = 240°

x = 20°

b. m(KAB) + m(BAC) = 180° (Linear Pair Postulate)

(5 20°) + m(BAC) = 180°

m(BAC) = 80°

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122 Geometriy 7

Check Yourself 141. The two acute angles in a right triangle measure 0.2x + 6.3° and 3.8x – 2.7. Find x.

2. The measures of the interior angles of a triangle are in the ratio 4 : 6 : 8. Find the degree

measures of these angles.

3. The vertex angle of an isosceles triangle measures 42°. An altitude is drawn from a base

angle to one of the legs. Find the angle between this altitude and the base of the triangle.

4. In an isosceles triangle, the angle between the altitude drawn to the base of the triangle

and one leg of the triangle measures 16° less than one of the base angles of the triangle.

Find the measure of the vertex angle of this triangle.

5. Two points E and F are drawn on the extension of the side MN of a triangle MNP such

that point M is between the points E and N and point N is between points M and F.

State which angle is the smallest angle in EPF if EM = MP, NF = NP, m(PMN) = 30°

and m(PNM) = 40°.

Solution m(A) + m(A) = 180° (Linear Pair Postulate)

m(A) = 180° – m

180° – m + n + k = 360° (Triangle Exterior Angle-Sum Theorem)

n + k = 180° + m (1)

Also, m + n + k = 280° (Given)

m + 180° + m = 280° (Substitute (1))

2m = 100°

m = 50°.

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m(A) = m,

m(B) = n and m(C) = k.

Find the value of m if m + n + k = 280°.

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6. In a triangle DEF, point M lies on the side DF such that MDE and DEM are acute

angles. Decide whether each statement about the figure is possible or impossible. If it is

possible, sketch an example.

a. FME is an acute triangle.

b. FME is a right isosceles triangle.

c. FME and DME are both acute triangles.

d. DME is equilateral and EMF is isosceles.

e. DME is isosceles and DEF is isosceles.

7. In the figure, KLN is an isosceles triangle in a plane,

m(KLN) = 120°, and L is the midpoint of the segment KM.

A point P is taken in the same plane such that MP = KL. Find

the measure of LPM when

a. the distance between N and P is at its maximum.

b. the distance between N and P is at its minimum.

8. One of the exterior angles of an isosceles triangle measures 85°. Find the measure of the

vertex angle of this triangle.

9. State whether each triangle is a possible or impossible figure. If it is possible, sketch an

example. If it is impossible, give a reason why.

a. A triangle with two obtuse exterior angles.

b. A triangle with one acute exterior angle.

c. A triangle with two right exterior angles.

d. A triangle with two acute exterior angles.

10.Find the value of x in each figure, using the information given.

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124 Geometriy 7

Answers1. 21.6° 2. 40°, 60° and 80° 3. 21° 4. 74° 5. PEF

6. aa. possible b. possible c. not possible d. possible e. possible

7. aa. 30° b. 60° 8. 95°

9. a. possible b. possible

c. impossible because the third exterior angle would be 180°

d. impossible because the third exterior angle would have to be more than 180°

10. aa. 45° b. 25° c. 27° d. 80°

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Properties 31. For any triangle, the following statements are true:

a. The measure of the angle formed by the bisectors of two interior angles of the triangle

is 90° more than half of the third angle, i.e. in the figure,

m(BKC) =

b. The measure of the angle formed by

the bisectors of two exterior angles of a

triangle is 90° minus half of the third

angle, i.e. in the figure,

m(BTC) = 90° –

c. The measure of the angle which is

formed by the bisector of one interior

angle and the bisector of a second

exterior angle is the half the measure

of the third interior angle.

We can refer to properties 1a, 1b and 1c as the Angle BBisector RRelations TTheorem.

( ).

2m BAC

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2m BAC

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So far we have looked at some basic properties of angles. Now we will study some other

useful and important properties.


2. In any triangle, the measure of the angle

formed by the altitude and the angle

bisector which both extend from the

same vertex is equal to the half the

absolute value of the difference of the

other two angles of the triangle.

3. In any triangle KLM, if N is any point in

the interior of KLM then

a. m(LNM) = m(LKM) + m(KLN)

+ m(KMN).

b. m(KNM) = m(KLM) + m(LKN)

+ m(LMN).

c. m(KNL) = m(KML) + m(MLN) + m(MKN).

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Solution Since the incenter is the intersection of the

angle bisectors, both AO and CO are bisectors.

By Property 3.1a,

m(AOC)= ( ) 80°90°+ = 90°+ =130 .

2 2m B

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EXAMPLE 56 The triangle ABC at the right has incenter O.

Find m(AOC).

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EXAMPLE 57 In the figure, K is the intersection point of

the bisectors of the exterior angles at

vertices A and B with m(A) = 120°

and m(B) = 40°. Find m(BKA).

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126 Geometriy 7

Solution m(A) + m(B) + m(C) = 180° (Triangle Angle-Sum Theorem in ABC)

120° + 40° + m(C) = 180°

m(C) = 20° (1)

m(BKA)= (Property 3.1b)

= (Substitute (1))

= 80°

20°90° –

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2m C

EXAMPLE 58 Find the value of x in the figure. �

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Solution m(C) + m(C) = 180° (Linear Pair Postulate)

m(C) = 180° – 3x (1)

m(AEB)= (Property 3.1c)

x = (Substitute (1))

5x = 180°

x = 36°

180° – 32

x

( )2

m C

EXAMPLE 59 In the figure at the right,

AN is an angle bisector,

m(ANC) = 100° and

m(B) = 2m(C).


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Solution m(C) = x is given, so m(B) = 2x.

Let us draw the altitude AH BC.

Since ANC is an exterior angle of AHN,

m(HAN) + m(AHN) = m(ANC)

m(HAN) + 90° = 100°

m(HAN) = 100° – 90° = 10°

m(HAN) = (Property 3.2)

10° =

10° =

x = m(C) = 20°.2x

|2 |2

x x

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m B m C

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EXAMPLE 60 One of the acute angles in a right triangle measures 20°. Find the angle between the altitude

and the angle bisector which are drawn from the vertex of the right angle of the triangle.

Solution Let us draw an appropriate figure. In the

figure at the right, A is the right angle, AN

is the angle bisector and

m(NAB) = m(NAC) = 45°.

Let m(C) = 20°, then m(B) = 70° and

m(HAB) = 20°.

Therefore, m(HAN) = m(NAB) – m(HAB) = 45° – 20° = 25°. This is the required angle

measure.

Note that we can also solve this example by using Property 3.2. This is left as an exercise for

you.

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EXAMPLE 61 In a triangle KLM, prove that if N is a point in the interior of KLM then

a. m(LNM) = m(LKM) + m(KLN) + m(KMN),

b. m(KNM) = m(KLM) + m(LKN) + m(LMN) and

c. m(KNL) = m(KML) + m(MLN) + m(MKN).

128 Geometriy 7


Given: N is an interior point of KLM

Prove: m(1) = m(2) + m(3) + m()

Proof:

Let us extend segment MN through N and

label the intersection point T of ray MN and segment KL.

m(LTN) = m(3) + m(4) (Triangle Exterior Angle Theorem)

m(1) = m(2) + m(3) + m(4) (Triangle Exterior Angle Theorem)

This means m(LNM) = m(LKM) + m(KLN) + m(KMN).

The proofs of b. and c. are similar. They are left as an exercise for you.

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EXAMPLE 62 ABC is an equilateral triangle and a point D int ABC such that AD DC and

m(DCA) = 42°. Find m(BAD).


m(B) = 60° (ABC is equilateral)

m(BCD) = 60° – 42°

= 18° (m(BCA) = 60°)

m(ADC) = m(B) + m(BAD) + m(BCD) (Property 3.3)

90° = 60° + m(BAD) + 18°

m(BAD) = 12°

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Check Yourself 151. Each figure shows a triangle with two or more angle bisectors. Find the indicated angle

measures in each case.

2. In the triangle ABC at the right, AN is an angle bisector

and AH is an altitude. Given m(C) – m(B) = 36°, find

m(HAN).

3. A student draws the altitude and the angle bisector at the vertex of the right angle of a right

triangle. The angle between them is 18°. Find the measure of the larger acute angle in the

right triangle.

4. Find the value of x in the figure.

Answers

1. aa. 110° b. 80° c. 35° d. 40° e. p f. 80° 2. 18° 3. 63° 4. 15°

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130 Geometriy 7

B. RELATIONS BETWEEN ANGLES AND SIDESTheorem

If one side of a triangle is longer than another side of the triangle then the measure of the angle

opposite the longer side is greater than the measure of the angle opposite the shorter side. In

other words, if two sides of a triangle have unequal lengths then the measures of the angles

opposite them are also unequal and the larger angle is opposite the longer side.

Proof Given: ABC with AB > AC

Prove: m(C) > m(B)

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longer sside oopposite llarger aangle

We begin by locating K on AB such that AK = AC. We then draw CK and continue with a two-

column proof.

Statements Reasons

1. AB > AC 1. Given

2. AKC is isosceles 2. Definition of isosceles triangle (AK = AC)

3. 3 2 3. Base angles in an isosceles triangle are congruent.

4. m(ACB) = m(2) + m(1) 4. Angle Addition Postulate

5. m(ACB) > m(2) 5. Definition of inequality

6. m(ACB) > m(3) 6. Substitution property

7. m(3) > m(B) 7. Triangle Exterior Angle Theorem

8. m(ACB) > m(B) 8. Transitive property of inequality

EXAMPLE 63 Write the angles in each triangle in order of

their measures.

Solution a. Since 7 > 5 > 3, m(A) > m(B) > m(C).

b. Since 5 = 5 > 4, m(E) = m(F) > m(D).

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Theorem

If two angles in a triangle have unequal measures then the sides opposite them have unequal

lengths and the longer side is opposite the larger angle.

Proof Given: ABC with m(B) > m(C)

Prove: AC > AB

We will give a proof by contradiction in

paragraph form.

According to the trichotomy property, exactly

one of three cases holds: AC < AB, AC = AB

or AC > AB.

Let us assume that either AC = AB or AC < AB and look for a contradiction.

If AC < AB then m(B) < m(C) by the previous theorem. Also, if AC = AB then

m(B) = m(C) by the definition of an isosceles triangle.

In both cases we have a contradiction of the fact that m(B) > m(C). That means that

our assumption AC AB must be false. By the trichotomy property, it follows that AC > AB.

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Trichotomy ppropertyFor any two real numbersa and b, exactly one ofthe following is true:a < b, a = b, a > b.

EXAMPLE 64 Order the sides of triangle in the figure

according to their length.

Solution m(A) + m(B) + m(C) = 180°

2x + 40° + 20° + 3x – 10° = 180°

5x = 130°

x = 26°

So m(A) = (2 26°) + 40° = 92° and m(C) = (3 26°) – 10° = 68°.

Since m(A) > m(C) > m(B), by the previous theorem we have a > c > b.

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EXAMPLE 65 In the figure at the right, KN = KM.

Prove that KT > KM.�

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132 Geometriy 7

Solution Given: KN = KM

Prove: KT > KM

Proof:

Statements Reasons

1. KN = KM 1. Given

2. 2 3 2. Base angles of isosceles triangle KNM

3. m(2) = m(1) + m(4) 3. Triangle Exterior Angle Theorem

4. m(2) > m(1) 4. By 3

5. m(3) > m(1) 5. Substitute 2 into 4.

6. KT > KM 6. By the previous theorem

EXAMPLE 66 Prove that in any triangle ABC, a + b + c > ha + hb + hc, where ha, hb and hc are the altitudes

to the sides a, b and c, respectively.

Solution Given: ABC with altitudes ha, hb and hc

Prove: (a + b + c) > (ha + hb + hc)

Proof:

Look at the figure. By the previous theorem,

in right triangle BCD, BC > BD, i.e. a > hb; (1)

in right triangle AEC, AC > CE, i.e. b > hc; (2)

in right triangle ABH, AB > AH, i.e. c > ha. (3)

Adding inequalities (1), (2) and (3) gives (a + b + c) > (ha + hb + hc).

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Solution In ABC, m(B) > m(A) > m(C) so AC > BC > AB. (1) (By the previous theorem)

In ADC, m(C) > m(A) > m(D) so AD > CD > AC. (2) (By the previous theorem)

Combining (1) and (2) gives us AD > DC > AC > BC > AB. So AD is the longest segment inthe figure.


Check Yourself 161. Write the measures of the angles in each triangle in increasing order.

2. Write the lengths of the sides of each triangle in increasing order.

3. Find the longest line segment in each figure using the given angle measures.

Answers

1. aa. m(B) < m(A) < m(C) b. m(M) < m(P) < m(N) c. m(N) < m(K) < m(M)

2. aa. c < b < a b. n = m < k c. k < s = t 3. aa. CD b. PK c. BC

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Activity For this activity you will need a piece of string and a ruler.

Cut the string into eight pieces of different lengths. Measure the lengths and label ormark each piece with its length.

Take any three pieces of string and try to form a triangle with them.

Make a table to note the lengths of the pieces of string and whether or not they formeda triangle.

Repeat the activity until you have two successes and two failures at making a triangle.

Look at your table. Which lengths of string together made a triangle? Which lengthsdidn’t make a triangle? What conjecture can you make about the sides of a triangle?

TTrriiaannggllee IInneeqquuaalliittyy

134 Geometriy 7

Properties 4 Triangle IInequality TTheorem

In any triangle ABC with sides a, b and c, the following inequalities are true:

|b – c| < a < (b + c),

|a – c| < b < (a + c),

|a – b| < c < (a + b).

The converse is also true. This property is also called the Triangle IInequality TTheorem.

Solution In ABC, |10 – 5| < x < (10 + 5) (Triangle Inequality Theorem)

5 < x < 15. (1)

In DBC, |7 – 4| < x < (7 + 4) (Triangle Inequality Theorem)

3 < x < 11. (2)

The possible values of x are the elements of the common solution of inequalities (1) and (2),

i.e. 5 < x < 11.

So x {6, 7, 8, 9, 10}.

EXAMPLE 69 Find all the possible integer values of x in the

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EXAMPLE 68 Is it possible for a triangle to have sides with the lengths indicated?

a. 7, 8, 9 b. 0.8, 0.3, 1 c. 1 1, , 1

2 3

Solution We can check each case by using the Triangle Inequality Theorem.

a. |9 – 8| < 7 < (8 + 9)

|8 – 9| < 8 < (7 + 9)

|7 – 8| < 9 < (7 + 8).

This is true, so by the

Triangle Inequality

Theorem this is a

possible triangle.

b. |0.8 – 0.3| < 1 < (0.8 + 0.3)

|1 – 0.3| < 0.8 < (1 + 0.3)

|1 – 0.8| < 0.3 < (1 + 0.8).

This is true, so by the

Triangle Inequality Theorem

this is a possible triangle.

c. This is impossible,

since

1<1 1

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EXAMPLE 70 Find the greatest possible integer value of m

in the figure, then find the smallest possible

integer value of n for this case.

Solution In ABD, |9 – 6| < m < (9 + 6) (Triangle Inequality Theorem)

3 < m < 15.

So the greatest possible integer value of m is 14.

In ADC, |8 – m| < n < (m + 8) (Triangle Inequality Theorem)

|8 – 14| < n < (14 + 8) (m = 14)

6 < n < 22.

So when m = 14, the smallest possible integer value of n is 7.

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Properties 51. In any triangle ABC,

a. if m(A) = 90° then

b. if m(A) < 90° then |b – c| < a <

c. if m(A) > 90° then < a < (b + c).

2. In any triangle ABC, if P int ABC

then (BP + PC) < (BA + AC).

3. In any triangle ABC, if m(B) > m(C) or m(B) < m(C) then ha < nA < Va.

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136 Geometriy 7

EXAMPLE 71 In a triangle ABC, m(A) > 90°, c = 6 and b = 8. Find all the possible integer lengths of a.

Solution Since m(A) > 90°, < a < (b + c) by Property 5.1.

Substituting the values in the question gives < a < (8 + 6), i.e.

10 < a < 14. So a {11, 12, 13}.

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2 2+b c

EXAMPLE 72 In the triangle ABC shown opposite,

P int ABC, AB = 10, AC = 8 and BC = 9.

Find the sum of all the possible integer val-

ues of PB + PC.

Solution In PBC, BC < (BP + PC) by the Triangle

Inequality Theorem.

So 9 < BP + PC. (1)

In ABC, (PB + PC) < (AB + AC) by Property 5.2.

So PB + PC < 10 + 8. (2)

Combining (1) and (2) gives 9 < (PB + PC) < 18.

So the possible integer values for PB + PC are 10, 11, 12, 13, 14, 15, 16 and 17.

The required sum is therefore 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 = 108.

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EXAMPLE 73 Prove that the sum of the lengths of the medians of a triangle is greater than half of the

perimeter and less than the perimeter.


Given: ABC with centroid G

Prove:

Proof:

We need to prove two inequalities.

+ +<( + + )<( + + ).

2 a b c

a b cV V V a b c

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The centroid of atriangle is the point ofintersection of itsmedians.


Proof tthat We will use the Triangle Inequality Theorem three times.

In CEB, (CE + EB) > BC, i.e. (Vc + ) > a. (Triangle Inequality Theorem)

In ADC, (AD + DC) > AC, i.e. (Va + ) > b. (Triangle Inequality Theorem)

In ABF, (BF + FA) > AB, i.e. (Vb + ) > c. (Triangle Inequality Theorem)

So (Vc + Va+ Vb + + + ) > (a + b + c). (Addition Property of Inequality)

So (1) (Subtraction Property of Inequality)

Proof tthat Vaa + Vbb + Vcc < a + b + c:

For the second part of the inequality, let us

draw another figure as shown at the right

and extend the median AD through D to a

point K such that

AD = DK. (2)

Then join K and B. Now,

BD = DC (AD is a median)

m(BDK) = m(ADC) (Vertical angles)

AD = DK. (By (2))

So by the SAS Congruence Postulate,

DBK DCA and so

|BK| = |CA| = b.

Then, in ABK,

2Va < b + c. (3) (Triangle Inequality

Theorem)

By considering the other medians in a similar way

we get 2Vb < (a + c) and 2Vc < (a + b). (4)

Adding the inequalities from (3) and (4) side by side gives us

2(Va + Vb + Vc) < 2(a + b + c). So (Va + Vb + Vc) < (a + b + c). (5)

Finally, by (1) and (5), as required.+ +

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a b cV V V a b c

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138 Geometriy 7

EXAMPLE 75 Prove that for any triangle ABC,

if P int ABC and x, y and z are as shown

in the figure then

(x + y + z) < (a + b + c) < 2(x + y + z), i.e.

+ +<( + + )<( + + ).

2a b c

x y z a b c

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Solution Let M and N be as in figure 1, and let be a line representing the river. Then we can use thefollowing method to locate A:

1. Draw a ray from N perpendicular to , intersecting at point B.

2. Locate point C on the extension of NB such that NB = BC.

3. Draw KC.

4. Locate A at the intersection of KC and , as shown in figure 2.

Now we need to show that A is really the location which makes AN + AK as small aspossible. Figure 3 shows an alternative location X on l. Notice that in KXC, (KX + XC) > KCby the Triangle Inequality Theorem. So (KX + XC) > (KA + AC) (1) by the Segment AdditionPostulate. Since AB NC and NB = BC, NXC is isosceles with XC = NX (2). By the samereasoning, NAC is isosceles with NA = AC (3). Substituting (2) and (3) into (1) gives us(KX + XN) > (KA + AN). So A is the best location for the station.

The result we have just proved does not mean that for a given triangle, the sum of themedians can be anything between the half perimeter and full perimeter of the triangle. Thisis because the lengths of the medians are directly related to the lengths of the sides. As wewill see in the next chapter, once we know the lengths of the three sides of a triangle thenwe can calculate the lengths of its medians. Their sum is a fixed number.

Remark

EXAMPLE 74 Two towns K and N are on the same side of the river Nile. The residents of the two townswant to construct a water pumping station at a point A on the river. To minimize the cost ofconstructing pipelines from A to K and N, they wish to locate A along the Nile so that thedistance AN + AK is as small as possible. Find the corresponding location for A and show thatany other location requires a path which is longer than the path through A.

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Check Yourself 171. Two sides of a triangle measure 24 cm and 11 cm respectively. Find the perimeter of the

triangle if its third side is equal to one of other two sides.

2. Determine whether each ratio could be the ratio of the lengths of the sides of a triangle.

a. 3 : 4 : 5 b. 4 : 3 : 1 c. 10 : 11 : 15 d. 0.2 : 0.3 : 0.6

3. The lengths of the sides DE and EF of a triangle DEF are 4.5 and 7.8. What is the greatest

possible integer length of DF?

4. The base of an isosceles triangle measures 10 cm and the perimeter of the triangle is an

integer length. What is the smallest possible length of the leg of this triangle?

5. In an isosceles triangle KLM, KL = LM = 7 and m(K) < 60°. If the perimeter of the

triangle is an integer, how many possible triangle(s) KLM exist?

6. In a triangle ABC, AB = 9 and BC = 12. If m(B) < 90°, find all the possible integer

lengths of AC.

Answers1. 59 cm 2. aa. yes b. no c. yes d. no 3. 12 4. 5.5 cm 5. six triangles

6. AC {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

The example that we have just seen shows an application of triangle inequality. But the resultwe obtained does not mean that the value of x + y + z can be any number less thana + b + c. In other words, the maximum value of x + y + z may be a lot less than a + b + c.In fact, the maximum value of x + y + z is always less than the sum of the lengths of thetwo longer sides of the triangle, because as the interior point moves towards one of thevertices, two distances increase but the third distance decreases. When this interior pointreaches the vertex point, the distance to this point becomes zero and the sum of the distancesbecomes the sum of the two sides which include this vertex. So the maximum value ofx + y + z will always be less than the sum of the length of the two longer sides.

Remark

Solution In ABP, c < (x + z). (Triangle Inequality Theorem)

In APC, b < (y + z). (Triangle Inequality Theorem)

In BPC, a < (x + z). (Triangle Inequality Theorem)

So (a + b + c) < 2(x + y + z). (1) (Addition property of inequality)

Also, (x + y) < (c + b), (Property 5.2)

(y + z) < (a + c) and (Property 5.2)

(x + z) < (b + a). (Property 5.2)

So (x + y + z) < (a + b + c) (2) (Addition property of inequality)

As a result, (x + y + z) < (a + b + c) < 2(x + y + z), (By (1) and (2))

or equivalently, + +<( + + )<( + + ).

2a b c

x y z a b c

140 Geometriy 7

88.. In the triangle ABC at

the right,

AB = AD = BE,

m(A) = 114° and

m(B) = 60°. Find

m(EDC).

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11.. Each figure below shows triangles with two or

more congruent sides. Find the value of x in each

figure, using the information given.

22.. An angle in a triangle measures 20° less than the

measure of the biggest angle in the triangle. The

measure of the third angle is half the measure of

the biggest angle. Find the measures of all three

angles.

44.. In a triangle ABC, the angle bisector of the interior

angle C makes an angle of 40° with the side AB.

Find the angle between the bisector of the exterior

angle C and the extension of the side AB.

66.. In the triangle MNP

opposite,

MS = MP,

ST = TP,

m(M) = 94° and

m(N) = 26°. Find m(MST).

55.. In a triangle KMN, the altitudes to sides KM and

MN intersect each other at a point P. Find

m(KPN) if m(KNM) = 72° and m(NKM) = 64°.

33.. The two acute angles in a right triangle measure

and respectively. Find x and

the measures of these angles.

( – 25)4x2

( +5)3x

EXERCISES 3.3

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B.

If m(ACD) = 18°,

find m(ABC).

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1177.. Each figure shows the lengths of two sides of a

triangle. Write an interval for the possible length

of the third side.

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1111.. The bisectors of the interior angles D and F in

a triangle DEF intersect at the point T. Find the

measure of DEF if its measure is one-third of

m(DTF).

1122.. a, b and b are the measures of the interior angles

of an isosceles triangle such that a and b are

integers and 24° < b < 38°. Find the smallest

possible value of a.

1133.. In a triangle DEF, point M is on the side DF and

MDE and DEM are acute angles.

Draw an appropriate figure for each of the

following, if it is possible.

a. FME is obtuse

b. FME is equilateral

c. DME is equilateral and DEF is isosceles

d. DME is isosceles and EMF is isosceles

e. DME is isosceles and DEF is equilateral

1144.. x, y and z are the exterior angles of a triangle.

Determine whether each ratio is a possible ratio

of x : y : z.

a. 2 : 3 : 5

b. 1 : 2 : 3

c. 6 : 11 : 19

d. 12 : 15 : 21

99.. In the figure,

PQ = PS = PR and

m(SPR) = 24°. Find

m(SQR).

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m(TMP) = 44°,

m(P) = 38° and

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1155.. Find the value of x in

the figure.

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1166.. In an isosceles triangle KMN, the bisectors of the

base angles K and M intersect each other at a

point T. Prove that m(KTM) = m(K).

1188.. For each figure, state the interval of possible

values for the length x.

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142 Geometriy 7

2288.. For each figure, order the numbered angles

according to their size.

2277.. A triangle has side lengths 2x + y, 2y + 3x and 2x.

Which one is bigger: x or y?

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a. b.

c. d.

1199.. A triangle ABC has sides a, b and c with integer

lengths. How many triangles can be formed such

that b = c and a b = 18?

2200.. In the figure, a, b

and c are integers.

Calculate the smallest

possible value of

a + b + c, using the

information given.

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m(A) > 90° and

m(C) > 90°.

If AB = 6 cm,

AD = 10 cm,

BC = 12 cm and

CD = 5 cm, find

the sum of the all the possible integer lengths of

the side BD.

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2233.. In the triangle ABC

at the right,

AD = 9 cm,

BD = 6 cm,

DC = 8 cm,

AC = x cm and

AB = y cm.

Find the sum of the smallest and largest possible

integer values of x + y.

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AB = 8 cm,

AC = 10 cm,

BD = 3 cm,

CD = 7 cm and

BC = 2x + 1 cm.

Find the sum of

all the possible integer values of x.

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AB = 8 cm,

BC = 12 cm,

CD = 6 cm and

DA = 4 cm.

Find the number

of possible integer lengths of AC.

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AC = 9 cm,

BC = a,

AB = c and

m(BAC) > 90°.

Find the smallest

possible value of a + c if a, c .

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a triangle to have sides with the lengths given.

a. 13, 9, 5 b. 5, 5, 14

c. 8, 8, 16.1 d. 17, 11, 6

e. 0.5, 0.6, 1 f. 18, 18, 0.09


3311.. A student has five sticks, each with an integer

length. He finds that he cannot form a triangle

using any three of these sticks. What is the shortest

possible length of the longest stick, if

a. the lengths of the sticks can be the same?

b. all the sticks have different lengths?

(Hint: Use the Triangle Inequality Theorem.)

3322.. How many distinct isosceles triangles have

integer side lengths and perimeter 200 cm?

3300.. State the longest line segment in each figure.

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a. b.

c. d.

2299.. Determine whether each statement is true or

false.

a. In a triangle ABC, if the measure of A is 57°

and the measure of B is 64° then the shortest

side of ABC is AB.

b. In a triangle KMN, if the measure of K is 43

and the measure of M is 47 then the shortest

side of KMN is KM.

c. In a triangle ABC, if B is an obtuse angle and

AH BC then HA < AB.

d. If an isosceles triangle KTA with base KA has

TA < KA then the measure of T is always less

than 90°.

e. An angle bisector in an equilateral triangle is

shorter than any of the sides.

f. All obtuse triangles are isosceles.

g. Some right triangles are equilateral.3333.. How many triangles can be drawn if the length of

the longest side must be 11 units and all side

lengths must be integer values?

3355.. In a triangle ABC, AB = 8 cm, BC = x and AC = y.

If m(A) > 90° and x, y , find the

smallest possible value of x + y.


AD = 5 cm,

AB = 12 cm,

BC = 9 cm and

DC = 8 cm.

If m(A) > 90° and

m(C) < 90°, find

all possible integer

values of BD.

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144 Geometriy 7

4400.. In a triangle DTF, m(D) = 90° and m(F) < 45°.

a. 2 m(T), m(D)

b. FT, 2DF

c. FD, DT

d. m(F), 2m(T)

4411.. In a triangle DEF, m(D) > m(E) = m(F).

a. m(D), 60°

b. m(E), 60°

4422.. In a triangle DEF, m(E) = 120° and EF > DE.

a. 120°, 3 m(D)

b. 2 m(E), 3 m(D)

3399..

a. (x – 10)°, (y + 20)°

b. MB + MC, AB + AC

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3366.. How many distinct triangles have integer side

lengths and perimeter 11?

3377.. Prove each theorem.

a. Hinge Theorem:

If two sides of one triangle are congruent to

two sides of another triangle, and if the included

angle of the first triangle is larger than the

included angle of the second, then the third

side of the first triangle is longer than the third

side of the second.

b. Converse of the Hinge Theorem:

If two sides a and b of one triangle are

congruent to two sides d and e of another

triangle, and if the third side of the first

triangle is longer than the third side of the

second, then the angle between a and b is

larger than the angle between d and e.

3388..

aa.. 130°, x b. y, 90°

c. y, x d. KM, MN

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Instructions ffor qquestions 338 tto 442

Each question gives two quantities to be compared,

separated by a comma. In each case, use the figure or

extra information to compare the quantities. Write

A if the first quantity is greater than the second,

B if the first quantity is smaller than the second,

C if the quantities are equal, or

D if the extra information is not enough for you to

be able to compare the quantities.

All variables represent real numbers. Figures are

generally not drawn to scale.


A. DISTANCE FROM A POINT TO A LINE

Let A(x1, y1) be a point and d: ax + by + c = 0 be a line, then the distance from A to the

line d is

.1 1

2 2

| + + |=

+

ax by cl

a b

Theorem

Proof Let the distance of A(x1, y1) to the line

d: ax + by + c = 0 be l = AH.

Take C(x2, y2) = AD d. x2 = x1 and y2 = CD

C is a point on the line ax + by + c = 0, so

ax1 + b CD + c = 0

b CD = –a x1 – c

So we have the coordinates of C,

Now, is the inclination of d and = m(CBD) = m(CAH) (angles with perpendicular sides).

In the right triangle ACH, and AH = AC cos ...(1)

Now, let’s find the equivalent expressions for AC and cos.

AC = AD – CD =

We know sec2 = 1 + tan2

so , and

so 2 2

1 1cos = = ...(3)

1+( ) 1 ( )a ab b

tan =a

mb

2

1cos =

1+tan

1 1 1 1( ), so = ...(2)a c a c

y x AC y xb b b b

cos =AHAC

1 1( , – – ).a c

C x xb b

1 .a c

CD xb b

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distance ffrom aa ppoint tto aa lline


1. Find the distance from a point to a line.

2. Find the distance between two parallel lines.

Objectives

146 Geometriy 7

Substituting (2) and (3) in (1),

l = AH = AC cos, and since l is the distance,

1 11 1

1 1 2 2 22 2

2

+ + + +1= ( ) = = .

1 ++1+

a cy x ax by ca c b bl y x

b b a a ba bbb

O(0, 0) = O(x1, y1). Using the formula,

1 1

2 2 2 2

| + + | |0 0+4| 4 4 2= = = = = 2 2.

22+ 1 +( 1)

ax by cl

a b

Solution

EXAMPLE 76 Find the distance from the point O(0, 0) to the line x – y + 4 = 0.

A A(x1, y1). Using the formula,

1 1

2 2 2 2

| + + | |3 5 4 2+5| 12= = = = 2.4.

5+ 3 +4

ax by cl

a b

Solution

EXAMPLE 77 Find the distance from A(5, 2) to the line 3x – 4y + 5 = 0.

1 1

2 2

| + + | |5 12 – 12 5+5 |= = =10

25+144+

|5 |= =10, so |5 |=130, i.e. 5 = 130, and so = 26.

13

ax by c kl

a b

kl k k k

Solution

EXAMPLE 78 The distance from A(12, 5) to the line 5x – 12y + 5k = 0 is ten units. Find the possible values

of k.

Check Yourself 18

1. Find the distance from the point P(–2, 3) to the line 3x + 4y + 9 = 0.

2. Find the distance from the point A(1, 4) to the line y = 3x – 4.

3. The distance between the point P(k, 3) and the line 4x – 3y + 5 is 4 units. Find k.

Answers

1. 3 2. 3. k {–4, 6}102


Let d1: a1x + b1y + c1 = 0

d2: a2x + b2y + c2 = 0 be two parallel lines.

Since d1 d2, we can write , so a1 = k a2 and b1 = k b2.

Now, let’s substitute these values into d1:

k a2x + k b2y + c1 = 0

k(a2x + b2y + ) = 0.

k 0, so we get d1: a2x + b2y + = 0.

When we compare d1 with d2, we see that their difference is a constant number.

In general, we can write two parallel lines d1 and d2 as:

d1: ax + by + c1 = 0

d2: ax + by + c2 = 0.

1ck

1ck

1 1

2 2

a bk

a b

Let d1: ax+ by + c1 = 0 and d2: ax + by + c2 = 0 be two parallel lines. Then the distance

between d1 and d2 is .2 1

2 2

| |=

+

c cl

a b

Theorem

B. DISTANCE BETWEEN TWO PARALLEL LINES

distance bbetween ttwo pparallel llines

Solution

EXAMPLE 79 Find the distance between the parallel lines x – 2y + 5 = 0 and 3x – 6y + 9 = 0.

Proof The distance of any point A(x, y) on line d1 to the line d2 is

In the equation of d1: ax + by + c = 0

ax + by = – c1, and so 2 1

2 2

| |= .

+

c cl

a b

2

2 2

| + + |.

+

ax by c

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It is important to notice that to find the distance between two parallel lines, first of all we

need to equalize the coefficients of x and y.

Remark

1

1

2

1 1 2

2 22

: 2 +5 = 0 3 6 +15 = 0was multiplied by 3.

: 3 6 +9 = 0 3 6 +9 = 0

Now, we have =15 | | 15 – 9 6 6 2 5= = = = = .

59+36 45 3 5= 9

d x y x yd

d x y x y

c c cl

a bc

148 Geometriy 7

There is also another way to solve the problem:

The distance between d1 and d2 is the same as the distance of any point on d1 or d2 to the

other line.

For example, A(0, –4) is one point on d2, and the distance of A to d1 is

The solution is the same.1 1

2 2

|3 2 +5| |3 0 2(– 4)+5| 13= = = 13.

13 133 +(–2)

x yl

1 1

2 2

1 1 2

2 2 2 22

: 3 2 +5 = 0 : 3 2 +5 = 0

: 3 +2 +8 = 0 : 3 2 8 = 0

So = 5 | | |5+8| 13= = = = 13.

133 +(–2)= –8

d x y d x y

d x y d x y

c c cl

a bc

Solution

EXAMPLE 80 Find the distance between the parallel lines 3x – 2y + 5 = 0 and –3x + 2y + 8 = 0.

Check Yourself 19

1. Find the distance between the lines 4x – 3y – 5 = 0 and –12x + 9y + 4 = 0.

2. The lines x + 2y + 1 = 0 and 3x + 6y + k = 0 are parallel and the distance between

them is ñ5. Find k.

3. Find the area of the square whose two sides are on the parallel lines 2x + y – 2 = 0 and

4x + 2y + 6 = 0.

Answers

1. 2. k {–12, 18} 3. 51115


A. Distance from a Point to a Line11.. Find the distance from the point A(–2, 3) to the

line 8x + 6y – 15 = 0.

44.. The distance between P(1, –2) and the line

7x – y + k = 0 is 4ñ2 units. Find k.

55.. The points A(1, 3), B(–2, 1) and C(3, –1) are the

vertices of the triangle ABC. Find the length of

the altitude of BC.

6.. The distance from P( , k) to the line

12x + 9y – 1 = 0 is 2 units. Find k.

12

8.. The distance between the parallel lines

12x + 9y – 2 = 0 and ax + 3y + c = 0 is three

units. Find the ratio , if 0.a

cc

B. Distance Between Two ParallelLines

77.. Find the distance between each pair of parallel

lines.

a. –2x + 3y – 4 = 0 and –2x + 3y – 17 = 0

b. x – y – 4 = 0 and –2x + 2y – 7 = 0

c. y = 2x + 1 and 2y = 4x – 3

33.. The distance from a line with equation

y – 4 = m(x + 2) to the origin is 2. Find m.

22.. The distance between B(2, 3) and the line

12y – 5x = k is . Find k.5

13

Write the equations of the lines which are four

units away from the line 3x + 4y + 10 = 0.9..

10.. The distance between the parallel lines

3x + 4y – 6 = 0 and 4x – ky + 4 = 0 is p. Find

k + p.

EXERCISES 3.4

Angles Chapter 3 Review Test A

11.. In the triangle ABC in

the figure,

m(A) = 4x,

m(B) = x and

m(C) = 30°. Find the value of x.

A) 10° B) 15° C) 20° D) 25° E) 30°

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66.. Which is the longest

side in the figure,

according to the given

angle measures?

A) BC B) AB C) BD D) CD E) BE

77.. In a triangle DEF, DE = EF and DF > EF.

Which statement is true?

A) DE < (DF – EF) B) m(E) > m(F)

C) m(E) < m(D) D) m(E) = 60°

E) m(E) = m(D)

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88.. What is the sum of

the smallest and

largest possible integer

values of x in the

figure?

A) 26 B) 24 C) 22 D) 20 E) 17

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44.. In the figure,

m(P) = 45°,

m(N) = 36° and

m(R) = 25°. Find

the value of x.

A) 260° B) 256° C) 254° D) 248° E) 244°

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22.. In a triangle MNP, the interior angle bisectors of

M and P intersect at the point S. Given that

N measures 40°, find m(PSM).

A) 95° B) 100° C) 105° D) 110° E) 120°

55.. Two sides of a triangle have lengths 8 and 12.

What is the sum of the minimum and maximum

possible integer values of the length of the third

side?

A) 24 B) 22 C) 19 D) 18 E) 16

33.. In the triangle STK

opposite, N TK and

SN is the interior angle

bisector of S.

If m(T) – m(K) = 40°,

find m(SNK).

A) 110° B) 105° C) 100° D) 95° E) 90°

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CHAPTER 3 REVIEW TEST A

Chapter Review Test 1A 151

99.. In a triangle ABC, D BC and AD bisects A. If

AB = 6 cm, BD = 3 cm and DC = 2 cm, find the

length of AD.

A) 5ñ2 cm B) 4ñ3 cm C) 3ñ2 cm

D) 2ñ3 cm E) ñ3 cm


BD = DC,

AD = AE and

m(C) = 20°. Find

m(EDC).

A) 70° B) 65° C) 60° D) 45° E) 30°

1111.. If MNP STK, which of the following

statements is false?

A) MN ST B) MP TK C) NP TK

D) PNM STK E) KT PN

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BD bisects B,

BD = BE and

DE = EC.

If m(A) = 80° and,

m(ACD) = 20°, what

is m(BDC)?

A) 100° B) 110° C) 120° D) 140° E) 150°

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m(BAC) = 90°,

m(C) = 60° and

BD = DC.

Find BC if

AD = 2x + 3 and

AC = 6x – 1.

A) 6 B) 8 C) 10 D) 12 E) 14

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m(BAC) = 90°,

m(BAD) = 12°,

BC = 8 cm and

AD = 4 cm. What is m(ABC)?

A) 52° B) 54° C) 58° D) 60° E) 64°

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1133.. a, b and c are the lengths of the sides of a triangle

ABC. Given that a, b and c are integers and

a2 – b2 =17, what is the sum of the minimum and

maximum possible values of c?

A) 7 B) 13 C) 17 D) 18 E) 23


ND = DP and

.

What is ?

A) B) C) D) E)3

63

53

43

33

2

MHNP

3=

3HDMH

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Angles Chapter 3 Review Test B

CHAPTER 3 REVIEW TEST B

11.. In the figure,

AB = AD,

AC = BC and

m(DAC) = 15°. Find

m(C).

A) 40° B) 45° C) 50° D) 60° E) 65°

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in the figure,

MS = NS and

KN = KP.

If m(MRP) = 117°,

what is m(MNP)?

A) 39° B) 41° C) 43° D) 45° E) 47°

33.. In a triangle ABC, D is a point on the side AB and CD

is the interior angle bisector of C. If AB = 15 cm

and 3 AC = 2 BC, find the length of DB.

A) 2 cm B) 3 cm C) 4 cm D) 6 cm E) 9 cm

55.. In a triangle ABC, points D and E are the

midpoints of the sides AB and AC respectively.

DE = (x + 5)/4 and BC = 8x – 5 are given. What is

the value of x?

A) 1 B) 2 C) 3 D) 4 E) 5

66.. ABC is a right triangle with m(A) = 90°, and AH

is the altitude to the hypotenuse. If m(C) = 30°

and BH = 2 cm, find HC.

A) 4 cm B) 5 cm C) 6 cm

D) 7 cm E) 8 cm

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88.. In the figure,

MS = SN and

MP = PN.

If m(P) = 20°,

m(KMP) = 40° and

m(KNP) = 30°,

what is m(SKN)?

A) 50° B) 45° C) 40° D) 35° E) 30°

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44.. In the figure,

MN = MP and

ML = MK.

If m(PLK) = 12°,

what is m(LMN)?

A) 18° B) 20° C) 24° D) 30° E) 36°

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77.. In the figure,

AB = AD.

What is ?

A) B) 1 C) D) E) 43

32

23

12

DEEC

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99.. In the figure,

HK = KN,

m(DAC) = 40° and

m(HKB) = 20°.

Find m(BKD).

A) 20° B) 30° C) 40° D) 60° E) 70°

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shown opposite, point

O is the center of the

inscribed circle of

MNP.

If KS NP,

KN = 6 and

SP = 8, what is the length of KS?

A) 10 B) 12 C) 14 D) 16 E) 18

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ACDE is a square,

m(ABC) = 60° and

BD = 2 cm. Find the

length of one side of

the square.

A) (3 – ñ3) cm B) (ñ3 – 1) cm C) (ñ3 + 1) cm

D) (4 – 2ñ3) cm E) (2ñ5 – 3) cm

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MK = KL,

MN = m,

KN = 2m and

NL = 3m. Find

m(KNL).

A) 45° B) 50° C) 60° D) 70° E) 75°

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1100.. In the figure, BD

bisects angle B. Given

m(ADB) = 90°,

DE BC,

AB = 12 and

BC = 16, find the length of DE.

A) 1 B) C) 2 D) E) 352

32

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1122.. The lengths of the sides of a triangle ABC are

integers a, b and c such that b = c and

(a + b + c) (a + b – c ) = 15.

Find the value of a.

A) 1 B) 2 C) 3 D) 5 E) 7


AE = BD = DC and

AB = AC.

What is m(FDC)?

A) 45° B) 50° C) 60° D) 62.5° E) 67.5°

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CH = HB,

AD = 3 and

DB = 8. What is the

sum of the all

possible integer values

of the length AC?

A) 30 B) 34 C) 40 D) 42 E) 51

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Angles Chapter 3 Review Test C

CHAPTER 3 REVIEW TEST C

11.. In the figure,

m(KBC) = m(KCA).

and m(LKB) = 80°.

What is the measure of

ACB?

A) 40° B) 60° C) 70° D) 75° E) 80°

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55.. In the figure, ABC,

CDE, and FEG are

equilateral triangles.

If BG = 16, what is

the sum of the

perimeters of the

three triangles?

A) 32 B) 36 C) 42 D) 46 E) 48

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the figure, CD is the

bisector of C, AE is

the median to BC and

DE AC.

If m(B) = 50°, what is m(BAC)?

A) 30° B) 35° C) 40° D) 45° E) 50°

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77.. Which of the

following is a possible

sum of the lengths of

AB and BC in the

figure?

A) 11 B) 12 C) 13 D) 14 E) 37

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22.. Which is the largest

angle in the figure,

according to the given

lengths?

A) M B) N C) S D) SPK E) K

33.. In an isosceles triangle XYZ, m(Y) = m(Z) and

m(X) < m(Y). What is the largest possible

integer measure of the angle Y?

A) 59° B) 60° C) 89° D) 90° E) 110°

44.. In a triangle ABC, points B, C and D are collinear

and AD is the angle bisector of the exterior angle

A. If AC = BC, DB = 12 and AB = 4, find the

length of BC.

A) 2 B) 3 C) 4 D) 5 E) 8

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88.. In the figure,

MP = PS = SN = PT

and ST = TN.

What is m(NMP)?

A) 36° B) 60° C) 72° D) 84° E) 108°

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99.. In the figure, ABC is


If BD = AE, what is

the measure of EFC?

A) 45° B) 60° C) 75° D) 90° E) 120°

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O is the incenter of

ABC, AB OT and

AC OV.

If BT = 6 cm,

TV = 7 cm and

VC = 5 cm, what is

the perimeter of the triangle OTV?

A) 12 cm B) 15 cm C) 16 cm

D) 18 cm E) 20 cm

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AF = FB and

AE = EC.

If EH + FH = 12,

what is AB + AC?

A) 16 B) 18 C) 22 D) 24 E) 36

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the figure, BH is the

exterior angle bisector

of B and

AE = EC.

If m(BHC) = 90°,

BC = 8 and

EH = 7, what is the length of AB?

A) 6 B) 7 C) 8 D) 10 E) 12

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1166..

In the figure, m(M) = 90°, m(MST) = 150°

and PM = MS = ST. What is m(N)?

A) 5° B) 10° C) 15° D) 22.5° E) 30°

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AC = BC and

AB = AD.

If m(CAD) = 18°

and m(EBD) = 12°,

what is m(AEB)?

A) 82° B) 80° C) 78° D) 72° E) 42°

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If PB = 16 and

PN = 10, what is the

length of AH?

A) 2ñ3 B) 3ñ3 C) 4ñ3 D) 5ñ3 E) 6ñ3

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MK = NK = PK.

What is x + y + z?

A) 270° B) 180° C) 90° D) 60° E) 45°

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Angles Chapter 3 Review Test D

CHAPTER 3 REVIEW TEST D


the figure, BN is the

bisector of ABC and

H is the intersection

point of the altitudes

of ABC.

If m(AHC) = 110° and m(HBN) = 20°,

what is m(BAC)?

A) 50° B) 55° C) 65° D) 75° E) 80°

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22.. In the figure,

MN = MP, KP = KT,

m(NMP) = m,

m(PKT) = k, and

points N, P and T are

collinear.

If m + k = 130°, what is m(MPK)?

A) 50° B) 55° C) 60° D) 65° E) 70°

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opposite, MK = TK,

NS = TS and

m(KTS) = 50°.

What is m(MPN)?

A) 70° B) 65° C) 60° D) 55° E) 50°

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figure, what is

the value of

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A) 5 B) C) D) 1 E) 12

12

21

22

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77.. The measure of one angle in a triangle is equal to

the sum of the measures of the other two angles.

Which statement about this triangle is always

true?

A) The triangle is equilateral.

B) The triangle is acute.

C) The triangle is a right triangle.

D) The triangle is obtuse.

E) The triangle is isosceles.

88.. In the triangle XYZ in

the figure,

m(YZX) = 90°,

XZ = PK and

XP = PY.

What is m(PKZ)?

A) 120° B) 135° C) 140° D) 150° E) 160°

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33.. In the figure,

AB = AC = b,

BC = a, and a < b.

What is the largest

possible integer value

of m(A)?

A) 59° B) 60° C) 44° D) 30° E) 29°

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44.. In the figure,

AD = BD,

m(DAC) = x and

m(BCE) = 2x.

If m(EAB) = 110°,

what is the value of x?

A) 30° B) 35° C) 40° D) 45° E) 50°

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99.. In the figure, PM is the

angle bisector of NPK,

MN = MP, NS = SP

and m(MKP) = 90°.

What is m(STP)?

A) 90° B) 85° C) 80° D) 75° E) 60°

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m(DAC) = m(B)

and

m(EAB) = m(C).

If m(AEC) = 130°, what is m(ADE)?

A) 50° B) 55° C) 60° D) 70° E) 80°

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KS = KN,

m(M) = 70°,

m(P) = x and

m(MKS) = 2x.

What is the value of x?

A) 55° B) 60° C) 65° D) 70° E) 75°

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AD and CB bisect

angles A and C,

respectively.

If m(AEC) = 75° and

m(B) = 30°,

what is m(ADC)?

A) 5° B) 10° C) 15° D) 20° E) 25°

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triangle and BD = CE.

If AD = 6ñ3, what is

the length of DE?

A) 6 B) 8 C) 4ñ3 D) 13 E) 6ñ3

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1122.. In the equilateral

triangle ABC in the

figure, AF = FC and

AH = BD. What is the

measure of EDC?

A) 5° B) 10° C) 15° D) 20° E) 30°

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AB = BC,

DE = BE

m(ABD) = 36° and

m(EDC) = 48°.

What is m(ACB)?

A) 76° B) 72° C) 68° D) 58° E) 52°

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KL = LM and

LH = MH.

If NH = 5 and

m(K) = 30°,

what is KM?

A) 15 B) 20 C) 25 D) 30 E) 40

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Angles Chapter 3 Review Test E

CHAPTER 3 REVIEW TEST E

11.. In the figure,

DE = DC and

DB = BF.

If m(A) = 45°,

what is m(ABC)?

A) 30° B) 45° C) 50° D) 60° E) 75°

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55.. In the figure, points K,

S, T, M, N and P are

the midpoints of the

sides on which they lie.

If AB = 12,

AC = 8 and

BC = 16, what is

P(MNP)?

A) 6 B) 8 C) 9 D) 10 E) 12

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66.. In the figure, KL RS

and KM bisects RKL.

If KL = 6,

KR = 4 and

MS = 8, what is the

length of PK?

A) 4 B) 5 C) 6 D) 7 E) 8

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77.. In the figure,

m(BAC) = 90°,

m(C) = 15° and

BC = 24. What is

the length of AH?

A) 4 B) 5 C) 6 D) 8 E) 12

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figure, what is the

smallest possible value

of a + b + c if a, b

and c are integers?

A) 7 B) 8 C) 9 D) 10 E) 11

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the right,

AB = AC,

m(A) = x + 13° and

m(B) = y – 38°.

What is the sum of the

minimum integer value of y and the maximum

integer value of x?

A) 248° B) 243° C) 240° D) 233° E) 204°

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figure, what is the

value of x?

A) 10° B) 15° C) 20° D) 25° E) 30°

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44.. In the figure,


triangle and

AD = EC = CF.

If BC = 12,

what is the length of

CF?

A) 2 B) 3 C) 4 D) 5 E) 6

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99.. In the figure,

AH = BH = HC.

If AC = 1,

what is HD?

A) B) C) D) E) 35

15

14

13

12

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the figure,

CD AB and

BE AC.

If m(BFC) = 140°,

what is m(A)?

A) 20° B) 30° C) 40° D) 45° E) 50°

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BD = DC,

CE = 3AE and

2AB = AC.

If m(A) = 90°, what

is m(DEC)?

A) 30° B) 40° C) 45° D) 50° E) 60°

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m(A) = 90°,

m(B) = 15° and

AB = 6 + 3ñ3.

What is the length

of AC?

A) 1 B) 2 C) 3 D) 4 E) 5

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MK = KP,

m(M) = 90°,

NS = 9 cm and

SP = 3 cm. Find the

length of KS.

A) 2 cm B) 3 cm C) 4 cm D) 5 cm E) 6 cm

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1166..In the figure, CD is the

bisector of C.

If m(BAE) = 15°,

m(EAC) = 60° and

m(B) = 45°, what is

m(DEA)?

A) 10° B) 15° C) 20° D) 22.5° E) 30°

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an equilateral triangle

and DEFH is a square.

Find the measure of

AKD.

A) 65° B) 67.5° C) 70° D) 75° E) 80°

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ABCD is a square and

ABF and BEC are

equilateral triangles.

What is m(FEC)?

A) 5° B) 10° C) 15° D) 20° E) 22.5°

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Angles 160

162 Geometriy 7

1. DefinitionYou can see many circular or ring-shaped geometric figures all around you. For example,wheels, gears, compact discs, clocks, and windmills are all basic examples of circles in theworld around us.

All radii of a circle are congruent. A circle is named by its

center. For example, the circle on the left is named

circle O.

We write a circle with center O and with radius r as

or C(O, r).

In this book, the point O in a circle is always the center

of the circle.

wheels compact disc gears

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It is easy to recognize a circle, but how can we define it as a shape? Let us look at a geometric definition.

A. BASIC CONCEPTS

NoteThe word ‘circle’ is derived from the latin word circus, which means ‘ring’ or ‘racecourse’.

Definition circle

A circle is the set of all the points in a plane that are at the same distance from a fixed point

in the plane. The distance is called the radius of the circle (plural radii), and the fixed point

is called the center of the circle.


1. Define the concept of a circle and its basic elements.

2. Describe and use the properties of chords.

3. Describe and use the properties of tangents.

4. Describe the possible relative positions of two circles in the same plane.

Objectives

163Circles

2. Regions Separated by a Circle in a PlaneA circle divides a plane into three separate regions. The set of points whose dis-tance from the center of a circle is less than the radius of the circle is called theinterior of the circle.

For example, if R is a point in the plane and|OR| < r, then R is in the interior of thecircle.

The set of points whose distance from thecenter is greater than the radius of the circle is called the exterior of the circle.

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For example, if Q is a point in the plane and |OQ| > r, then the point Q is in the exterior ofthe circle. The set of points whose distance from the center is equal to the radius is calledthe ccircle iitself, and the points are on tthe ccircle. For example, if P is a point in the plane and|OP| = r, the point P is on the circle.

To construct a circle, fix a pin on a piece of paper, connect a string of any length

to the pin, tie the other end of the string to your pencil, and turn your pencil

on the paper around the pin for one

complete revolution, keeping the string

taut. You will get a circle.

You can also use a compass to draw a circle.Mark a point O as the center and set your compass to thelength of the radius. Turn your compass around the center for one complete revolution. You will get a circle.

NoteThe union of a circle and its interior is called a circular closed region or a disc.

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EXAMPLE 1 Name the points in the figure which are

a. in the interior of the circle.

b. on the circle.

c. in the exterior of the circle.

Solution a. Since |OA| < r and |OB| < r, points A and B are in

the interior of the circle.

b. Since |OC| = |OD| = r, points C and D are on the circle.

c. Since |OE| > r, |OF| > r and |OG| > r, points E, F, and G are in the exterior of the circle.

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164 Geometriy 7

In the figure, chord [CD] passes through the center of the circle, so [CD] is a diameter.

We can see that the length of every diameter in a given circle is the same. For this reason,

we usually talk about ‘the diameter’ of a circle to mean the length of any diameter in the

circle.

The length of the diameter of a circle is twice the radius. For example, if r is the radius of a

circle and d is the diameter, then d = 22 r, or .

The diameter of a circle is the longest chord in the circle. 2dr =

Definition diameter

A chord which passes through the center of a circle is called a diameter of the circle.

3. Auxiliary Elements of a Circle

For example, [AB] and [CD] in the figure are chords.

Definition chord

A line segment which joins two different points on a

circle is called a chord. �

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EXAMPLE 2 1. Find the length of the diameter for each

given radius.

a. b. 3x cm

c. 2x + 5 cm d. 7x – 12 cm

2. The length of the diameter of a circle is

20 cm and the radius is 2x – 4. Find x.13 cm2

Solution1. a. d = 2 r d = = 7 cm

b. d = 2 (3x) = 6x cm

c. d = 2 (2x + 5) = 4x + 10 cm

d. d = 2 (7x – 12) = 14x – 24 cm

2. d = 2 r

20 = 2 (2x – 4)

2x – 4 = 10

2x = 14

x = 7 cm

12 32

165Circles

4. Relative Position of a Line and a Circle in the SamePlane

A line and a circle in the same plane can have one of

three different positions relative to each other.

If the distance from the center of the circle to the line

is greater than the radius of the circle, then the line

does not intersect the circle.

In the figure, [OH] l and |OH| > r,

and l C(O, r) = .


is equal to the radius, then we say that the line is tangent to the circle. In the figure, |OH| l,

|OH| = r, and l C(O, r) = {H}. H is the only point of

intersection of the line and the circle.

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is less than the radius, then the line intersects the

circle at two points.

In the figure, [OH] and |OH| < r,

and C(O, r) = {A, B}.

For example, line is a secant in the figure on the left.

Definition tangent

A line which intersects a circle at exactly one point is called a tangent of the circle. The inter-

section point is called the point oof ttangency.

Definition secant

A line which intersects a circle at two different points is called a secant of the circle.

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EXAMPLE 3 Name all the radii, diameters, chords, secants, and

tangents of the circle in the figure.

Solution [OF], [OC], and [OB] are radii.

[FC] is a diameter. l is a secant line.

[EF], [ED], and [FC] are chords. GH is a

tangent, and A is a point of tangency.

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166 Geometriy 7

Check Yourself 11. Define the terms center, radius, chord, diameter, tangent, and secant. Show them in a

figure.

2. How many regions does a circle divide the plane into?

3. Sketch all the possible relative positions of a circle and a line in the same plane.

4. Look at the figure on the right.

a. Name the tangents.

b. Name the secants.

c. Name the chords.

d. Name the radii.

e. Name the diameters.

Answers

1. center: a point inside the circle that is equidistant from all the

points on the circle.

radius: a distance from the center to a point on the circle.

chord: a line segment joining two different points of a circle.

diameter: a chord passing through the center of a circle

tangent: a line intersecting a circle at exactly one point.

secant: a line intersecting a circle at two different points.

2. three parts: the interior of the circle, the circle, and the exterior of the circle.

3.

4. a. EF, EB b. BC, DB c. [AB], [DB], [BC] d. [OD], [OA], [OB], [OC] e. [BD]

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167Circles

Remember that a chord is a line segment which joins two different points on a circle. In this

section we will look at the properties of chords.

B. CHORDS

Property

A radius that is perpendicular to a chord bisects the chord.

For example, in the figure, if [OH] [AB] then

|AH| = |HB|.

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EXAMPLE 4 A chord of length 10 cm is 12 cm away from the center

of a circle. Find the length of the radius.

Solution Look at the figure.

In AHO, r2 = 52 + 122

r2 = 25 + 144

r2 = 169

r = 13 cm.

210 5 cm2

| AB|

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Property

In the same circle or in congruent circles, two chords which are equidistant from the center

are congruent.

For example, in the figure, if |OM| = |ON|,

then |AB| = |CD|.

The converse of this property is also true:

if |AB| = |CD|, then |OM| = |ON|.�

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168 Geometriy 7

Property

In the same circle or in congruent circles, if two chords have different lengths, then the

longer chord is nearer to the center of the circle.

For example, in the figure,

if |CD| > |AB|, then |OF| < |OE|. The converse

of this property is also true: if |OF| < |OE|,

then |CD| > |AB|.�

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EXAMPLE 6 In the circle in the figure, |OM| < |ON| and r = 9 cm.|AB| = 3x + 2 cm and|CD| = 5x – 2 cm are given.Find the possible integervalues of x.

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EXAMPLE 5 In the figure, |AB| = 8 cm,

|CN| = 4 cm, and

|OM| = 3 cm. Find |OC| = x.

Solution |CD| = 8 cm, since |CN| = 4 cm. So |AB| = |CD|,

and by the property, |OM| = |ON| = 3 cm.

Let us use the Pythagorean Theorem to find the length of [OC]:

|OC|2 = |ON|2 + |NC|2

x2 = 32 + 42

x = 5 cm.

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Solution If |OM| < |ON|, then |CD| > |AB|.

5x – 2 > 3x + 2

2x > 4

x > 2 (1)

Since the longest chord is the diameter, the greatest possible value of |CD| is the diameter.

d = 2r, d = 2 9 = 18 cm

|CD| 18

5x – 2 18

5x 20

x 4 (2)

From (1) and (2), the possible integer values of x are 3 and 4.

169Circles

Check Yourself 21. In the figure, the radius of the circle is 5 cm and

|AB| = |CD| = 8 cm. Find|OE|.

2. In the figure, |AB| = |CD|, [OM] [AB], [ON] [CD], and

|ON| = |OM| = 4 cm. Given |AB| = 5x + 1 cm and

|CD| = 4x + 2 cm, find the radius of the circle.

3. In the figure, |AP| = 12 cm, |PB| = 4 cm, and

|OP| = 11 cm. Find the radius of the circle.

4. In the figure, |AB| = 12 cm, |DC| = 2 cm,

[OD] [AB].

Find the radius of the circle.

Answers

1. 3 cm 2. 5 cm 3. 13 cm 4. 10 cm

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Remember that a tangent is a line in the plane which

intersects a circle at exactly one point. The point is

called the point of tangency. In this section we will look

at the properties of tangents.

� �C. TANGENTS

Property

If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point

of tangency.

For example, in the figure, if l is tangent to the circle C

at point H, then [OH] l. �

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170 Geometriy 7

Property

If two segments from the same exterior point are tangent to a circle, then they are congruent.

For example, in the figure, if [PA and [PB are tangent to

the circle at points A and B respectively, then

|PA| = |PB|.

Property

Two tangent line segments from the same external point determine an angle that is bisected by

the ray from the external point through the center of the circle.

For example, in the figure, if [PA and [PB are tangent to

the circle then [PO is the angle bisector of APB, i.e.

mAPO = mBPO. �

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EXAMPLE 7 The circle in the figure is inscribed in the triangle ABC.

|AK| = x + 5 cm,

|BM| = 2x + 3 cm,

|CL| = 2x + 5 cm, and the perimeter of triangle ABC

is 46 cm.

Find |MC|.

Solution |AK| = |AL|, |BK| = |BM|, and |CM| = |CL|.

P(ABC) = |AB|+|BC|+|AC|

= |AK|+|KB|+|BM|+|MC|+|CL|+|LA|

= 2 |AK| + 2 |BM| + 2 |CL|

= 2 (x + 5) + 2 (2x + 3) + 2 (2x + 5)

= 2x + 10 + 4x + 6 + 4x + 10

= 10x + 26

P(ABC) = 10x + 26 = 46 x = 2 cm

So |MC| = 2x + 5 = 9 cm.

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171Circles

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D. RELATIVE POSITION OF TWO CIRCLES IN THE SAMEPLANE

Tangent circles can be externally tangent or internally tangent, as shown in the figure.

Definition nonintersecting ccircles

Two circles which have no common point are called nonintersecting ccircles.

Definition tangent ccircles

Two circles which have only one common point are called tangent ccircles.

If two or more circlesshare the same center,then they are called concentric circles.

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172 Geometriy 7

r1 + r2 + r3 =21

16+ r3 =21

r3=5 cm.

r1 + r2 + r3 =21

r1 +12=21

r1 =9 cm.

r1 + r2 =16

9+ r2 = 16

r2 =7 cm.

|AB| = r1 + r2 = 16

|BC| = r2 + r3 = 12

|CA| = r1 + r3 = 14

2 (r1 + r2 + r3) = 42

r1 + r2 + r3 = 21

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EXAMPLE 8 The circles in the figure with centers A, B, and C are

externally tangent to each other.

|AB| = 16 cm, |BC| = 12 cm, and |AC| = 14 cm are

given. Find the radii of the circles.

Solution Let the radii of circles A, B, and C be r1, r2 and r3

respectively. Then we can write,

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Definition intersecting ccircles

Two circles which have two common points are called intersecting ccircles.

173Circles

11.. Describe each line and

line segment in the

figure as an element of

the circle.

22.. The points in the figure

are in the same plane as

the circle. State the

position of each point

with respect to the

circle.

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55.. In the figure, X, Y, and Z are

points of tangency.

|AX| = 6 cm,

|CZ| = 4 cm, and

|BY| = 2 cm.

Find the perimeter of

ABC.

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66.. In the figure,

[OA] [BC],

|AK| = 2 cm, and

|KC| = 4 cm.

Find |OK|.

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|O1O2| = 3 cm and

r1 + r2 = 11 cm.

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|AB| = 3x + 4,

|CD| = 2x + 9, and

|OM| > |ON|.

Find the greatest possible

integer value of x.

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77.. In the figure,

|AC| = 6 cm and

|AB| = 3 cm.

Find |OB| = r.��

88.. In the figure,

|BC| = 12 cm and

|AD| = 8 cm.

Find the radius of the

circle.�

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99.. In the figure,

|AP| = 6ñ3 cm and

mAPB = 60°.


circle.

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33.. In the figure, the radius

of circle O is 15 cm,

|CD| = 24 cm, and

|OH| = 12 cm.

a. Find |OI|.

b. Find |AB|.

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ment about the figure

with a suitable symbol.

a. If |OE| = |OF|, then

|AB|...|CD|

b. If |OE| > |OF|, then

|AB|...|CD|.

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EXERCISES 4.1

174 Geometriy 7

We use the ï sign over two or more points to denote the

arc which includes the points. For example, in the fig-

ure, we write AïB to denote the arc between A and B, and

AùCB to denote the arc ACB.

Notice that any two points of a circle divide the circle into

two arcs. If the arcs are unequal, the smaller arc is called

the minor aarc and the larger arc is called the major aarc.

In the figure on the right, AïB is the minor arc and AùCB

is the major arc.

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After studying this section you will beable to:

1. Describe the concepts of arc andcentral angle.

2. Name inscribed angles and calcu-late their measure.

3. Use the propertiesof arcs, central angles, andinscribed angles tosolve problems.

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A. ARCS AND CENTRAL ANGLES

Definition arc oof aa ccircle

An arc of a circle consists of two points on the circle and the unbroken part of the circle

between these two points.

We use the ï sign over two or more points to denotethe arc which includes the points. For example, in thefigure, we write AïB to denote the arc between A and B,and AùCB to denote the arc ACB.

Notice that any two points of a circle divide the circleinto two arcs. If the arcs are unequal, the smaller arc iscalled the minor aarc and the larger arc is called themajor aarc.

In the figure on the right, AïB is the minor arc and AùCB is the major arc.

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Definition central aangle oof aa ccircle

An angle whose vertex is at the center of a circle is called a central aangle of the circle.

175Circles

EXAMPLE 9 Find the measure of the

indicated central angle of

each circle.

Solution Remember that the measure

of a minor arc is equal to the

measure of its central angle.

a. mAOB = mAïB = 50°

b. mCOD = mCïD = 120°

c. mAOB = mAïB = 180°

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B. INSCRIBED ANGLES

Property

In the same circle or in congruent circles, if two chords

are congruent, then their corresponding arcs and cen-

tral angles are also congruent.

For example, in the figure, if [AB] [CD] then AïB CïDand mAOB mCOD.

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Property

If a line through the center of a circle is perpendicular

to a chord, it bisects the arcs defined by the endpoints

of that chord.

For example, in the figure, if [PK] [AB] then

[AH] [HB]

AïP PïB

AïK KïB.

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For example, angle ABC in the figure is an inscribed

angle. [AB] and [BC] are both chords of the circle.

The arc AïC in the figure is called the intercepted aarc of

the inscribed angle ABC.

Definition inscribed aangle oof aa ccircle

An angle whose vertex is on a circle and whose sides

contain chords of the circle is called an inscribed aangle.

a. b. c.

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176 Geometriy 7

mOBC = mOCB = a°

mOAB = mOBA = b°

mABC = mOBA + mOBC

= a° + b°

mCOE = mCBO + mOCB = 2a°

mAOE = mOAB + mOBA = 2b°

mAOC = mAOE + mEOC

= 2 (a° + b°)

So mABC = 2

m AOC .

Property

The measure of an inscribed angle is half of the measure

of the central angle which intercepts the same arc.

ProofIn the figure, let mBCO = a° and mBAO = b°.

Since the triangles BOC and AOB are isosceles

triangles, we can write

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Now remember that the measure of a minor arc is the same as the measure of its central

angle. So we can write the property in a slightly different way:

Property

The measure of an inscribed angle is equal to the half

the measure of its intercepted arc.

For example, in the figure, mABC = .2ïmAC

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EXAMPLE 10 Find the measure of x in each figure.

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a. b. c.

177Circles

Solution a. mABC =

50 =

mAïC = 100°

mx = 100°

2ïmAC

2ïmAC

b. mABC =

mx =

mx = 60°

1202

2ïmAC c. mABC =

=

mx = 45°

902

2m AOC

Solution a. mCAD = mCBD =

x° = y° =

x = y = 20

402

2mCDï

b. mBAC =

x° =

x = 25 and y = 50

° 50=2 2y

2 2m BOC mBC= ï

Property

If two inscribed angles intercept the same arc of a

circle, then the angles are congruent.

For example, in the figure, ABC ADC, because

they both intercept AïC.

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EXAMPLE 11 Find the value of x and y in each figure.

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178 Geometriy 7

Property

An angle inscribed in a semicircle is a right angle.

For example, in the figure,

if mAùLB = mAùMB = mAùNB = 180°, then

mALB = mAMB = mANB = 90° or

mL = mM = mN = 90°.

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c. mBAC =

20 =

y = 40

°2y

2m BOC

mBDC =

x° =

x = 20

40°2

2m BOC

EXAMPLE 12 Find the value of x in each figure.

Solution a. Since AC is the diameter, the arc AùBC is a semicircle.

So ABC is inscribed in a semicircle, and therefore mABC = x° = 90°, x = 90.

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b. mDAC =

=

= 30°

60°2

mDC2ï y = 2 mBAC

= 2 20

= 40

mAïD + mDïC + mCïB = 180

x° + 60° + 40° = 180°

x = 80

179Circles

Property

The measure of the angle formed by a tangent and a chord is equal to the half of the meas-

ure of its intercepted arc.

For example, in the figure, mCAB = mAOB.12

Proof

Let us draw the diameter [AD] and the chord [BD].

[AC] [AD] (a radius is perpendicular to a tangent at

the point of tangency)

[AB] [BD] (definition of a semicircle)

mDAB + mBAC = 90°

mDAB + mADB = 90° (in triangle ABD)

mADB = mBAC

mADB = (inscribed angle rule)

mBAC = (inscribed angle rule)2ïmAB

2ïmAB

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Rule

Let [AB] and [CD] be two chords of a circle.

If [AB] [CD], then

mABC = mBCD (alternate interior angles).

So mAïC = mBïD.

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c. Let us draw the chord [BD].

mADB = 90°

mCAB = mCDB = 10°

mADC + mCDB = 90°

x + 10 = 90

x = 80

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180 Geometriy 7

EXAMPLE 13 Find the value of x and y in each figure.

Solution

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a. Let us draw the radius [OA].

Then [AC] [AO]. AOB is isosceles triangle.

mOAB = 30° and

mOAB + mBAC = 90°

30° + x° = 90°

x = 60

y = 2 60 = 120

b. [AB] [CD] and mBAD =

mBïD = 2 15° = 30°

mAïC + mCïD + mDïB = 180°

30 + mCïD + 30 = 180°

mCïD = 120°

mDCE = x° =

= = 60120°

2

2mCDï

mBD2ï

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a. bb. cc.

[AD] [BC] and [AF] [AD]. So [AF] [BC]. Therefore, mABC = mBAE and

mEAB = x° = 30°, x = 30.

c. mADC =

mAïC = 2 30° = 60°

mABC =

mABC = = 30°60°2

ïmAC2

ïmAC2

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181Circles

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Rule

The measure of an angle formed by two secants, a

secant and a tangent, or two tangents drawn from a

point in the exterior of a circle is equal to half of the

difference of the measures of the intercepted arcs. �

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Rule

The measure of an angle formed by two chords that

intersect in the interior of a circle is equal to half the

sum of the measures of the intercepted arcs.�

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For example, in the figure, mAED = mBEC = and

mAEB = mCED =

and2

=x+ y a+b= .

2

2ï ïmA B + mCD

2mBC + mADï ï

182 Geometriy 7

EXAMPLE 14 Find the value of x in each figure.

Solution a. mBED =

70° =

mAïC = 140 – 60 = 80

mADC =

b. mCAE =

mCAE = x° = = 35°, x = 35

c. mQPR + mQïR = 180°

60 + mQïR = 180

mQïR = 120°

mQïR + mQùTR = 360°

mQùTR = 240°

mQSR = x° = = 120°, x = 120240°=

2 2mQTRù

70°2

100 – 302 2

mCE – mBD = (mBD = 2×m BCD)ï ï ï

80= = 40°, = 402 2ïmAC x

+60 ( = 2 )2

ï ïmAC mBD m BAD

2ï ïmAC + mBD

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183Circles

44.. In the figure,

mBOC = 100° and

mACO = 20°.

Find mAOC.

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99.. In the figure,

mAOC = mABC = 3x°.

Find the value of x.��

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[AB] is a diameter and

mOCB = 40°.

Find mOAC.�

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mAOC = 100° and

mOAB = 70°.

Find mOCB.�

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mDPA = 50°.

Find mBCA. � �!*�

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mAPD = 30°,

mDKA = 60°,

mBAC = a°, and

mDCA = b°.

Find a and b.

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mAïD = and

mDPC = 75°.

Find mBAC.

2mBCï

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88.. In the figure,

[AE is tangent to the

circle at the point B,

and mEBC = 75°.

Find mA.

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mCDB = 10°

and mABD = 50°.

Find mP.� �!*

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11.. In the figure,

mAOC = 120°.

Find mABC.�

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mCBD = 120°.

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mBAC = 30° and

mBKC = 70°.

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EXERCISES 4.2

184 Geometriy 7

1. Circumference of a CircleRemember from chapter 4 that the distance around a polygon is called the perimeter of the

polygon.

C == 22r .

A. CIRCUMFERENCE AND ARC LENGTH

If you measure the circumference and diameter of a circle and divide the circumference by

the diameter, you always get the same constant. This constant is approximately equal to

3.14, and denoted by .

Definition circumference

The distance around a circle is called the circumference of the circle.

NotePi (, pronounced like the English word ‘pie’) is a Greek letter. It is the first letter of a Greek word that means ‘measure around.’

(

Property

For all circles, the ratio of the circumference to the diameter is always the same number. The

number is called (pronounced like‘pie’).

This is the formula for the measure of the circumference of a circle.

So if the circumference of a circle with a diameter d is C, then we can write or C = d or C =d


1. Describe the concepts of circumference and arc length.

2. Find the area of a circle, an annulus, a sector, and a segment.

Objectives

185Circles

1. Find five different circular objects. Use a piece of string to measure their

circumference (C), and use a ruler to measure their diameter (d). Write

the values in a table.

2. For each circular object calculate the ratio and then calculate the

average of all the ratios.

3. How do the number and the formula C = d relate to this activity?

Cd

2. Arc LengthRemember that an arc is a part of a circle. The measure of an arc is equal to the measure of

its central angle.

arc length of = ,

circumference of the circle 360°arc length of

so = .2ð r 360°

AB mAB

AB

ï ï

ï

arc llength oof AïB = 22 r

360°

In the above formula the measure of the arc is given in degrees. The length of the arc is given

in a linear unit such as centimeters.

We can rewrite this as .

EXAMPLE 15 a. Find the diameter of a circle with circumference 24 cm.

b. Find the circumference of a circle with radius 5 cm.

c. Find the circumference of a circle with diameter 9 cm.

Solution a. Let the diameter of the circle be d, then the circumference of the circle is C = d:24 = d

d = 24 cm.

b. C = 2 5 = 2 5 = 10cm

c. C = 2r = 2r = d = 9 cm

Rule

In a circle, the ratio of the length of a given arc AïB to

the circumference is equal to the ratio of the measure

of the arc to 360°.�

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186 Geometriy 7

a. The length of a semicircle is halfof the circumference. arc length of AïB

= 2r = 2 6

= 2 6 = 6cm12

180°360°360

b. The length of a 90° arc is a quarter ofthe circumference. arc length of CïD

= 2r = 2 10

= 2 10 = 5cm14

90°360°360

c. arc length of EïF

= 2r = 2 12

= 2 12 = 4cm16

60°360°360

d. arc length of GùTH

= 2r = 2 18

= 2 18 = 21cm2136

210°360°360

Solution

Check Yourself 31. Find the circumference of the circle with the given radius.

a. r = 3 cm b. r = 5 cm c. r = 7 cm d. r = 10 cm

2. Find the radius of the circle with the given circumference.a. 12 cm b. 24 cm c. 36 cm d. cm

3. Find the length of the minor arc in each figure.

Answers1. a. 6 cm b. 10 cm c. 14 cm d. 20 cm

2. a. 6 cm b. 12 cm c. 18 cm d. cm

3. a. cm b. cm c. cm d. 8 cm8310

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2

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EXAMPLE 16 Find the length of each arc.

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187Circles

1. Area of a Circle

To understand why this property is true, let us divide a

circle into 16 equal parts, and rearrange them as

follows:

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As the number of equal parts increases, the area of the circle gets closer and closer to the

area of a parallelogram.

The area of a parallelogram is

So the area of a circle with radius r is A = r2.

2 2 2C 2 rA = r = r = r .

A == r22

B. AREA OF A CIRCLE, A SECTOR, AND A SEGMENT

Property

The area of a circle is times the square of the radius.

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b. Let the radius of

the circle be r, then

A = r2

16 = r2,

r2 = 16

r = 4 cm.

c. The formula for the

circumference of a

circle is C = 2 r:

10 = 2 r

r = 5 cm.

So the area of the

circle is

A = r2

= 52

= 25cm2.

EXAMPLE 17 a. Find the area of a circle with radius r = 6 cm.

b. Find the radius of a circle with area 16 cm2.

c. Find the area of a circle with circumference 10 cm.

Solution a. Let the area of the

circle be A, then

A = r2

A = 62

A = 36cm2.

188 Geometriy 7

2. Area of an Annulus

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A(annulus) == R22 – r22

= (R22 – rr22)

How can we find the area of an annulus? Look at the

diagram.

Definition annulus

An annulus is a region bounded by two concentric circles.

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EXAMPLE 18 Find the area of the annulus bounded by concentric

circles with radii 5 cm and 3 cm long.

Solution The radius of the big circle is R = 5 cm.

The radius of the small circle is r = 3 cm.

A = R2 – r2

A = (R2 – r2)

A = (52 – 32)

A = (25 – 9)

A = 16 cm2

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189Circles

EXAMPLE 19 Find the area of each shaded sector.

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| AB| rA =

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360

| |=

2 360

| |=

2 360

aA B r

A B a r

A B r a r

ï

ï

ï

In the figure,

(|AïB| = l ).

3. Area of a Sector

For example, in the figure, the smaller region AOB is a sec-

tor of the circle. If the degree measure of arc AB is

mAïB = a° then the area of sector

We can also calculate the area of a sector in a different

way:

2a= .360

AOB p r

Definition sector oof aa ccircle

A sector of a circle is the region bounded by two radii of the circle and their intercepted arc.

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In the above formula the measure of the arc is given in degrees. The length of the arc is given

in a linear unit such as centimeters.

Rule

The area of a sector of a circle is half the product of the length of the arc and the length of

its radius.

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a. b. c.

190 Geometriy 7

4. Area of a Segment

.

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area of segment = –

A = A(sector AOB) – A(AOB)

area of

triangle

area of

sector

2 b haA = ð r –

360 2

Solution a. r = 5 cm and ma = 72°.

b. r = 8 cm and l = 6 cm.

c. mBOC = 2 mBAC, so

mBOC = 30° and r = 6 cm.

2 2 230 1(sector )= ð = ð 6 = ð 36 = 3ð cm .360 360 12 m BOCA BOC r

26 8(sector )= = = 24 cm

2 2

l r

A POS

2 2 272 1(sector )= = 5 = 25= 5 cm360 360 5

aA AOB r

Definition segment oof aa ccircle

A segment of a circle is a region bounded by a chord and its intercepted arc.

EXAMPLE 20 Find the area of each shaded segment.

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191Circles

Check Yourself 41. Find the area of a circle with the given radius.

a. r = 3 cm b. r = 5 cm c. r = 12 cm d. r = 16 cm

2. Find the area of a circle with the given circumference.

a. 4 cm b. 12 cm c. 20 cm d. cm

3. Find the circumference of a circle with area 36 cm2.

4. The ratio of the radii of two circles is 5 : 3. What is the ratio of their areas?

5. The area of the shaded region in the figure is 32 cm2 and

R = 9 cm. Find r.�

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Solution a. Since mAOB = 90°,

A(segment) = 9 – 18 cm2.

b. |OH| = 6 cm and

|AB| = 12ñ3 cm.

A(sector AOB) =

=

= 48 cm2.

A(AOB) =

A(segment) = 48 – 36ñ3 cm2.

c. A(sector AOC) =

A(AOC) =

A(segment) = 9 – 18 cm2.

26 6 36 18 cm2 2

22ð ð 36

= =9ð cm4 4 r

2| | | | 12 3 636 3 cm

2 2

AB OH

1 1443

2120 ð 12360

2236( )= = =18 cm

2 2 rA AOB

2 290 1(sector )= = 36 = 9 cm360 4

A AOB r

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192 Geometriy 7

6. Find the area of the shaded region in each circle.

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Answers1. a. 9 cm2 b. 25 cm2 c. 144 cm2 d. 256 cm2

2. a. 4 cm2 b. 36 cm2 c. 100 cm2 d. cm2

3. 12 cm

4.

5. 7 cm

6. a. cm2 b. cm2 c. 36 cm2 d. cm2 e. 36 cm2 f. 25 cm2 g. 17 cm2 f. 16 cm225 50464

325

6

259

4

193Circles

33.. In the figure,

mAOB = 120° and

r = 6 cm.

Find the length of arc

AùXB.

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44.. In the figure, B is the

center of a circle and

ABCD is a square with

|AD| = 5 cm.

Find the area of the

shaded region.� �

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66.. In the figure,

|OB| = 5 cm,

mDOB = 60°, and

|BA| = 3 cm.


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55.. In the figure, mAOB = mCOD = mEOF = 20°

and r = 6 cm. Find the

sum of the areas of the

shaded regions. .

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EXERCISES 4.311.. In the figure,

mAOB = 30° and

r = 6 cm.


shaded region.

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22.. In the figure,

mAOB = 45° and

r = 10 cm.


shaded region.

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99.. In the figure, A, B, and Care the centers of threecongruent tangent circles.If the sum of the circumferences of the circles is 24 cm, findthe area of the shadedregion.

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B, C, P, and K are the centers of four circles in thefigure. Given |AB| = |BC| = |CD| = 4 cm,

find the area of shaded region.

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1111.. In the figure, ABCD is asquare with perimeter 64 cm. Find the area ofthe shaded region.

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194 Geometriy 7


A(AOB) = 48 cm2,

|OC| = 8 cm, and

[OC] [AB]. Find r.�

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mBCD = 130° and

mOAC = 40°.

Find mCBO. �

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mOAB = 45° and

mOCB = 60°.

Find mAOC. �

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mAOC = 160° and

mABC = x°.

Find mABC.

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mOAD = 40° and

mBOC = 50°.

Find mCOD.�

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B is the point of

tangency and

mOAB = 30°.

Find mABC.

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|AB| = |CD| = 8 cm and

|OH| = 3 cm.

Find the radius of

the circle.

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|CE| = 3x – 2,

|FB| = x + 4, and

|OE| = |OF|.

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mAOB = 60° and

|AB| = 5 cm.


circle.�

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|AB| = 9 cm,

|BC| = 8 cm, and

|CA| = 5 cm.

Find the radius of

circle A.

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mOAB = 50° and

mBCO = 35°.

Find mAOC.

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1122.. In the figure, B and Dare the centers of twocircles.

If ABCD is a square

and the shaded area

is 16 cm2, find

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195Circles


mBAD = 60° and

|AD| = |DC|.

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|OB| = r = 4 cm.


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|AB| = 8 cm and

|AC| = 6 cm.


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3333.. ABCD is a square with sides 10 cm long. Find the

area of each shaded region.


mAOE = 60° and

|OA| = |DC|.

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mBAD = 30°.

Find mACD.

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mBAC = 20° and

mDFE = 30°.

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2299.. In the circle in the figure,

|OA| = 6 cm,

mAOB = 50°,

mCOD = 30°,

and mEOF = 40°.

Find the sum of the areas of the shaded regions.

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of the circle is 6 cm and

the length of arc AùXB is

4 cm. Find the area of

the shaded region.

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2244.. A and C are points of tan-gency on the circle in thefigure.

Given mABC = 60° and

mBCD = 70°,

find mBAE.

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c. d.

e. d.

Angles 196

CHAPTER 4 REVIEW TEST

11.. In the figure,

|OA| = 4 cm and

|OC| = 7 cm.

What is |BC|?

A) 2 cm B) 3 cm C) 4 cm D) 5 cm

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66.. In the figure,

line l is tangent to the

circle at point C and

|OA| = |AB| = 5 cm.

Find |BC| = x.

A) 4 cm B) 5 cm C) 5ñ3 cm D) 6 cm

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77.. Find mABC in the figure.

A) 49° B) 50° C) 51° D) 52°

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88.. In the figure,

mABD = 60° and

mCED = 80°.

Find mCDE.

A) 10° B) 20° C) 25° D) 40°

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99.. In the figure,

mBDC = 70°.

Find mACB.

A) 20° B) 25° C) 30° D) 40°

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1100.. In the figure, line l is

tangent to the circle at

point A and

|AB| = |AC|.

Find mCAD.

A) 65° B) 55° C) 50° D) 45°

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22.. Find |AB|in the figure if

|CH| = 4 cm.

A) 8 cm B) 7 cm C) 6 cm D) 5 cm

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of the circle is 10 cm

and |OH| = 6 cm.

Find |AB|.

A) 8 cm B) 12 cm C) 16 cm D) 20 cm

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44.. In the figure,

|OC| = 3ñ2 cm,

|AC| = 1 cm, and

|BC| = 7 cm.

What is the length

of the radius?

A) 3 cm B) 3ñ3 cm C) 4ñ2 cm D) 5 cm

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55.. In the figure,

|OK| = |OH| = 5 cm,

|AB| = 2a + 2 cm, and

|CD| = a + 13 cm.

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the radius?

A) 13 cm B) 12 cm C) 11 cm D) 10 cm

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[PE and [PD are

tangent to the

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A and B, respectively.

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mAPB = 50°.

A) 60° B) 65° C) 70° D) 75°

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mDCE = 30° and

mAïB = 80°.


A) 65 B) 70 C) 75 D) 80

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|AB|= 2 cm and

|AC| = 2ñ3 cm.

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the circumference?

A) 3 cm B) 4 cm C) 6 cm D) 8 cm

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the perimeter of the

circle is 10 cm and

|OH| = 3 cm.

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A) 5 cm B) 6 cm C) 7 cm D) 8 cm

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mAPC = 35° and

mBïD = 100°.

Find mADC.

A) 15° B) 20° C) 30° D) 40°

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mABC = 35°,

mACB = 55°, and

|BC| = 4 cm.

What is the area of the

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A) 2 cm2 B) 3 cm2 C) 4 cm2 D) 8 cm2

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1199.. In the figure, the circle

has radius 6 cm and

mABC = 75°.


shaded region.

A) 6 cm2 B) 9 cm2 C) 12 cm2 D) 15 cm2

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2200.. In the figure, ABCD is a

square. |BE| = 4 cm,

|DF| = 6 cm, and B and

D are the centers of two

circles. Find the area of

the shaded region.

A) 40 – 10 cm2 B) 50 – 13 cm2

C) 36 – 12 cm2 D) 64 – 20 cm2

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1177.. Find the length of the

arc AïB in the figure if

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mACB = 60°.

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1188.. In the figure, ABCD is a

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A) 9 – 2 cm2 B) 16 – cm2

C) 36 – 9 cm2 D) 49 – 12 cm2

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196 Geometriy 7

1. Because there are no simpler concepts for us to buid on. Therefore, we need to understand these concepts

without a precise definition.

3. A ray has closed enpoint but a half line has an open endpoint.

4. 2 5. 3 7. a. size, length, width, thickness b. line c. plane d. skew lines

8. a. true b. true c. true d. false e. true

9. a. 10 b. 21 c. 210 d. 5050

10. lines: HL, HG rays: [LC, [LH, [HL, [HG, [GH half lines: ]LC, ]LH, ]HL, ]HG, ]GH

11. a. line segment CD b. half open line segment PQ c. open line segment AB d. ray KL e. half line MN

f. line EF

12. a. l (E) = l b. d (F) = {C} c. n (G) = 13. ‘M, N and P’, ‘R and S’, and ‘L and K’, are coplanar

points.

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15. (D) (E) = m(D) (F) = l(E) (F) = dm d l = {O}

16. a. 5 b. (P) (Q) = EB, (P) (S) = EA, (P) (T) = AB, (Q) (T) = BC, (Q) (R) = EC, (T) (R) = DC,

(S) (R) = ED, (T) (S) = AD c. 3 lines pass through point A, B, C, and D, 4 lines pass through point E.

EXERCISES 1.1

1. a. b. c. {K, O, M} d. e. {N} f. g. h. {P} i. j. k.

2. a. {A} [CD b. ]AC[ ]AD[ c. ]CD[ d. ]CE ]DF e. ]AB] [BC[ ]AH ]DG

5. a. acute angle b. right angle c. obtuse angle d. straight angle e. complete angle

6. a. 20° b. 12° c. 20° 7. a. 32° b. 20° c. 10°

8. a. 115° b. 65° c. 115° d. 65° e. 115° f. 65° g. 65°

9. 130° 10. 40° 11. 25° 12. 50° 13. 100° 14. 70° 15. x = y + z 16. 160° 17. 35° 18. 140° 19. 80°

20. 90° 21. 35°

EXERCISES 2.1

197Answers to Exercises

1. ADE, DEK, DKF, BDF, CKF, CKE, DEC, ADC, DFC, BDC, CEF, ABC

2. eight triangles: GDT, DTE, ETF, FTG, GDE, GFE, GDF, DEF 3. 51 cm 4. 10 cm

5. 28.2 cm 6. a. B, E, F, C b. F c. segment AC and point E d. segment FC without

endpoints

7. 8. 12 9. 7 cm

10. a. b. c. 11. a. Hint: Construct medians for each

side. b. Hint: Construct angle bisectors for

each angle. c. Hint: Construct altitudes

for each vertex.

d. Hint: Construct perpendicular bisectors for each side. 14. 15. a. BFC b. CEF, BEF,

ABC c. BFC d. ABF e. ABF 17. a. yes b. no c. yes d. yes e. no 18. a. x {4, 9, 14} b. none 19.

20. a. in the interior b. in the interior c. in the interior d. in the interior e. on the triangle f. in the interior

g. in the interior h. on the triangle i. in the exterior j. in the interior k. in the interior l. in the exterior

21. a. sometimes b. always c. never d. sometimes e. never f. never g. always h. always

2121cm

2

56 168= ; =

5 13b ch h

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EXERCISES 3.1

22. a. b. c. d. e. f.

23. a. b.

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198 Geometriy 7

1. a. 36° b. 114° c. 54° d. 20° e. 100° f. 50° g. 90° h. 64° 2. 40°, 60°, 80° 3. x = 120°, acute angles: 85°, 5°4. 50° 5. 136° 6. 60° 7. 72° 8. 117° 9. 12° 10. 8° 11. 36° 12. 106°13. a. b. c. not possible d. e. not possible

14. a. no b. no c. no d. yes 15. 55° 17. a. 2 < a < 14 b. 4 < p < 20 c. 1 < m < 7 18. a. 4 < x < 12 b. 4 < x < 11 c. 3 < x < 10 19. three 20. 9 21. 12 22. 29 23. 36 24. 9 25. five 26. a. yes b. no c. no d. no e. yes f. yes 27. x 28. a. 1 = 2 > 3 b. 1 > 2 > 3 c. 3 > 2 > 1 d. 1 > 3 > 2 29. a. false b. falsec. true d. false e. true f. false g. false 30. a. AC b. AC c. DC d. BC 31. a. 5 b. 8 32. 49 33. 2534. there are no values 35. 11 36. four triangles with side lengths (1, 5, 5), (2, 4, 5), (3, 4, 4), (3, 3, 5) 38. a. Ab. D c. D d. D 39. a. A b. B 40. a. A b. B c. A d. A 41. a. A b. B 42. a. D b. A

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EXERCISES 3.3

2. a. 70° b. 1 c. 60° d. 6 e. 3 3. A K; D L; E N; AD KL; DE LN; AE KN

4. a. 6 b. 20° c. 22° d. 5. a. BC = 3, MN = 8 9. 10 cm 11. 2 12. m(OKM) = 10°, m(OML) = 60°,

m(OLK) = 20° 13. 8 14. 15 15. 84° 16. 17. 84° 18. 38° 19. 9.6 cm 20. 20° 21. 22. 3ñ3 23. 8 cm

24. 3ñ3 25. 4 cm 26. 14 27. 15 28. 8ñ3 29. 3 cm 30. 16 31. 2 cm 32. 6 cm 33. 2 cm 34. 2 cm

35. 4ñ3 cm 36. 70° 37. 12 38. 99° 39. 9 cm 40. 3ñ3 – 3 41. 24 cm 42. yes 43. 200 km 44. 25° 45. 7 – ñ5

46. 6 47. 70° 48. 57° 49. 150° 50. 8 cm 51. 16 cm 52. 18° 53. 8 and 12 54. 8 55. 45° 58. 4 cm 59. 10°

60. 3ñ2 60. 72 62. 6 cm 63. 12 cm 64. 12 cm 65. 8 66. 6 cm 67. 6 cm 68. 6 cm 69. 8 cm 70. 9 cm

71. 72. 6 73. 2 74. 18 75. 1532

14

52

118

EXERCISES 3.2

EXERCISES 3.4

1. 2. {21, 31} 3. 4. {–49, 31} 5. 6. 7. a. ò13 b. c. 8.

9. 3x + 4y – 10 = 0 ; 3x + 4y + 30 = 0 10. 53–

15

1243

52

15 24

35 25

{– , }9 9

16

29

3–

41310


EXERCISES 4.11. radii: [OF], [OC], [OA], [OB] diameter: [FC] chords: [ED], [FC], [GB] tangent: AH secant: GB center: O

2. Points C, O, and D are in the interior region of the circle. Point E is on the circle. Points A, B, G and F are in the

exterior region of the circle 3. a. 9 cm b. 18 cm 4. a. = b. > 5. 24 cm 6. 3 cm 7. cm 8. 4ñ6 cm

9. 6 cm 10. r1 = 4 cm, r2 = 7 cm 11. 4 cm

92

EXERCISES 4.21. 120° 2. 10° 3. 120° 4. 30° 5. 50° 6. 40° 7. 70° 8. 60° 9. 40 10. 60° 11. a = 15° b = 45°

12. 20°

EXERCISES 4.3

1. 33 cm2 2. cm2 3. 4 cm 4. cm2 5. 6 cm2 6. cm2 7. (6 – 9ñ3) cm2

8. (72 – 18) cm2 9. (16ñ3 – 8) cm2 10. 12 cm2 11. (256 – 64) cm2 12. (8ñ2 – 8) cm

13. 10 cm 14. 5 cm 15. 3 16. cm 17. 3 cm 18. 170° 19. 10° 20. 150° 21. 100° 21. 30° 22. 60° 23. 50°

24. 120° 25. 20° 26. 120° 27. 80° 28. 12 cm2 29. 12 cm2 30. 4 cm2 31. cm2 32. a. 50 cm2

b. (50 – 100) cm2 c. (100 – 25) cm2 d. (50 – 100) cm2 e. cm2 f. cm225

(75 – )2

25( +25)

4

25 - 482

5 33

13225

(25 – )425

2

200 Geometriy 7

TEST 11. D2. A3. B4. C5. C6. D7. C8. B

9. D10. D11. C12. D

TEST 21. C2. A3. D4. C5. B6. B7. B8. B

9. C10. C11. D

TEST 2A1. E2. D3. A4. C5. A6. E7. B8. C

9. C10. C11. B12. C13. D14. C15. E16. C

TEST 2B1. C2. A3. E4. C5. A6. C7. B8. A

9. B10. C11. C12. A13. C14. C15. A16. E

TEST 2C1. E2. D3. C4. B5. E6. E7. D8. C

9. B10. A11. D12. C13. D14. A15. C16. C

TEST 2D1. D2. D3. A4. C5. E6. B7. C8. D

9. A10. E11. E12. C13. A14. A15. C16. D

TEST 2E1. D2. E3. B4. C5. C6. A7. C8. C

9. A10. C11. D12. C13. C14. C15. B16. B

SSyymmbbooll MMeeaanniinngg= is equal to

is not equal to

is greater than

is greater than or equal to

is less than

is less than or equal to

is approximately equal to

|x| absolute value of x

pi

ñ square root

A angle A

A exterior angle of A in a triangle

mA measure of angle A in degrees

degrees

minutes

seconds

right angle

mABC measure of angle ABC in degrees

AïB minor arc with endpoints A and B

mA ïB measure of minor arc AB in degrees

AùCB major arc with endpoints A and B

mA ùCB measure of major arc ACB in degrees

AB line AB, passing through the points A

and B

[AB] line segment AB or segment AB, withendpoints A and B

|AB| length of segment AB

[AB ray AB with initial point A, passingthrough B

]AB half line AB

]AB] half-open line segment AB, excludingpoint A and including point B

]AB[ open line segment

[AB] closed line segment

SSyymmbbooll MMeeaanniinngg is congruent to

is not congruent to

is parallel to

is not parallel to

is perpendicular to

is similar to

ABC triangle with vertices A, B and C

haa length of the altitude to side a

is an element of

is not an element of

union

intersection

is contained by

A B A is contained by B

A B A is not contained by B

A.S.A angle-side-angle

S.A.S side-angle-side

S.S.S side-side-side

A.A angle-angle

(E) plane E

(int ABC) interior of the triangle ABC

(ext ABC) exterior of the triangle ABC

A(ABC) area of the triangle ABC

P(ABC) perimeter of the triangle ABC

ABCD quadrilateral ABCD

ABCD paralelogram ABCD

O circle with center O

C circumference

sin sine

cos cosine

tan tangent

cot cotangent

sec secant

cosec cosecant

202 Geometriy 7

acute aangle: An acute angle isan angle with measure greaterthan 0° and less than 90°.

acute ttriangle: An acute triangle hasthree acute angles.

adjacent aangles: Two anglesare adjacent if they share a com-mon vertex and side, but haveno common interior points.

adjacent ssides: In a triangle orother polygon, two sides that share a common vertex areadjacent sides.

alternate eexterior aangles: Two anglesare alternate exterior angles if they lieoutside l and m on opposite sides of t,such as b and g.

alternate iinterior aangles: Two angles are alternateinterior angles if they lie between l and m on opposite sidesof t, such as d and e. (See figure for alternate exteriorangles.)

altitude oof aa ttriangle: An altitude of atriangle is a segment from a vertex thatis perpendicular to the opposite side orto the line containing the opposite side.An altitude may lie inside or outside the triangle.

angle: An angle consists of two differentrays that have the same initial point. Therays are the sides of the angle and the ini-tial point is the vertex of the angle.

angle bbisector: An angle bisector is aray that divides the angle into two con-gruent angles.

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angle bbisector oof aa ttriangle: An anglebisector of a triangle is a segment thatbisects one of the angles of thetriangle. Its endpoints are points on thetriangle.

angle oof ddepression: The angle formedby the horizontal and the line of sight toan object below the horizontal.

angle oof eelevation: The angle formedby the horizontal and the line of sight toan object above the horizontal.

area: The number of square units that cover a given surface.

base: The lower face or side of a geometric shape.

center oof aa ccircle: The center of a circle isthe point inside the circle that isequidistant from all the points on thecircle.

central aangle oof aa ccircle: A central angle of a circle is anangle whose vertex is the center of the circle.

circle: A circle is the set of all points in a plane that areequidistant from a given point, called the center of thecircle.

circumference oof aa ccircle: The circumference of a circle isthe distance around the circle.

collinear: Points, segments,or rays that are on the sameline are collinear.

complementary aangles: Two angles are complementary ifthe sum of their measures is 90°. Each angle is acomplement of the other.

concave ppolygon: See non-convex polygon.

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concurrent: Two or more lines or segments areconcurrent if they intersect at a single point.

congruent aangles: Two angles are congruent if they havethe same measure.

congruent aarcs: On the same circle or on congruent circles,two arcs are congruent if they have the same measure.

congruent ppolygons: Two polygons are congruent if there isa correspondence between their angles and sides such thatcorresponding angles are congruent and corresponding sidesare congruent. Congruent polygons have the same size andthe same shape.

congruent ssegments: Two segments are congruent if theyhave the same length.

consecutive iinterior aangles: Twoangles are consecutive interior angles ifthey lie between l and m on the same sideof t, such as b and e.

convex ppolygon: A polygon is convex ifno line that contains a side of thepolygon contains a point in the interior ofthe polygon.

coplanar: Points, lines, segments or rays that lie in the sameplane.

corresponding aangles: Two angles are

corresponding angles if they occupy cor-

responding positions, such as a and e

in the figure.

concentric ccircles: Circles that havedifferent radii but share the same centerare called concentric circles.

cone: A solid figure that has a circularbase and a point at the top.

cube: A square prism that has six equalsquare sides.

cylinder: A solid with circular ends andstraight sides.

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decagon: A decagon is a polygon that hasten sides.

degree: A unit of angle and arc measure.

diameter oof aa ccircle: A diameter of a circle is a chord thatpasses through the center. The diameter, d, is twice theradius: d == 22r.

diagonal: A line segment joining twonon-adjacent vertices of a polygon.

equiangular ttriangle: An equiangular triangle has threecongruent angles, each with a measure of 60°.

equilateral ttriangle: An equilateral triangle has three con-gruent sides.

exterior aangles oof aa ttriangle:When the sides of a triangle areextended, the angles that areadjacent to the interior angles of thetriangle are the exterior angles. Eachvertex has a pair of exterior angles.

exterior oof aan aangle: A point D is in theexterior of A if it is not on the angle or inthe interior of the angle.

half lline: A ray without an endpoint(initial point).

half pplanes: Two halves of a plane that areseparated by a line

hexagon: A hexagon is a polygon with sixsides.

hypotenuse: In a right triangle, the sideopposite the right angle is the hypotenuseof the triangle.

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204 Geometriy 7

inscribed aangle oof aa ccircle: An angle is

an inscribed angle of a circle if its vertex is

on the circle and its sides are chords of the

circle.

interior oof aan aangle: A point D is in the

interior of A if it is between points that lie

on each side of the angle.

intersecting llines: Coplanar lines which have only one

point in common.

intersecting pplanes: Planes which have one common line.

isosceles ttriangle: An isosceles triangle has at least two

congruent sides.

isosceles ttrapezoid: A quadrilateral with one pair of

parallel sides and at least two sides the same length.

kite: A convex quadrilateral with two pairs of

equal adjacent sides.

legs oof aa rright ttriangle: Either of the two

sides that form a right angle of a right tri-

angle.

legs oof aan iisosceles ttriangle: One of the

two congruent sides in an isosceles

triangle.

line: A line is an undefined term in geometry. In Euclidean

geometry a line is understood to be straight, to contain an

infinite number of points, to extend infinitely in two direc-

tions, and to have no thickness.

line ssegment: See segment.

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major aarc: On circle P, if mAPB < 180°,then the points A and B together with thepoints of the circle that lie in the exteriorof mAPB form a major arc of the circle.Major arcs are denoted by three letters, asin AùCB.

midpoint oof aa ssegment: The midpointof a segment is the point that divides thesegment into two congruent segments.

minor aarc: On circle P, if mAPB < 180°,

then the points A and B, together with the

points of the circle that lie in the interior of

mAPB form a minor arc of the circle.

Minor arcs are denoted by two letters, such

as AïB.

noncollinear: Points, segments, or rays that are notcollinear.

non-cconvex ppolygon: A polygon isnon-convex (concave) if at least oneline that contains a side of the polygoncontains a point in the interior of thepolygon.

non-ccoplanar: Not coplanar.

oblique llines: Lines are oblique if theyintersect and do not form right angles.

obtuse aangle: An obtuse angle is an angle with measuregreater than 90° and less than 180°.

obtuse ttriangle: An obtuse triangle has exactly one obtuseangle.

octagon: An octagon is a polygon witheight sides.

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parallel llines: Two lines are parallel if

they are coplanar and do not intersect.

parallel pplanes: Two planes are parallel

if they do not intersect.

parallelogram: A quadrilateral with

opposite sides parallel, and hence equal in

length.

pentagon: A pentagon is a polygon with

five sides.

perimeter oof aa ppolygon: The perimeter of a polygon is the

sum of the length of its sides.

perpendicular llines: Two lines are

perpendicular if they intersect to form a

right angle.

perpendicular lline aand pplane: A line

is perpendicular to a plane if it is

perpendicular to each line in the plane.

plane: A plane is an undefined term in geometry. In

Euclidean geometry it can be thought of as a flat surface that

extends infinitely in all directions.

point: A point is an undefined term in geometry. It can be

thought of as a dot that represents a location in a plane or in

space.

polygon: A polygon is a plane figure formed by three or more

segments called sides, such that the following are true:

1. each side intersects exactly two other sides, once at each

endpoint, and

2. no two sides with a common endpoint are collinear.

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postulate: A postulate is a statement that is accepted as truewithout proof.

proof: A proof is an organized series of statements that showthat the statement to be proved follows logically from knownfacts (given statements, postulates, and previously proventheorems).

protractor: A device used to determine themeasures of angles.

pythagorean ttriple: A set of three positiveintegers a, b, and c that satisfy the equationa2 + b2 = c2 is a pythagorean triple.

prism: A solid figure that has twobases that are parallel, congruentpolygons and with all other facesthat are parallelograms.

pyramid: A solid figure with apolygon base and whose otherfaces are triangles that share acommon vertex.

quadrilateral: A polygon with four sides. The sum of theangles is 360°.

radius oof ccircle: A radius of a circle is a segment that hasthe center as one endpoint and a point on the circle as theother endpoint.

ray: The ray AB, or [AB, consists of theinitial point A and all points on line thatlie on the same side of A as B lies.

rectangle: A rectangle is a parallelogram that has four rightangles.

rectangular pprism: A solid figurethat with two bases that are rectanglesand with all other faces that areparallelograms.

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regular ppolygon: A polygon whosesides are equal and whose angles areequal.

right pprism: A prism that has twospecial characteristics: all lateraledges are perpendicular to the basesand all lateral faces are rectangular.

rhombus: A rhombus is a parallelogram that has fourcongruent sides.

right ttriangle: A triangle with exactly one right angle.

scale ffactor: In two similar polygons or two similar solids,the scale factor is the ratio of corresponding linearmeasures.

scalene ttriangle: A scalene triangle is a triangle that has nocongruent sides.

segment: A segment AB, or [AB],consists of the endpoints A and B andall points on the line AB that liebetween A and B.

similar ppolygons: Two polygons are similar if theircorresponding angles are congruent and the lengths oftheir corresponding sides are proportional.

sine: The ratio of the length of the sideopposite an angle to the length of thehypotenuse in a right triangle.

surface aarea: The sum of all the areas of the surfaces of asolid figure.

skew llines: Two lines are skew if theydo not lie in the same plane.

space: The set of all points.

sphere: A sphere is the set of all pointsin space that are a given distance r froma point called the center. The distance ris the radius of the sphere.

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square: A square is a parallelogram thatis both a rhombus and a rectangle; thatis, it has four congruent sides and fourright angles.

straight aangle: A straight angle is anangle that measures 180°.

supplementary aangles: Two angles are supplementary ifthe sum of their measures is 180°. Each angle is asupplement of the other.

tangent: The ratio of the length of theside opposite an angle to the length of theside adjacent to the angle in a right trian-gle.

tangent tto aa ccircle: A line is tangent toa circle if it intersects the circle at exact-ly one point.

theorem: A theorem is a statement that must be proved tobe true.

transversal: A transversal is a line thatintersects two or more coplanar lines atdifferent points.

trapezoid: A quadrilateral with exactlyone pair of opposite parallel sides. Thesum of the angles is 360°.

vertex oof aa ppolygon: A vertex of a polygon is a commonendpoint of two of its sides.

vertical aangles: Twoangles are vertical if theirsides form two pairs ofopposite rays.

volume: The number of cubic units needed to occupy agiven space.

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7 Geometry