Basic Triangles

8/6/2019 Basic Triangles

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David W. Sabo (2003) Triangles Page 1 of 5

A

B C

c b

a

600

60 0 60 0

C

A

B

bc

aleg

The Mathematics 11Competency Test

Triangles

Triangles are plane geometric figures with three straight line

sides and three vertices. Typically, the three vertices arelabelled with upper case characters from the beginning of the alphabet, and the sides opposite each vertex arelabelled with the corresponding lower case letters, as shownin the sketch to the right.

Often, the symbol is used for the word triangle. Specifictriangles are identified by listing the characters labellingtheir vertices. Thus ABC is a shorthand notation to denote the triangle sketched above to theright.

Types of Triangles

There are quite a number of terms to describe triangles of variousspecial types. You should be familiar with at least the following types:

equilateral triangles : all three sides have the same length. All threevertices are the same size: 60 0. Notice how sometimes small cross-lines are used to indicate which sides of the triangle have equallengths.

isosceles triangles : two of the three sides have the same length. Thetwo angles opposite the equal-length sides have the same measure.

scalene triangle : this term is used to refer to a general triangle, having no specific relationshipsbetween the lengths of any sides. One thing we can say about the lengths of the sides of anytriangle in general is that no one side can be longer than the sum of the lengths of the other twosides.

right triangle : has one vertex which forms a rightangle. The vertex with the right angle isconventionally labelled C. The two shorter perpendicular sides are called legs of the righttriangle. The longest side, opposite the right angle,is called the hypotenuse .

A very important property of right triangles isPythagorass formula or theorem:

2 2 2c a b= +

This formula will be exploited later when we introduce trigonometry.

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A

B38 0

Properties of Triangles

For every plane triangle, the vertex angles always add up to 180 0:

0180 A B C + + =

Example: Two angles in a triangle are 63.6 0 and 42.1 0. Determine the third angle.

solution: Let x be the third angle which we are to determine. Then, since all three angles must add up to180 0, we must have that

x + 63.6 0 + 42.1 0 = 180 0.So

x = 180 0 (63.6 0 + 42.1 0) = 74.3 0.

The third angle must be 74.3 0.

Example: Determine the measure of angles A and B in thediagram to the right.

solution: The three angles inside the triangle must add up to 180 0.The angle at the lower right is indicated to be a right angle,so it must measure 90 0. Thus

38 0 + 90 0 + A = 180 0

Therefore,

A = 180 0 (38 0 + 90 0) = 52 0

Angle B, together with the 38 0 angle form a straight angle which as a measure of 180 0. Thus

B + 38 0 = 180 0 So

B = 180 0 38 0 = 142 0.

For any right triangle, C = 90 0, so A + B = 90 0 the two acute angles are complementary.

In any triangle, the longest side is always opposite the largest angle, and the shortest side isalways opposite the smallest angle.

Perimeter of a Triangle

The perimeter, P, or distance around any triangle, is simply the sum of the lengths of its sides:

P a b c = + +

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altitude(h)

base (b)

base (b)

Example: A triangular plot of land is sketched in thefigure to the right. Compute its perimeter and the cost of putting a fence around it if the fence material costs$15.95 per meter.

solution: The perimeter is just the sum of the lengths of the three sides:

P = 42.6 m + 81.3 m + 102.7 m

= 226.6 m

To get the cost of the fence, just multiply the number of meters of fence required (the perimeter of the triangle) by the cost in dollars per meter. Thus

$226.6 15.95 $3614.27Cost mm

= = .

Area of a Triangle

Imagine rotating a triangle so that one of its sides ishorizontal, as shown in the sketch to the right. We willcall this side the base of the triangle. Now, draw in avertical line from the top vertex of the triangle to thisbase. Such a line is called an altitude of the triangle.Obviously, since this altitude is vertical and this baseis horizontal, they are perpendicular. If we symbolizethe length of the base of the triangle by the letter b,and symbolize the length of the corresponding altitudeby the letter h (for height), then the area , A, of the triangle, the amount of surface it occupies, is

given by

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A bh= (one-half base times height)

You will get exactly the same answer if you use any of the three sides of the triangle as the base. In somecases, the altitude may not be inside the triangle, aswas the situation in the sketch above. Instead, it mayhave to be a vertical line from the top vertex drawn tointersect an extension of the base, and shown to theright. The base is still just the actual horizontal side of the triangle and not its extension.

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c

15 m

8 m

C B

A

Example: Calculate the perimeter and the area of the triangleshown in the sketch to the right.

solution: This triangle is a right triangle with the length of the twolegs given. To calculate the perimeter, we need the lengths of allthree sides. Fortunately, since this is a right triangle, Pythagorasstheorem applies, giving us a way to calculate the length of the thirdside (in this case, the hypotenuse).

So, letting c stand for the length of the longest side of this triangle,we have by Pythagoras that

c2 = (8 m) 2 + (15 m) 2 = 64 m 2 + 225 m 2 = 289 m 2

Therefore,

2289 17c m m= =

So,perimeter = p = sum of the lengths of the three sides

= 8 m + 15 m + 17 m = 40 m.

To calculate the area, imagine rotating the triangle so that theside of length 15 m is on the bottom, horizontal. Then the sideof length 8 m will be vertical (since it is perpendicular to the sideof length 15 m). In this orientation, the side of length 15 m formsa base of the triangle, (so b = 15 m) and the side of length 8 mforms the corresponding altitude of the triangle (h = 8 m). Thus,we get

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Area A bh= =

( ) ( ) 21 15 8 602

m m m= =

Example: With reference to the figure to the right:

(i) compute the perimeter and area of the triangle drawn withheavy solid lines, and,

(ii) determine the measures of the angles labelled A, B,and C.

solution:

To calculate the perimeter and areas, we need to dealwith lengths. So, well resketch this triangle, showing alllengths and rotating it so that the lower right hand sideis horizontal.

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12.3 mb46.8 m

To compute the perimeter, we need the lengths of all threesides. Two are already given. To get the length of thethird side, labelled b in this rotated sketch, we note that

b + 12.3 m = 46.8 m

Therefore

b = 46.8 m 12.3 m = 34.5 m.

So, now,

perimeter = p = 34.5 m + 22.4 m + 50.4 m = 107.3 m.

To calculate the area, we note that the vertical length, 18.7 m, forms an altitude (h = 18.7 m)corresponding to the base, b = 34.5 m, formed by the horizontal side in the rotated sketch. Thus

( ) ( ) 21 1 34.5 18.7 322.5752 2

Area A bh m m m= = = =

To determine the measures of the angles A, B, and C, we make use of the fact that the sum of the angles inside a triangle is always 180 0, and angles combining to form a straight angle alsosum to 180 0. Thus

B + 90 0 + 56.7 0 = 180 0 B = 33.3 0

A + 56.7 0 = 180 0 A = 123.3 0 and

C + A + 21.8 0 = 180 0 so

C = 180 0 21.8 0 A

= 1800

21.80

123.30

= 34.9 0

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Basic Triangles