Geometry Book

Geometry

Mathematics

Curriculum Framework

Revised 2004 Amended 2006

Course Title: Geometry Course/Unit Credit: 1 Course Number: Teacher Licensure: Secondary Mathematics Grades: 9-12

Geometry

This course will help students develop communication skills, enhance reasoning, and make connections within mathematics to other disciplines and the real world. Students will use physical models and appropriate technology to investigate geometric concepts in problem solving situations. In this course, students are engaged in problematic situations in which they form conjectures, determine the validity of these conjectures, and defend their conclusions to classmates. Strand Standard Language of Geometry 1. Students will develop the language of geometry including specialized vocabulary, reasoning, and application of

theorems, properties, and postulates. Triangles 2. Students will identify and describe types of triangles and their special segments. They will use logic to apply the

properties of congruence, similarity, and inequalities. The students will apply the Pythagorean Theorem and trigonometric ratios to solve problems in real world situations.

Measurement 3. Students will measure and compare, while using appropriate formulas, tools, and technology to solve problems

dealing with length, perimeter, area and volume. Relationships between two- and three- dimensions

4. Students will analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.

Coordinate Geometry and Transformations

5. Students will specify locations, apply transformations and describe relationships using coordinate geometry. * denotes amended changes to the framework

Geometry Mathematics Curriculum Framework Revision 2004 Amended 2006

Arkansas Department of Education

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Language of Geometry

Content Standard 1. Students will develop the language of geometry including specialized vocabulary, reasoning, and application of theorems, properties, and postulates.

LG.1.G.1 Define, compare and contrast inductive reasoning and deductive reasoning for making predictions based on real world

situations • venn diagrams • matrix logic • conditional statements (statement, inverse, converse, and contrapositive) • *figural patterns

LG.1.G.2 Represent points, lines, and planes pictorially with proper identification, as well as basic concepts derived from these undefined terms, such as segments, rays, and angles

LG.1.G.3 Describe relationships derived from geometric figures or figural patterns

LG.1.G.4 Apply, with and without appropriate technology, definitions, theorems, properties, and postulates related to such topics as complementary, supplementary, vertical angles, linear pairs, and angles formed by perpendicular lines

LG.1.G.5 Explore, with and without appropriate technology, the relationship between angles formed by two lines cut by a transversal to justify when lines are parallel

LG.1.G.6 Give justification for conclusions reached by deductive reasoning *State and prove key basic theorems in geometry (i.e., the Pythagorean theorem, the sum of the measures of the angles of a triangle is 180° , and the line joining the midpoints of two sides of a triangle is parallel to the third side and half it’s length

2 Geometry: Language of Geometry

Mathematics Curriculum Framework Revision 2004 Amended 2006 Arkansas Department of Education

Key: LG.1.G.1 = Language of Geometry. Standard 1. Geometry. 1st Student Learning Expectation

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Triangles

Content Standard 2. Students will identify and describe types of triangles and their special segments. They will use logic to apply the properties of congruence, similarity, and inequalities. The students will apply the Pythagorean Theorem and trigonometric ratios to solve problems in real world situations.

T.2.G.1 Apply congruence (SSS …) and similarity (AA ...) correspondences and properties of figures to find missing parts of geometric figures and provide logical justification

T.2.G.2 Investigate the measures of segments to determine the existence of triangles (triangle inequality theorem)

T.2.G.3 Identify and use the special segments of triangles (altitude, median, angle bisector, perpendicular bisector, and midsegment) to solve problems

T.2.G.4 Apply the Pythagorean Theorem and its converse in solving practical problems

T.2.G.5 Use the special right triangle relationships (30˚-60˚-90˚ and 45˚-45˚-90˚) to solve problems

T.2.G.6 Use trigonometric ratios (sine, cosine, tangent) to determine lengths of sides and measures of angles in right triangles including angles of elevation and angles of depression

T.2.G.7 *Use similarity of right triangles to express the sine, cosine, and tangent of an angle in a right triangle as a ratio of given lengths of sides

3 Geometry: Triangles


Key: T.2.G.1 = Triangles. Standard 2. Geometry. 1st Student Learning Expectation

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Measurement

Content Standard 3. Students will measure and compare, while using appropriate formulas, tools, and technology to solve problems dealing with length, perimeter, area and volume.

M.3.G.1 Calculate probabilities arising in geometric contexts (Ex. Find the probability of hitting a particular ring on a dartboard.)

M.3.G.2 Apply, using appropriate units, appropriate formulas (area, perimeter, surface area, volume) to solve application

problems involving polygons, prisms, pyramids, cones, cylinders, spheres as well as composite figures, expressing solutions in both exact and approximate forms

M.3.G.3 Relate changes in the measurement of one attribute of an object to changes in other attributes (Ex. How does changing the radius or height of a cylinder affect its surface area or volume?)

M.3.G.4 Use (given similar geometric objects) proportional reasoning to solve practical problems (including scale drawings)

M.3.G.5 *Identify and apply properties of and theorems about parallel and perpendicular lines to prove other theorems and perform basic Euclidean constructions

4 Geometry: Measurement


Key: M.3.G.1 = Measurement. Standard 3. Geometry. 1st Student Learning Expectation

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Relationships between two- and three- dimensions

Content Standard 4. Students will analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.

R.4.G.1 Explore and verify the properties of quadrilaterals

R.4.G.2 Solve problems using properties of polygons:

• sum of the measures of the interior angles of a polygon • interior and exterior angle measure of a regular polygon or irregular polygon • number of sides or angles of a polygon

R.4.G.3 Identify and explain why figures tessellate

R.4.G.4 Identify the attributes of the five Platonic Solids

R.4.G.5 Investigate and use the properties of angles (central and inscribed) arcs, chords, tangents, and secants to solve problems involving circles

R.4.G.6 Solve problems using inscribed and circumscribed figures

R.4.G.7 Use orthographic drawings ( top, front, side) and isometric drawings (corner) to represent three-dimensional objects

R.4.G.8 Draw, examine, and classify cross-sections of three-dimensional objects

R.4.G.9 *Explore non-Euclidean geometries, such as spherical geometry and identify its unique properties which result from a change in the parallel postulate

5 Geometry: Relationships between two- and three- dimensions


Key: R.4.G.1 = Relationships between two- and three- dimensions. Standard 4. Geometry. 1st Student Learning Expectation

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Coordinate Geometry and Transformations

Content Standard 5. Students will specify locations, apply transformations and describe relationships using coordinate geometry.

CGT.5.G.1 Use coordinate geometry to find the distance between two points, the midpoint of a segment, and the slopes of parallel, perpendicular, horizontal, and vertical lines

CGT.5.G.2 *Write the equation of a line parallel to a line through a given point not on the line CGT.5.G.3 *Write the equation of a line perpendicular to a line through a given point CGT.5.G.4 *Write the equation of the perpendicular bisector of a line segment CGT.5.G.5 Determine, given a set of points, the type of figure based on its properties (parallelogram, isosceles triangle, trapezoid)

CGT.5.G.6 Write, in standard form, the equation of a circle given a graph on a coordinate plane or the center and radius of a circle

CGT.5.G.7 Draw and interpret the results of transformations and successive transformations on figures in the coordinate plane

• translations • reflections • rotations (90˚, 180˚, clockwise and counterclockwise about the origin) • dilations (scale factor)

6 Geometry: Coordinate Geometry and Transformations


Key: CGT.5.G.1 = Coordinate Geometry and Transformations. Standard 5. Geometry. 1st Student Learning Expectation

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GEOMETRY Glossary

Adjacent angles Two coplanar angles that share a vertex and a side but do not overlap Alternate interior angles

Two angles that lie on opposite sides of a transversal between two lines that the transversal intersects

Altitude of a triangle A perpendicular segment from a vertex of a triangle to the line that contains the opposite side Angle Two non-collinear rays having the same vertex Angle of depression

When a point is viewed from a higher point, the angle that the person’s line of sight makes with the horizontal

Angle of elevation

When a point is viewed from a lower point, the angle that the person’s line of sight makes with the horizontal

7 Geometry Glossary


Apothem The distance from the center of a regular polygon to a side

Arc An unbroken part of a circle Area The number of square units needed to cover a two-dimensional space. Attributes A quality, property, or characteristic that describes an item or a person (Ex. color, size, etc.) Biconditional

A statement that contains the words “if and only if” (This single statement is equivalent to writing both “if p, then q” and its converse “if q then p.)”

Bisector A segment, ray or line that divides into two congruent parts Center of a circle The point equal distance from all points on the circle Central angle

An angle whose vertex is the center of a circle (Its measure is equal to the measure of its intercepted arc.)

8 Geometry Glossary


Centroid The centroid of the triangle is the point of congruency of the medians of the triangle.

Chords A segment whose endpoints lie on the circle Circle The set of all points in a plane that are an equal distance (radius) from a given point (the center) which is also in the

plane Circumcenter A circumcenter is the point of concurrency of the perpendicular bisectors of a triangle.

Circumference The distance around a circle Circumscribed

A circle is circumscribed about a polygon when each vertex of the polygon lies on the circle. (The polygon is I inscribed in the circle.)

Collinear points Points that lie on the same line Complementary angles Two angles whose measures add up to 90 degrees Concentric circles Concentric circles lie in the same plane and have the same center

9 Geometry Glossary


Conditional statements

A statement that can be written in the form “if p, then q” (Statement p is the hypothesis and statement q is the conclusion.)

Cone

A three dimensional figure with one circle base and a vertex

Congruent Having the same measure for angles and segments. Having the same measure and shape for shapes Conjecture Something believed to be true but not yet proven (an educated guess) Consecutive angles

In a polygon, two angles that share a side

Consecutive sides In a polygon, two sides that share a vertex Contrapositive The contrapositive of a conditional statement (“if p, then q” is the statement “if not q, then not p”) Converse

The converse of the conditional statement interchanges the hypothesis and conclusion (“if p, then q, becomes “if q, then p”)

Convex polygon A polygon in which no segment that connects two vertices can be drawn outside the polygon Coordinate geometry Geometry based on the coordinate system

10 Geometry Glossary


Coordinate plane A grid formed by two axes that intersect at the origin (The axes divide the plane into 4 equal quadrants.) Coplanar points Points that lie in the same plane Corollary A corollary of a theorem is a statement that can easily be proven by using the theorem. Corresponding parts

A side (or angle) of a polygon that is matched up with a side (or angle) of a congruent or similar polygon

Cosine In a right triangle, the ratio of the length of the leg adjacent to the angle to the length of the hypotenuse Cross-section A cross-section is the intersection of a solid and a plane. Cylinder

A three-dimensional figure whose bases are circles of the same size

Deductive reasoning

Using facts, definitions, and accepted properties in a logical order to reach a conclusion or to show that a conjecture is true

Dilations Transformations producing similar but not necessarily congruent figures



Exterior angle of a polygon

An angle formed when one side of the polygon is extended (The angle is adjacent to an interior angle of the polygon.)

Geometric mean If a, b, and x are positive numbers, and a/x = x/b, then x is the geometric mean of a and b. Incenter The incenter of a triangle is the point of congruency of the angle bisectors of the triangle.

Inductive reasoning A type of reasoning in which a prediction or conclusion is based on an observed pattern Inscribed angle

An angle whose vertex is on a circle and whose sides are chords of the circle



Inscribed circle A circle is inscribed in a polygon if the sides of the polygon are tangent to the circle.

Inscribed polygon A polygon is inscribed in a circle if the vertices of the polygon are on the circle.

Interior angles of a polygon The inside angle of a polygon formed by two adjacent sides Inverse statement The inverse of the conditional statement (“if p, then q” is the statement “if not p, then not q”) Irregular polygon A polygon with at least two non-congruent sides or angles Isometric drawings Drawings on isometric dot paper used to show 3-dimensional objects Isosceles triangle A triangle with at least two sides congruent Line of symmetry

The line over which a figure is reflected resulting in a figure that coincides exactly with the original figure

Linear pair of angles

Two adjacent angles form a linear pair if their non-shared rays form a straight angle.

Matrix logic Using a matrix to solve logic problems



Median of a triangle

A segment that has as its endpoints a vertex of the triangle and the midpoint of the opposite side

Midpoint of a segment The point that divides a segment into two congruent segments Midsegment

A segment whose endpoints are the midpoints of two sides of a polygon

Orthocenter The orthocenter is the point of concurrency of the altitudes of a triangle.

Orthographic drawings An orthographic drawing is the top view, front view and right side view of a three-dimensional figure. Parallel lines Lines in a plane that do not intersect Parallelogram A quadrilateral with both pairs of opposite sides parallel



Perimeter The distance around a polygon Perpendicular bisector

The perpendicular bisector of a segment is a line, segment or ray that is perpendicular to the segment at its midpoint.

Perpendicular Two lines, segments, rays, or planes that intersect to form right angles Planes A flat surface having no boundaries Platonic solid

A polyhedron all of whose faces are congruent regular polygons, with the same number of faces meeting at every vertex

Point A specific location in space Polygon

A closed plane figure whose sides are segments that intersect only at their endpoints, with each segment intersecting exactly two other segments

Postulates A mathematical statement that is accepted without proof



Prism

A three-dimensional figure with two congruent faces called bases--that lie in parallel planes (The other faces called lateral faces are rectangles that connect corresponding sides of the bases.)

Pyramid

A three-dimensional figure with one base that is a polygon (The other faces, called lateral faces, are triangles that connect each side of the base to the vertex.)

Quadrilateral A four-sided polygon Radius A line segment having one endpoint at the center of the circle and the other endpoint on the circle Reflections Mirror images of a figure (Objects stay the same shape, but their positions change through a flip.) Regular heptagon A heptagon with all sides and angles congruent Regular octagon A octagon with all sides and angles congruent Regular octagon An octagon with all sides and angles congruent Regular pentagon A pentagon with all sides and angles congruent Regular polygon A polygon with all sides and angles congruent Rotations A transformation in which every point moves along a circular path around a fixed point called the center of rotation Scale drawings Pictures that show relative sizes of real objects



Secant

A line, ray or segment that intersects a circle at two points

Similarity The property of being similar Similar polygons

Two polygons are similar if corresponding angles are congruent and the lengths of corresponding sides are in proportion.

Sine

In a right triangle, the ratio of the length of the leg opposite the angle to the length of the hypotenuse

Slope The ratio of the vertical change to the horizontal change Slope-intercept form A linear equation in the form y = mx + b, where m is the slope of the graph of the equation and b is the y intercept Special right triangles

A triangle whose angles are either 30-60-90 degrees or 45-45-90 degrees



Sphere

The set of all points in space equal distance from a given point

Standard form of a linear equation

The form of a linear equation Ax + By = C where A, B, and C are real numbers and A and B are not both zero Ex. 6x + 2y = 10

Supplementary angles Two angles whose measures add up to 180 degrees Surface area The area of a net for a three-dimensional figure Tangent In a right triangle, the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle Tangent to a circle

A line in the plane of the circle that intersects the circle in only one point

Tessellate

To cover a plane with a pattern of polygons without gaps or overlaps

Theorem A conjecture that can be proven to be true Transformation A change made to the size or position of a figure Translation A transformation that slides each point of a figure the same distance in the same direction



Transversal

A line that intersects two or more other lines in the same plane at different points

Triangle Inequality Theorem The sum of the lengths of any two sides of a triangle is greater than the lengths of the third side. Trigonometric ratios The sine, cosine and tangent ratios Venn diagram A display that pictures unions and intersections of sets Vertical angles

Non-adjacent, non-overlapping congruent angles formed by two intersecting lines (They share a common vertex.)

∠1 and ∠3 are vertical angles. ∠2 and ∠4 are vertical angles.

Volume The number of cubic units needed to fill a three-dimensional space



L. Marizza A. Bailey Arkansas School of Mathematics, Sciences and the Arts

Hot Springs, Arkansas [email protected]

Summer Workshop

Outline Content Standard 1

1. Teaching Students How to Argue. a. The language of logic. (LG.1.G.1) 1. Identifying hypotheses and conclusions 2. Identifying Conditional statements 3. Translating into logical language c. The Law of Detachment 1. Which statements are logically equivalent? 2. How to check the validity of a statement, d. How to argue and always win (LG.1.G.6)

1. Students will follow arguments to their conclusion 2. Identifying valid arguments and identifying flaws in weak arguments. 3. Constructing valid arguments. 2. What is a proof? The axiomatic approach to mathematics. a. Euclid’s Axiomatic Approach to Geometry revolutionized mathematics. (10:45 – 11:45) 1. A history 2. Definitions 3. Axioms (Common Notions) 4. Postulates b. How is a proof different than an answer or solution? c. Teaching students to conjecture and prove. d. Two-Column Proofs and Paragraphs 1. Pythagorean Theorem Discovery and Proof (Geometry Sketchpad in Boole Lab (LG1.G.6) 2. Sum of Angles of a Triangle (LG.1.G.6) 3. Triangle Midsegment Theorem

Introduction to Arguments

Learning how to argue well is essential for any occupation. We all use conditional statements when we speak of a cause, an effect, or an implication. Below is an example of an argument one might use in everyday language.

The clouds cause the rain.

The rain causes the grass to grow. The grass needs to grow to feed the cows.

This language we usually use for arguing is imprecise and can be confusing. People who do not understand logic well will be unclear of whether a cause is a necessary condition, or a sufficient condition. A necessary condition is one that is required for the effect, but not necessarily the only required ingredient. Example: Flour is a necessary ingredient for cake but not the only ingredient necessary. Therefore, flour is not sufficient for cake. A sufficient condition is one that is enough to make the effect happen. Example: If it is raining, there must be clouds in the sky to make the rain. Therefore, the knowledge of rain is sufficient to deduce that clouds are in the sky. However, it is not necessary that it rain for there to be clouds. If we can make this language precise, we not only will be able to argue more efficiently, but will be able to find the error in the arguments of others. The precise language for a conditional statement requires a hypothesis and a conclusion. It is of the form: If (hypothesis), then (conclusion.

Example: A) If there is a fire, then the fire alarm will sound. This is a true statement. However B) If the fire alarm sounds, then there is a fire. is not a true statement. Write the letter corresponding to the statement above which is logically equivalent? 1) Fire alarm means fire. _____ 2) Fire is needed for fire alarms. _____ 3) Fire is the only thing needed for fire alarms._____ Here are some more examples of conditional statements in layman’s terms. Let’s see if we can translate them into the more refined language of logic.

1. Only women have babies.

2. Joe needs at least a C on the test to pass the class.

3. All Joe need is a B on the test to pass the class.

4. Give me $10 and I’ll baby-sit for you.

5. I’ll take out the garbage when pigs fly.

6. All babies wear diapers.

7. All squares are rectangles.

8. Hard work means success.

9. All bubble blowing babies shall be beaten senseless by all patrons in

the bar. (Sponge Bob the Movie)

10. To get an ‘A’ for the day, you must finish this worksheet.

Logic Activity Title of Lesson: Translating Conditional Statements into the Language of Logic (LG.1.G.1) Activity : (15 points) 1. Assign each person in the group one of the jobs: Recorder, Timekeeper, Facilitator, and Presenter. 2. Translate each of the statements into If-Then form. (5 pts) 3. Identify hypothesis and conclusion. (5 points) 4. Assess validity of statement (5 points) 4. Present findings to class (5 pts) Purpose of Activity: This activity is designed to help you analyze conditional statements and to use the more precise language of logic. 1. There are always clouds when it’s raining outside. 2. We need to study to get good grades. 3. A good night’s sleep is needed for me to be energetic tomorrow. 4. Clouds are sufficient for rain. 5. Organization leads to success. 6. All rectangles are squares. 7. The world will be at peace in 50 years 8. Teachers will be millionaires when pigs fly. 9. Only furry monkeys go to school. 10. When I am Albert Einstein, all triangles will have three sides .

Hypothesis T/F Conclusion T/F Statement

T/F 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Hands on Conditional Statements

Lesson: Students will compare, contrast conditional statements (inverse, converse and contrapositive). (SLE LG.1.G.1) Purpose: Students will kinesthetically identify and construct the above conditional statements. By assigning a truth value to the hypothesis and conclusion, they can assess the validity of each statement. Using their previous knowledge of the Law of Detachment, they will investigate it’s implications on the truth value of the converse, contrapositive and inverse. Materials:

1. Scissors 2. Construction Paper 3. Glue or tape 4. Writing Utensil Activity:

1. Students will work in groups of two. 2. They will cut all of the copies on the hypothesis and conclusion and

the word “not”. 3. They will construct the inverse, converse, and contrapositive on the

construction paper given. Each one will be labeled and glued (or taped) to the construction paper.

4. They will identify the truth value of the hypothesis, conclusion, and statement.

5. They will write a short reflection as to their conjecture on a generalized relationship between these truth values.

Not I finish my work

I will leave early


I will leave early


I will leave early


I will leave early


I will leave early


I will leave early


I will leave early

LAW OF DETACHMENT AND COROLLARIES

L. MARIZZA A. BAILEY

Now that we are familiar with construction of conditional statements in thelanguage of logic, we can begin to identify which conditional statements arelogically equivalent. The ”If, then” statements we have been working on arecalled conditional statements , or implications. The reason why they are calledimplications, is because the hypothesis implies the conclusion. Let us look atan example.

Example 1. *If you are 30, then you are too old to party.

The hypothesis is you are 30The conclusion is you are too old to party

Another way to say this is:You are 30 implies that you are too old to party.

In symbolic logic, we try to condense the notation, so we can see in general,when two logical statements are equivalent.

Letp = You are 30.q = You are too old to party.

Then the statement, *, above is written

p =⇒ q

In the previous activity, we started to study the correlation between the truthvalue of the hypothesis, conclusion and the entire statement. In order to assessthe validity of any arbitrary logical statement, it is necessary that we create arule for this relation.

0.1. Law of Detachment. The Law of detachment says thatif the hypothesis is true,then the only way that an implication can be true,is if the conclusion is also true.A false hypothesis has no bearing on the truth value of the implication.

In symbolic logic, the above statement is as follows:

Date: June 4, 2007.

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Hyp: p Conc: q Implication: p =⇒ q

T T TT F FF T TF F T

From this table we see if the hypothesis is true, then the conclusion has to betrue, or the whole implication is false.

Example 2. Let us look at a simple mathematical statement:

If you multiply two negatives together, their product will be positive.

Suppose I give you the numbers −1 and −2.Do they satisfy the hypothesis?Is their product positive?

Suppose I give you the numbers 1 and 4.Do they satisfy the hypothesis? Is their product positive?

Suppose I give you the numbers −1 and 4.Do they satisfy the hypothesis? IS their product positive?

Now let us look at an example which fails the truth table.

Example 3. If a number is odd, then it is prime.

Suppose I gave you the number 3.Does it satisfy the hypothesis?Is it prime?



This last number is a counterexample to the statement and therefore showsthat the statement is not true for all numbers, and therefore, is not a truestatement.

0.2. Converse. Let p = It is raining.Let q = There are clouds in the sky.

We have already discussed the validity of p =⇒ q, rain does imply clouds.However, if I switch the hypothesis and conclusion, does the statement remaintrue?

3

There are clouds in the sky implies there is rain.or

q =⇒ p

is not necessarily a true statement because there have been cloudy days whereno rain has fallen.If we look at the truth value of this particular statement compared to the truthof p and q, we will see why:

Hyp: p Conc: q Implication: p =⇒ q Converse: q =⇒ p

T T T TT F F TF T T FF F T T

As you can see, in the cases where the hypothesis and conclusion have dif-ferent truth values, the original implication and it’s converse also have differenttruth values.

It is a common mistake to think that the converse of an implication is alwaystrue.

Here are some examples of implications and their converse. Write the conversefor those implications where one is not already written.

(1) If it doesn’t snow, the Saints win.

If the Saints win, it didn’t snow on Sunday.(2) If you keep the flint in one drawer and the steel in the other, you’ll never

make a fire.(The Music Man)

If you never make a fire, you kept the flint in one drawer and the steelin the other.

(3) If you lay down with your dog, you wake up with ticks.

(4) If wishes were fishes, I could start a tuna company.

(5) If I had a nickel for every time someone said ”AYP”, I would be amillionaire.

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0.3. Contrapositive. If you, first, negate the hypothesis and conclusion beforeswitching them, you get the contrapositive.

Let p = It is raining.Let q = There are clouds in the sky.

¬q =⇒ ¬pIf there are no clouds in the sky, it is not raining.

This seems to be a true statement, so let us look at the truth table to verify.Hyp: p Conc: q Implication: p =⇒ q ¬q ¬p ¬q =⇒ ¬pT T T F F TT F F T F FF T T F T TF F T T T T

As you can see the column under the implication p =⇒ q and ¬q =⇒ ¬pare the same. Therefore, the two statements are logically equivalent.

Below are the same implications, with the contrapositive written below. Writethe contrapositive for those implications where there is not one written.

(1) If it doesn’t snow, the Saints win.If the Saints don’t win, it snowed on Sunday.

(2) If you keep the flint in one drawer and the steel in the other, you’ll nevermake a fire.(The Music Man) If you make a fire, you did not keep theflint in one drawer and the steel in the other.

(3) If you lay down with your dog, you wake up with ticks.

(4) If wishes were fishes, I could start a tuna company.

(5) If I had a nickel for every time someone said ”AYP”, I would be amillionaire.

Here are some examples of statements you may see in the business world ofscams. (I got them off the internet) Translate them into formal logic.

(a) Every time We Close Your Sale, You Make Money!(B) So, If you’re like me, frustrated and tired of searching for the right op-

portunity. Sick of all the false hopes and unkept promises. Well look nofurther.

(C) IF YOU ARE ONE OF THE 95% WHO HATE SELLING... THENTHIS IS THE PERFECT BUSINESS FOR YOU!

(D) When you refer friends and family, you can earn bonus cash when theyare approved.

(E) With a MoneyGram location just around the corner from you, returningcustomers now have another option from MyCashNow to receive theirloan. Have your loan sent to you in minutes!

Arkansas School of Mathematics, Sciences and the ArtsE-mail address: [email protected]

LAW OF SYLLOGISM


0.1. Law of Syllogism. Now we know the in’s and out’s of logic. Let’s see ifwe can make a good argument. We need to make a string of implications, sothat they tie all together to make a conclusion.

In the book, they describe it like this:

p =⇒ q

andq =⇒ r

thenp =⇒ r

Let us look at an example

Example 1. If I am over 30, then I am too old to party.If I am too old to party, then I am boring.

If I assume both statements to be true, then I can conclude thatIf I am over 30, then I am boring is a true statement.

Now using the Law of Syllogism, we can conclude that since Mrs. Bailey isover 30, then she is boring.

Here are some more examples, that illustrate the use of the Law of Syllogism.

Example 2. If we find his fingerprints on the counter, then he was at the mur-der scene.If he was at the murder scene, he is our prime suspect.

From these two sentences I can conclude that:

We found his finger prints on the counter. What can you conclude?We found someone else’s finger prints on the counter. What can you conclude?We didn’t find his fingerprints? What can you conclude?

Here is a trickier one:

If he is fails the lie detector test, then the lawyer will not defend him.If he hears that his mother is in the hospital, he will fail his lie detector test.

What can you conclude from the sentence above?

Date: June 4, 2007.

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Another tricky one:

If we go to the store, we will spend too much money.If we don’t spend too much money, we will go on vacation.

I went to the store. What can you conclude from this?

Below are some famous logical fallicies. See if you can find the error in theargument. This is for fun.

Example 3. Proof that 1 = −1.

−1 = −1 By reflexivity of equality1−1

=−11

Converting into fractions√

1−1

=

√−11

Taking square root of both sides√

1√−1=√−1√

1Distributing square roots

√1√

1 =√−1

√−1 Clearing Fractions1 = −1 Square root is inverse of square

Example 4. Proof that 2 = 1.

Let x = y.Then

2x = x+ y

2x− 2y = x− y2(x− y) = x− y

2 =x− yx− y2 = 1


Geometry Quiz

Law of Syllogism Study the following statements. If you are a geometry student, you must take an End of Course Geometry exam. If you are a geometry student, you must learn how to prove. If you learn how to prove, you will win every argument.

1. If Susie has to take an End of Course Geometry Exam, what can you conclude about her?

2. If Leo is a geometry student, what can you conclude about him?

3. If Dana wins every argument, what can you conclude about her?

4. If Paul learns how to prove, what can you conclude about him?

The Process of ProofThe Process of Proof

A Heuristic ApproachA Heuristic Approach

Triangle Sum TheoremTriangle Sum TheoremTriangle Sum TheoremTriangle Sum Theorem

Identify HypothesisIdentify HypothesisIdentify Hypothesis.Identify Hypothesis.Identify Conclusion.Identify Conclusion.Fi d d fi iti f ll th ti l tFi d d fi iti f ll th ti l tFind definitions of all mathematical terms Find definitions of all mathematical terms associated with theorem.associated with theorem.Construct any other necessary figures.Construct any other necessary figures.

The mazeThe mazeA figure is a triangle

A

C

A B

The sum of the m∠ABC +m∠BCA + m∠CAB = 180°

The sum of the interior angles is 180º

The mazeThe maze The Fifth Postulate

A figure is a triangleA

C

m

n D

A B

AC // m and AB // nConsecutive Interior Angles

Alternate Exterior Interior Angles Theorem

Angles Theorem

Alternate Interior Angles Theorem

Corresponding Angles Theorem



The mazeThe maze The Fifth Postulate


C

m

n D

A B

AC // m and AB // nAngle Sum Postulate

m∠ACB + m ∠BCD = m ∠ACD

m∠ACD + m ∠CAB = 180 °

Consecutive Interior Angles Theorem

m∠ACB + m ∠BCD + m ∠CAB =180 °



The mazeThe maze C

The Fifth Postulate


C

m

n D

A B

AC // m and AB // nAngle Sum Postulate

m∠ACB + m ∠BCD = m ∠ACD

m∠ACD + m ∠CAB = 180 °

Consecutive Interior Angles Theorem

m∠ACB + m ∠BCD + m ∠CAB =180 °

The sum of the

m∠BCD = m∠ABC Alternate Interior Angles Thm

The sum of the interior angles is 180ºm∠ACB + m∠ABC + m∠CAB = 180°

Proof Activity

Attached to this sheet are three conditional statements. Identify the

hypothesis and write it next to the word “Given”. Write the conclusion next

to the word “Prove”. For each proof you will be given a set of statements

and a set of reasons. Find the given statement(s) and glue them on the first

row. The conclusion must be glued to the last row under the statement

column. Place them in order on the labeled spaces so that the Law of

Syllogism applies. If you have done the activity correctly you should have a

coherent proof in the end. Most of the time, you will need to draw a picture

to see what is going on.

Here is a sample:

Conditional Statement: If ABD EBC, then EBD ABC.

Given: ABD EBC

Prove: EBD ABC

Statement Reason

ABD EBC

Given

ABC + CBD ABD

Angle Addition Postulate

EBD + DBC EBC


CBD DBC

Reflexive property of congruence

CBD ABD - ABC

CBD EBC - EBD


ABD - ABC EBC - EBD

Transitive Property of Congruence

EBC - ABC EBC - EBD Substitution

ABC EBD


A

B

D

E

C

Assume X,Y and Z are collinear and Q,R, and S are collinear.

Conditional Statement: If XZ QS and XY RS, then YZ RS.

Given:

Prove:

Statement Reason

If AB and CD intersect at point O, then AOD BOC.

Statement Reason

Proof 1

XY + YZ = QR + RS Substitution

XY RS GIVEN

XZ QS GIVEN

YZ QR Segment Subtraction Postulate

XY + YZ = XZ and QR + RS = QS Segment Addition Postulate

XY + YZ = QR + XY Substitution from Given

Proof 2

m AOD + m DOD = 180 AOD and DOB are linear pairs

m DOB + m BOC = 180 DOB and BOC are linear pairs

m BOD = 180 - m BOC Subtraction property of equality

m BOD = 180 - m AOD

180 - m BOC = 180 - m AOD Transitive property of equality

m BOC = m AOD Subtraction Property of Equality

BOC AOD Definition of Angle Congruence

PARALLEL LINES


1. Definitions

Definition 1 (Corresponding Angles). Two angles are called corresponding an-gles if they lie on the same side of two lines and a transversal.Ex. ∠1 and ∠5 are corresponding angles

Definition 2 (Alternate Interior Angles). Two angles are called alternate in-terior angles if they lie in between two lines lines and on alternate sides of thetransversal.Ex. ∠4 and ∠6 are alternate interior angles

Definition 3 (Alternate Exterior Angles). Two angles are called alternate ex-terior angles if they lie outside two lines lines and on alternate sides of thetransversal.Ex. ∠1 and ∠7 are alternate exterior angles

Definition 4 (Consecutive Angles). Two angles are called consecutive anglesif they are on the interior of the parallel lines, and on the same side of thetransversalEx. ∠4 and ∠5 are consecutive angles

2. Theorems about parallel lines

Theorem 1 (Corresponding Angle Theorem). If the two lines are parallel, thentheir corresponding angles are congruent.

Theorem 2 (Alternate Interior Angle Theorem). If the two lines are parallel,then their alternate interior angles are congruent.

Theorem 3 (Alternate Exterior Angle Theorem). If the two lines are parallel,then their alternate exterior angles are congruent.

1

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2

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3

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2

Theorem 4 (Consecutive Angle Theorem). If the two lines are parallel, thentheir consecutive angles are supplementary.

3. Theorems to prove lines are parallel

Theorem 5 (Corresponding Angle Converse). If the two corresponding anglesare congruent, then their lines are parallel.

Theorem 6 (Alternate Interior Angle Theorem). If the two lines alternate in-terior angles are congruent, then their are parallel.

Theorem 7 (Alternate Exterior Angle Theorem). If the two lines alternateexterior angles are congruent , then their are parallel.

Theorem 8 (Consecutive Angle Theorem). If the two consecutive angles aresupplementary , then their lines are parallel.

4. Proofs

Problem 1. Show that if l1 is parallel to l2, then ∠1 is supplementary to ∠8

Proof:Statement Reasons1. ∠1 ∼= ∠5 1.2. 2. Linear Pairs are Supplementary3. ∠1 is supplementary to ∠8 3.

Problem 2. Show that if ∠2 supplementary ∠5, then l1 is parallel to l2

Proof:Statement Reasons1. m∠2 + ∠5 = 180o 1.2. 2. Definition of Linear Pair3. 3. Substitution4. 4. Corresponding Angle Converse.

Can you find another way to prove the claim? Find one other proof for eachof the Problems above.The less the number of lines one requires to prove a claim, the more elegant theproof.

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A History of Geometry Marizza Bailey

Euclid is one of the most revered mathematicians of all time, and sometimes known as the Father of Mathematics. His greatest work was the creation of Euclid’s Elements, an immense collection of 13 books and 465 propositions (Journey Through Genius, William Dunham). His legacy is not the creation of new mathematics in these books, but rather, his thorough and organized attempt to present mathematics in a logical fashion. His attempt succeeded and revolutionized mathematics around the world. He began the Elements with 23 definitions, 5 postulates and 5 axioms (and a partridge in a pear tree). These were givens of mathematics; the framework for the mathematical model given by these assumptions gave way to Euclidean Geometry.

The beauty and significance of this approach was that he could only prove the first propositions with those axioms, postulates and definitions that were already given. Once he had proved a proposition, he could use it in his next proof, thereby building the mathematics on a solid but growing foundation. This eliminated circular reasoning and assigning to each proposition an “unambiguous string of predecessor leading back to the original axioms. (Journey Through Genius, William Dunham) It is the fifth postulate, known as the parallel postulate that has been the object of much discussion throughout the centuries. A generalization of this postulate has led to the creation of a new geometry known as Non-Euclidean geometry. The fathers of this new field are Lobachevski (1829) and Bolyai (1832), who founded Hyperbolic and Elliptical Geometry; Euclidean geometry is now known as parabolic geometry.

Proofs are not Answers

We have discovered the ease in which someone can fall prey to a fallacious argument, and hopefully this has motivated you to want to prove all claims that are made. It is important, then, for use to distinguish when an answer with justification is necessary, and when a proof is necessary. The following activity is designed to motivate students to identify the difference between an answer and a proof. Activity 1: Add the following sequence of number pairs: 1+1 = 1+2 = 1+3 = 1+4 = 2+2= 2+3= 2+4= 2+5= 3+3= 3+4= 3+5= 3+6= From the answers above, what can you conjecture about the sum of two odd numbers? Two even numbers? One odd number and one even number? Odd + Odd = Even + Even = Even + Odd = Now Prove Your Conjecture: Let a,b be odd numbers. Then a = 2n+1 and b = 2m+1, by definition of odd. By substitution, a + b = 2n + 1 + 2m + 1 = 2n + 2m + 2 = 2( n + m + 1) is even. Now construct a similar proof for the other cases.

Activity 2: I used this activity in my Algebra 2 and Pre-Calculus course the first week of class to instill a healthy skepticism in my students. It can be used in any course, and will probably have the same effect. Students are already familiar with the concept of probability. If |E| is the number of times the even E can occur, and |O| is the number of possible outcomes, then the probability of an event E occurring is

|E| = P(E) |O| We would like to find the probability of rolling a 2, 3, 4, 5, 6, 7, ..., 12 on a pair of dice. In order to do this, we must count the number of possible ways to roll each value. Values of Each

Dice Sum of values Number of Occurrences (1,1) 2 1

(1,2), (2,1) 3 2 (1,3), (2,2), (3,1) 4 3

5 4 6 5

As you can see this is very exhausting, and we don’t want to compute this for every value. However, a pattern seems to be emerging on the very right column. As the sum of the values increases, the number of occurrences is increasing. Make a conjecture as to what the number of occurrences is for a given value of the pair of die. We can use the calculator to emulate this situation by asking the calculator to generate 30 random integers from 1 to 6 for L1 and L2.

Calculator Procedure

2nd Stat: Edit Cursor up to L1 until it is highlighted Enter

Math Prob 5: randomint(1,6,30) Enter

Repeat this process for L2. Highligh L3. L3 should be the sum of the values of L1 and L2. To do this, type 2nd 1 + 2nd 2 enter

To analyze the outcome, create a STAT PLOT by going to 2nd Y = A bar graph on the list L3 would be the most helpful

The best viewing window would be

X∈[1,13] and Y ∈[0,10] because the value of the sum of the two die will be between 1 and 12, and the frequency should not be more than 10 if you have only 30 trials.

Then number of occurrences divided by the number of trials is called the experimental probability. The experimental probability of 8 is 6/30 = 1/5. We can repeat this simulation with 100 trials to get a more accurate analysis.

Note that the experimental probability of 8 is still the highest at 1/5. Does this graph support your conjecture? Did you conjecture that the higher values would have the highest probability? Why do you think that conjecture was erroneous? Let us finish the chart and see we were mistaken.

Dice Sum of values Number of Occurrences (1,1) 2 1

(1,2), (2,1) 3 2 (1,3), (2,2), (3,1) 4 3

(1,4), (2,3), (3,2) , (4,1) 5 4 (1,5), (2,4), (3,3), (4,2), (5,1) 6 5

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) 7 6 (2,6), (3,5), (4,4), (5,3), (6,2) 8 5

(3,6), (4,5),(5,4), (6,3) 9 4 (4,6), (5,5), (6,4) 10 3

(5,6), (6,5) 11 2 (6,6) 12 1

The experimental probability of 8 is 1/5, but the theoretical probability is 5/36. The highest outcome should be that of a 7. Why do you think this wasn’t the case with the simulation? Hopefully, students will be more careful to generalize their answers to an all encompassing umbrella statements without further investigation. Note also that even analyzing 100 simulated trials can lead us to an incorrect conclusion. Mathematical truths can only be achieved through proof. Once we have constructed a proof, it is a TRUTH and cannot be made untrue. That is the beauty of mathematics.

Pythagorean Theorem Discovery Approach

We can use what is already known about Cartesian plane to correlate geometry and algebra.

If we look at random points in the Cartesian plane, and then measure their distance to the origin, we might notice a pattern. We will use Geometry Sketchpad to construct some arbitrary points on the Cartesian plane, and measure their distance to the origin.

Step 1: Graph: Define Coordinate System

Step 2: Click the point button on the left. (under the arrow) Construct two points and a line between them. Step 3: Construct points on this line. Repeat this process to your satisfaction. Measure: Coordinates Drag coordinates to corresponding labeled point.

Step 4: Click the line segment button (under the circle). Click the origin and then one of your constructed points. Do this repeatedly until you have exhausted all constructed points. You can now enter all the data in the calculator in a table. 2nd Stat (List) Edit L1 x-coordinate L2 distance to origin L3 (L2)2

Distance to origin Points on the line x=1.01

m HN = 2.06 cm

m HM = 1.42 cm

m HL = 2.25 cm

m HK = 3.18 cm

m HJ = 5.10 cm

m HI = 3.15 cm

I: (1.01, -2.99)N: (1.01, -1.80)

O: (1.00, 0.00)

M: (1.01, 1.01)

L: (1.01, 2.01)

K: (1.01, 3.02)J: (1.01, 5.00)

Stat Plot: Plot 1: x L1 , yL3 Quadratic Regression: L1, L3, Y1

It seems very close to the function d2 = (1.01)2 + x2 Remember, y = 1.01 so we get . d2 = y2 + x2

This is not a proof, merely a conjecture!

A Proof of the Pythagorean Theorem. Here is one proof of the Pythagorean Theorem that is based on an algebraic approach to the problem. Since most students have taken Algebra before Geometry, this might make them more comfortable. Have them cut out each of the figures below. They will see how the areas are related by superimposing the squares on each other.

a

b

c

Solution: (a + b)2 = 2ab + c2

a2 + 2ab + b2 = 2ab + c2 a2 + b2 = c2

Note to teacher:

These are the hypotheses from which you may read aloud. You may change any of them.

You may construct a power point for each statement, transparencies, or just read them aloud.

Your choice should depend on the ability of the students.

There may be many conclusions implied by one statement.

Please make a list of all the possible conclusions implied by each statement below so that you may have a key.

Procedure:

1. Break up your students in pairs. 2. Every time you read a statement, have the each student write it down. 3. They should list all statements they feel can be concluded from the hypothesis, and justify

their response. 4. After discussion amongst the pair, they can place the markers on corresponding squares. 5. The first pair to get Bingo wins. 6. Have them turn in their work for credit.

This method will be time consuming, but more educational. They will be forced to come up with some justification for their response, rather than guessing.

Statements:

1. If angle A and angle B are linear pairs and angle A is obtuse, then what can you say about the third angle ….

2. If line AB is perpendicular to lineCD and ACB are collinear, then angle ACD …

3. If two lines, AB and CD , intersect at point P, then APC∠ and BPD∠ …

4. If two lines intersect so the linear pairs are congruent, then …

5. If a line bisects line segment AB at point M to form segments con AM MB , then ..

6. If the sum of two angles of a triangle is less than 90 degree, thenthe third angle …

7. If two coplanar lines are not parallel, then

8. If two lines have the same slope, but different y‐intercepts, then …

9. If four points are the vertices of a tetrahedron, then …

10. If one line represents the intersection of the ceiling and a wall and another line represents the intersection of the adjacent wall and the floor, then …

11. If two lines are parallel and angle A and angle B form consecutive interior angles, then …

12. If line l is the graph of the equation y = 2x + 3 and line m is the graph of the equation 4y ‐ 8x = 12, then …

Logical Conclusion Bingo

Match each hypothesis the teacher reads aloud with a statement from it which can be concluded.

The angles are congruent

The segments are congruent

The angle is a right angle

The angle is obtuse

The angle is acute

The angles are supplementary

The lines are perpendicular

The lines are parallel

The lines are skew

The point is the midpoint of the segment

The line segments form a triangle

The line segments do not form a triangle

The three points are collinear

The four points are coplanar

The Corresponding Angles are Congruent

The Consecutive angles are congruent

The Consecutive Angles are Supplementary

The Alternate Interior Angles are congruent


The Angles are Vertical Angles

The lines intersect at exactly one point

The lines are the same

The endpoints are equidistant to the point P

The three points are not collinear

The four points are not coplanar




The angle is acute



The angle is obtuse















The lines are skew









The angle is acute



The angle is obtuse





The lines are skew


















The angle is acute




The angle is obtuse





The lines are skew


















The angle is acute




The angle is obtuse





The lines are skew


















The angle is acute




The angle is obtuse





The lines are skew
















Employee

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The conclusion must not be stated in the hypothesis.



The angle is acute




The angle is obtuse










The lines are skew














The angle is acute



The angle is obtuse










The lines are skew














The angle is acute



The angle is obtuse















The lines are skew









The angle is obtuse

The angle is acute














The lines are skew












The angle is obtuse

The angle is acute



















The lines are skew







The angle is obtuse

The angle is acute




The lines are skew























The angle is obtuse

The angle is acute




The lines are skew




















Summer Workshop

Outline

Content Standard 2 1. Triangle Theorems a. Congruent Triangles (T.2.G.1) 1. Proving the congruence theorems 2. Counterexamples of AAA and SSA 3. Developing Lemmas from Congruence Theorems. 4. Using Congruence Theorems to Prove lemmas b. Special Segments and associated theorems (T.2.G.3) 1. Proofs for isosceles triangles and intersection 2. Corollaries to properties 3. Geometry Sketchpad c. Similar Figures (AAA Thm) (T.2.G.1) 2. Trigonometric Ratios (Spaghetti Trig) (T.2.G.7) 2. Real World Applications a. Special Segments and their Applications (T.2.G.2) b. Right Triangle Geometry Applications (T.2.G.5) c. Similar Figures Applications (T.G.2.6) d. Pythagorean Theorem and it’s applications( T.2.G.4)

Congruence of Triangles

The fifth postulate states:

If a straight line crossing two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if extended indefinitely, meet on that side on which are the angles less than the two right angles.

Since any two non parallel lines intersect at exactly one point, the third vertex is uniquely determined. If you consider, this further, you will see that if you are given a line segment and two angles whose vertices are the endpoints of the line segment, they will be able to form a triangle if and only if those angles add up to less than 180° (two right angles). By the fifth postulate, the intersection of those two lines must be on this side of the transversal. Since there is only one possible intersection point, the triangle is uniquely determined. It is important to understand the difference between

1. Axioms: A starting assumption from which other statements are logically derived

2. Definitions: Assigns properties to a mathematical object

3. Postulates: The more modern word for axioms

4. Propositions, Theorems, Lemmas, Corollaries: A statement that can be proved

logically using previously proved statements:

5. Conjectures, Hypothesis: Claims yet to be proven

The following are propositions: SSS, SAS, ASA, AAS. We will work in geometry

sketchpad to illustrate the clarity of these propositions in Euclidean Space to the

students.

Definition: Two triangles are congruent if all their corresponding sides and angles

are congruent.

SAS Theorem:

Rather than show that two triangles are congruent if they have two sides which

are congruent and the angle between them congruent, we will show that given two

segments of fixed length and a fixed angle between them, there is only one triangle

which can be generated from these parts in the corresponding order.

This is a corollary from the first postulate which states that given any two points,

there exists a line between them. The third postulate states that given any point and

distance, there exists a circle with that center and radius equal to that distance.

Construct a circle with center A and radius AB. Using these two postulates, one can

conclude that given any two points, there exists a line segment with whose endpoints

are those two points. This uniquely determines the triangle ∆ABC.

A

B

C

SSS Theorem

Suppose DE = AB, EF = BC and DF =AC. Show the two triangles are congruent.

Without changing the geometry of these objects we can assume they are joined at

AB and EF.

Statement Reason

1. ABCD is a parallelogram 1. Opposite Sides Congruent

Converse

2. AB // FD and ED // AC 2. Definition of a parallelogram

3. ∠ DEF ≅ ∠ ACB 3. Alternate Interior Angle Theorem

∠ DFE ≅ ∠ABC

4. m∠A + m∠ B + m∠C 4. Triangle Sum Theorem and

= m∠D + m∠E + m∠ F Transitivity

5. m∠A = m∠D 5. Right Cancellation Law

A

D

E

B C

F

D A

E = B

F = C

AAS Theorem:

Suppose ∠A ≅ ∠D

∠C ≅ ∠E

AC = DE.

Show ABC DEFΔ ≅ Δ .

Proof:

Statement Reason

1. m∠A + m∠B + m∠C = 180º 1.

2. 2. Triangle Sum Theorem

3. m A m B m C m D m E m F∠ + ∠ + ∠ = ∠ + ∠ + ∠ 3.

4 4 Definition of Congruence

5. m∠B = m∠F 5.

6. 6. ASA Proposition

A

D

E

B C

F

A

May 22 2007.GWB - 1/14 - Tue May 22 2007 18:16:35

Marizza Bailey

Text Box

In order to animate the point along the path of the arc, it is necessary to define the point as "a point on the circle", rather than construct the circle relative to the point.

Marizza Bailey

Text Box

In this case, the AB < AC sin(C), as you can see in the highlighted text above, therefore, there cannot exist a triangle with the given sides and angle

May 22 2007.GWB - 2/14 - Tue May 22 2007 18:18:47

Marizza Bailey

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As you can see, CD is equal to AC sin(A) which means there is exactly one triangle with given sides and angle.

May 22 2007.GWB - 3/14 - Tue May 22 2007 18:20:18

Marizza Bailey

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I bet you can guess which case this is...

May 22 2007.GWB - 6/14 - Tue May 22 2007 18:23:23

Marizza Bailey

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Does the sin(A) have anything to do with this? Is it just the length of AC? Explain.

May 22 2007.GWB - 7/14 - Tue May 22 2007 18:23:35

May 22 2007.GWB - 8/14 - Tue May 22 2007 18:23:51

May 22 2007.GWB - 13/14 - Tue May 22 2007 18:25:19

May 22 2007.GWB - 14/14 - Tue May 22 2007 18:25:30

Angle Bisector Angle Bisector of a segment AB Definition: AD is the angle bisector of ∠BAC if m∠BAD = m∠CAD Lemma: Every point on the angle bisector of ∠BAC is equidistant to sides BA and AC.

Let C be a point on the line MC. We can construct a triangle, „ABC. The distance from D to AB is the length of the segment perpendicular to AB through D. A similar definition can be made for a distance from D to AC. Intuitively; we define it this way because it is the shortest distance from D to the line AC. Proof:

Statement Reason 1. m ∠ DBA = m∠DCA 1. Definition of perpendicular

2. m ∠CAD = m ∠BAD 2.

3. 3. Reflexive property of Congruence

4. ∆ADC ≅ ∆ADB 4.

5. DB = DC 5.

A

B

C

D

Employee

Polygon

Theorem: The intersection of all angle bisectors of a triangle is the center of the circumscribed circle. The point D is the intersection of the angle bisectors of ∠ABC, ∠ACB and∠ BAC. By the lemma above, the point is equidistant to AC, CB, and AB. If we assume that three points uniquely determine a circle, then D must be the center of the circle tangent to the points AB, BC and AC. This is called the inscribed circle. We can use Cabri Jr to illustrate. The following Activities will help the students use Cabri to verify the above lemmas and theorems. If Cabri Jr. is unavailable, or you would prefer hands on constructions, the following activity will help students “see” the incenter.

A C

B

D

Perpendicular Bisector Perpendicular Bisector of a segment AB Definition: Let M be the midpoint of AB. The perpendicular bisector of AB is the line through M perpendicular to AB.

1) Passes through the midpoint of AB 2) Perpendicular to AB Lemma : Every point on the perpendicular bisector of AB is equidistant to A and B Let C be a point on the line MC. We can construct a triangle, ∆ABC. Proof:

Statement Reason 1. AM = MB 1.

2. m ∠ΑΜC = m ∠BMC 2.

3. 3. Reflexive property of Congruence

4. ∆AMC ≅ ∆BMC 4.

5. 5. Corresponding Parts of Congruent

Triangles are Congruent (CPCTC)

A

B

C

M

Theorem: The intersection of all perpendicular bisectors of a triangle is the center of the circumscribed circle. The point D is the intersection of the perpendicular bisectors of AB, AC and BC. By the lemma above, the point is equidistant to A, C, and B. If we assume that three points uniquely determine a circle, then D must be the center of the circle through points A, B and C. This is called the circumscribed circle. We can use Cabri Jr to illustrate. The following Activities will help the students use Cabri to verify the above lemmas and theorems. If Cabri Jr. is unavailable, or you would prefer hands on constructions, the following activity will help students “see” the circumcenter.

A C

B

D

Special Segments Verbal QuizSpecial Segments Verbal Quiz

Medians, Angle bisectors, Perpendicual Bisectorsp

and Altitudes

Median

• Passes through midpointA

• Connects to opposite corner

BC

2

Perpendicular Bisector


• Is perpendicular to side.

CB

3

Which one is it



• Median

BC

4

Angle Bisector

• Cuts angle in halfA

BC

5

Altit deAltitudeA

• Is perpendicular• Connects to opposite pp

corner• Height of a vertex

BC

6

Which one is it

• Cuts angle in half• Angle bisectorg

7

Which one is it

• Passes through midpoint

• Is perpendicular• Perpendicular Bisector

8

Which one is it

• Passes through midpoint


• Median

9

Which one is it

• Perpendicular• Connects to opposite pp

corner• Altitude

10

Which one is the Median

• Passes through midpointC


• LBL

MA B

11

Which one is the perpendicularWhich one is the perpendicular bisector

• Passes through midpointC

• Is perpendicular• CM

L

MA B

12

Which one is the Altitude

• Connects to opposite cornerC

• Is perpendicular• LB, CM

L

MA B

13

Identify the following segment• Is it the perpendicular• Is it the perpendicular

bisector? Why or why not.

C

• Is it the altitude? Why or why noty

• Is it the median? Why or why not.

L

60º 60º• Is it the angle

bisector? Why or why MA B

60º 60º

14

not.

Geometry Test FORM A

Special Segments of a Triangle

Name:____________________ For each of the following pictures, identify whether the segment, is a

A. perpendicular bisector B. angle bisector C. median D. altitude There may be more than one answer to each triangle. You may write the letter only, rather than the entire word

1

2

3

5 5

2

3

1

4

5

6

7

8

40° 40°

4

5

4

4

7

8

42° 42°

6

9

10

11) Which of the following represents a angle bisector of the triangle?

10

9

12) Which of the following represents the median of the triangle below?

13) Sketch the perpendicular bisector of AB.

14) Sketch the altitude of CA.

OPEN RESPONSE Use the information given in the triangle below to prove the statements below.

a) Name the theorem that gives ABM is congruent to CBM. b) Use part a) to show that ∠ ABM is congruent to ∠ CBM.

45° 45°

Proofs Worksheet Vocabulary Properties

Angle Bisector

Perpendicular Bisector

Median

Altitude

Midsegment

1. Let ∆ABC ≅ ∆DBC. Suppose m∠ACB = 90º. Prove BC is the perpendicular bisector of ∆ADB.

Statement Proof 1. ∆ABC ≅ ∆DBC 1. m∠ ACB = 90º 2. 2. Definition of Linear

Pair 3. 3. 4. 4. CPCTC 5. BC is the perpendicular bisector 5. of AB

A C

D

B

2) Let AB be the perpendicular bisector of ∆CDE. Show AE AC≅ . Statement Reason 1. 1. Definition of segment Bisector 2. 2. Def. of Perpendicular

3. 3.

4. 4. SAS Postulate 5. 5. CPCTC

D

E B

C

A

3) Let AD be the angle bisector of ∠CAB Show ∆CAD ≅ ∆DAB if and only if ∆ABC is isosceles. Write a

paragraph proof. By definition of isosceles, then __________________________. Furthermore, ∠CAD ≅ ∠BAD, by ____________________________ ____________________________________________. By the reflexive property of congruence, ___________________________. Therefore, ____________________________by SAS Theorem. Now suppose ∆CAD ≅ ∆DAB. Then ∆ABC is isosceles, by _______ __________________________.

D

B A

C

4) Let P be the center of the circle below. Show AP is the median of

∆ABC. 5) Show that the altitude, median and perpendicular bisector are all the same

segment of and isosceles triangle. a) Altitude if and only if Angle Bisector b) Angle bisector if and only if median c) Median if and only if Perpendicular bisector

A

C

B

P

Geometry Bisectors and stuff

Perpendicular Bisector of a segment AB

1) Passes through the midpoint of AB 2) Perpendicular to AB 3) Every point on it is equidistant to A and B 4) The intersection of all perpendicular bisectors of a triangle is the center of the

circumscribed circle. (around) 5) The intersection of all the perpendicular bisectors of a triangle is the center of the

circle which touches each vertex exactly once.

Angle bisector of a vertex

1) Splits the angle in two 2) Every point on the angle bisector is equidistant to adjacent sides 3) The intersection of all angle bisectors of a triangle is the center of the inscribed

circle (inside). 4) The intersection of all angle bisectors of a triangle is the center of the circle inside

the triangle which touches each side exactly once. Median of a segment AB with opposite vertex, C.

1) The median passes through the midpoint of AB 2) The median passes through the vertex , C, opposite of AB 3) The intersection of all the medians, M, of a triangle is called the centroid 4) The intersection of all the medians, M, of a triangle is the center of gravity of a

triangle with equally distributed mass. 5) The distance from the C to M is two thirds the length of the median.

Altitude of a segment AB with opposite vertex, C.

1) The altitude is perpendicular to AB 2) The altitude passes throught the vertex, C, opposite AB 3) The altitude measure the “height” of the triangle.

Midsegment of a segment AB with opposite vertex C.

1) Passes through the midpoint of AB 2) Passes through the midpoint of the side, AC, opposite AB. 3) Is parallel to the third side, BC. 4) Is half the length of the third side, BC.

Problems

1) Let ABC be an equilateral triangle. Show that the perpendicular bisector is also

the angle bisector. (Draw a picture) a) Is it also the altitude? Explain b) Is it also the median? Explain.

2) A cellular company want to build a satellite receptor station which will service

Stephens, Camden and McNeil. Stephens is 25 miles away from Camden and 15 miles away from McNeil. McNeil is about 35 miles away from Camden. The cellular company want to build it in a spot equidistant to all three cities. How should it go about doing it? Explain and draw a diagram to illustrate your theory.

3) You are a creative artist and want to create a unique mobile which does not hang,

but balances instead. You have a flat steel sheet in the shape of a traingle that you want perfectly balanced on a steel pin. The dimensions of the traingle are 3 meters, 4 meters, and 5 meters. How would you find the point on which you need to place the pin? Draw a diagram.

4) You are an engineer that is asked to build a new park. You need to build roads to

the park that will allow access to Stephens , McNeil, El Dorado, and Magnolia. The Transportation Department tells you you can only build 3 roads. Highways 82, 57, and 79 all intersect to form a triangle. You are told the park has to be same distance to all three highways. What do you do? Explain your answer and draw a diagram.

May 22 2007.GWB - 9/14 - Tue May 22 2007 18:24:09

May 22 2007.GWB - 10/14 - Tue May 22 2007 18:24:26

May 22 2007.GWB - 11/14 - Tue May 22 2007 18:24:46

May 22 2007.GWB - 12/14 - Tue May 22 2007 18:25:03

Dilation and Similar Triangles

The triangles above are similar. They have three congruent angles. We would like to discuss the slope of the hypotenuses without having to relate them to the x and y coordinates. Instead, we would like to relate the slope with the angle the hypotenuse makes with the positive x‐axis.

Call the angle the hypotenuses make with the positive x‐axis, θ. Then, the slope of the hypotenuse will be tan θ . We know this is well‐defined because the triangles are similar, and, hence, the ratio of y to x is the same.

Now, let us define

cos θ x‐coordinate of the point on the unit circle θ degrees from the positive x‐axis.

sin θ y‐coordinate of the point on the unit circle θ degrees from the positive x‐axis.

The following activity will help us see this better.

5

4

3

2

1

-1

-2

-3

-4

-5

-6 -4 -2 2 4 6

θ

Spaghetti Sine and Cosine Wave.

Steps: 1. Plot the points which correspond to the degree listed in the table below. 2. Find the x‐coordinate using triangles and fill in the table 3. Find the y‐coordinate using triangles and fill in the table 4. Cut a linguine the same length as the x‐coordinate. a. Place it on the cosine wave on the corresponding angle. b. Remember to take into account it’s direction (positive or negative) c. Glue it down. 5. Cut a linguine the same length as the y‐coordinate. a. Place it on the sine wave on the corresponding angle. b. Remember to take into account it’s direction (positive or negative) c. Glue it down. 6. Repeat this process until you have finished all of the angles.

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

-1.5 -1 -0.5 0.5 1 1.5

Degree x‐coordinate y‐coordinate

0 30 45 60 90 120 135 150 180 210 225 240 270 300 330 345

What is the pattern?

Do you see a relationship between the x and y coordinates?

Trigonometry Quiz 1 No Decimals !!

1)

Find y. sin(25°) cos(25°) tan(25°) cot(25°) sec(25°) csc(25°)

25°

5

5

2

y

2) A regular hexagon is inscribed in a circle with center, C.

a) Find the radius of the circle. b) Find the perimeter of the regular hexagon. c) Find the area of the regular hexagon.

C

5

3)

Suppose sin(α) = .25. What is cos(β)?

β

α

Name___________________________

Right Triangle Trig

1. An observer standing 32 feet from a point directly below a hot air balloon determines that the angle of elevation to the balloon is 24°. a. How high is the balloon at this time? b. If the angle of elevation changes to 36°, find the change in the

ACTUAL distance (not the distance along the ground) between the balloon and the observer.

2. The pilot of a plane traveling at a constant altitude of 24,000 feet spots his destination at an angle of depression of 12°. Four minutes later, the angle of depression to his destination is 53°. Find the speed of the plane in miles/hour.

PART III Released Open-Response Items — Geometry

35

RELEASED MATERIALS. MAY BE DUPLICATED.

GEOMETRY OPEN-RESPONSE ITEM F

3.4 feet 1.5 feet

wing span

1 foot

(Not drawn to scale.)

F. Kyle is building a model airplane as shown in the diagram above. The wings are congruent.

1. Determine the length of the wing span of the model airplane. Show or explain all of your workeven if you use mental math or a calculator. Include units in your answer. (Round to the nearesthundredth.)

2. Kyle wants to store his airplane in a box. The base of the box is 312feet wide by 61

2feet long.

Determine whether or not his airplane will fit in the box. Show or explain all of your work even

if you use mental math or a calculator. Include units in your answer.

BE SURE TO LABEL YOUR RESPONSES (1) AND (2).

0303--01160



Euclid

Archimedes

Summer Workshop


1. Polygons and Polyhedra a. Develop formula for area (M.3.G.2) 1. Discovering formula for area of parallelogram 2. Discovering formula for area of trapezoid 3. Discovering formula for area of rhombus b. Develop formula for area of circle 1. Geometric Probability Activity(M.3.G.1) 2. Data Analysis to CONJECTURE area of circle 3. Double reducto ad abdsurdium c. Develop formulas for volume and surface area (M.3.G.2) 1. Prisms 2. Cylinders and Spheres 3. Cones, Pyramids and Tetrahedrons. Oh my! 2. Euclidean Construction (M.3.G.5) a. Midpoint b. Parallel through point c. Perpendicular line through point d. Parallelogram e. Angle Bisector f. Reflection over a line l g. Angle Bisector 3. Using Cabri Jr. to construct Euclidean constructions

Discovering Area Activity

Purpose: The purpose of this activity is for the students to discover the formula for the area of various polygons SLE Materials:

1. Construction 2. Scissors 3. Tape or glue

Cut out the shapes below and make the parallelogram ABCD into rectangle BEFC. What is the relationship between the area of the two quadrilaterals? What is the area of triangle ABE and triangle DCF?

A

B C

D E F

Finish the proof that triangle ABE and DFC are congruent. 1. BE is the altitude of parallelogram ABCD 1. Definition of Altitude 2. 2. Definition of Altitude 3. BE ≅ CF 3. 4. AB ≅ DC 4. Corresponding Sides Thm 5. ∠ AEB ≅ ∠ DFC 5. 6. ∆AEB ≅∆ DFC 6. How does this prove that the areas of the quadrilaterals are congruent? Can you generalize this to give a formula for the area of a parallelogram with altitude, h, and base length, b?

Area of a Rhombus

Rhombus To find the area of the rhombus, we add the area of both triangles. Since both are congruent we need to multiply the area of one triangle by 2 Area of Triangle ABC = Area of rhombus ABCD=

2 in

3 in

A

B C

D

Make a conjecture of a formula for the area of a rhombus: Can you prove this conjecture? Find another method to partition the rhombus which would yield the same formula.

Discovering Area Activity

Purpose: The purpose of this activity is for the students to discover the formula for the area of various polygons SLE Materials:

1. Construction 2. Scissors 3. Tape or glue

Cut out the shapes below and make the trapezoid ABCD into one rectangle. What theorem proves that ∠ EBC and ∠ DCB are right angles?

A

B C

D E F

b1

b2

h

What is the area of rectangle BCDE in terms of b1 and h? AR = What is the area of ∆ ABE in terms of h and AE? AT1 = What is the area of ∆DCF in terms of h and DF? AT2 = What is the area of the trapezoid in terms of AR, AT1 and AT2? Note that DF + b1 + AE = b2. Use this fact to find a formula for the area of a trapezoid in terms of b1 + b2. Is there another way to partition the trapezoid to yield this formula?

OPEN RESPONSE ITEM GEOMETRY

Let

Let EF is the median of Isosceles Trapezoid ABCD.

Let MG be the perpendicular bisector of EF.

A. If AB = 12 and EF = 8, find DC.

B. If EM = 5, find MG.

C. Find the area of the trapezoid.

A B

D C

M

E F G

Geometric ProbabilityA Discovery Approach

A carnival game consists of throwing plastic balls into a square pool. If the balls land in the center island, you win a prize. A diagram of the game is illustrated below.

1 ft 6 ft

What is the probability of winning a prize? To answer this question, let's model this problem in the calculator and use data analysis to makea conjecture. A conjecture in this case will not lead us to a theorem, but a useful definition whichwill model reality as accurately as possible.

First we must construct some equations or inequalities that model the figure above. In order tofind algebraic functions whose graphs fit the above picture we must choose a coordinate system

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

Points that land in the pool will be in the set of points P = {(x,y) | -3 < x < 3 and -3 < y < 3 }Points that land in the island will be in the set of points I = {(x,y) | -.5 < x < .5 and -.5 < y < .5}

Create a list of 100 random integers between -3 and 3. Open L1| L2 icon. Double Click L1.Enter "6rand(100) - 3". This will generate 100 random numbers between 0 and 1. Therefore"6x+1" will translate and dilate so that we have 100 random numbers between -3 and 3.Repeat this process in L2.In order to count how many of the points land on the island, use the greatest integer functiontranslated to the right by 1/2.

L3 = int ( L1 + .5 ) L4 = int ( L2 + .5) will take all the values in the list, and 1/2 and truncatethe decimal portion. This will make all value between -.5 and .5 go to 0. So we need to count thenumber of 0's.

Let L5 = abs(L3) + abs(L4) be the sum of the absolute values of L3 and L4. This will count howmany points land in the island because both have to be zero, in order for the point to land in thesquare

1.23431 -2.0087 1 -2 3

2.72753 -2.6923 3 -3 6

0.44067 -1.2654 0 -1 1

-0.3063 -1.6888 0 -2 2

0.99979 -1.9309 1 -2 3

-2.446 1.51015 -2 2 4

-0.3029 0.2225 0 0 0

2.7866 -2.1359 3 -2 5

-1.6848 -2.7685 -2 -3 5

-0.5463 -1.9494 -1 -2 3

1.19586 1.62314 1 2 3

0.48945 -2.0612 0 -2 2

(0., 3.)

(1., 9.)

(2., 22.)

(3., 34.)

(4., 19.)

(5., 11.)

(6., 2.)

Which of the numbers represent the number of points which land on the island? 3 points satisfy both properties. Therefore, 3 out of 100 of the balls land on the island, or 3%If we look at the area of the island, 1 square foot. The area of the pool is 36 square feet.

136

= .027778

This is about 3%. It makes sense to model geometric probability of event, E, occuring as

P⎛⎝E⎞⎠ = Area⋅of ⋅Event⋅Occuring

Area⋅of ⋅all⋅Possible⋅Outcomes

Finding the Area of a Circle

Suppose the pool is a square with side length 2 yard and the area that will win a small prize is acircle of radius 1 yard. By the definition above,

P⎛⎝E⎞⎠ = Area⋅of ⋅Circle

Area⋅of ⋅square

Therefore, Area⋅of ⋅Circle⋅ = 4 P⎛⎜⎜⎝E⎞⎟⎟⎠ Once again, we generate 100 random real numbers between 0 and 1. To get them to be in thesquare, we must stretch them by a factor of 2 and translate down 1 unit.

.

-0.0667686462 0.601496935 0.605191387 0

0.2835254669 -0.71217525 0.766537851 0

-0.0159583092 0.385823131 0.386153021 0

-0.995085001 -0.03444695 0.995681049 0

0.440568924 -0.18997264 0.479781805 0

-0.5765093565 0.983042836 1.139621102 1

-0.5299412012 -0.42210555 0.677503338 0

0.2498081923 -0.0491395 0.254595411 0

-0.169580102 -0.9401387 0.955310516 0

0.01912176609 0.776006579 0.776242136 0

-0.8397688866 0.516974807 0.986141336 0

-0.2249773741 -0.49569345 0.544359082 0

To check if they are in the unit circle let L3 = (L1^{2} + L2^{2})^{.5} and L4 = int(L3)Then if the length of the point is less than 1, the integral part will be zero.

Next, we will make a histogram of the information in List 4 to count how many balls landed inthe circle.

1 2

55110165220275330385440495550605660715770825880

(0., 764.)

P(E) = .85 which implies the area of a circle is given by .764⋅⎛⎝4⎞⎠ = 3.056Therefore, the area of a circle is close to 3.056.

If we do this several times we should get that the average is approaching 3.14159.

Repeat this process with squares that have various lengths and inscribed circles.

Did you find a pattern?

Finding the Area of a Circle

Suppose the pool is a square with side length 2 yards and the area that you must hit to win asmall prize is a circle of radius 1 yard. By the definition above

-1 1

-1

1

P⎛⎝E⎞⎠ = Area⋅of ⋅Circle

Area⋅of ⋅Square

Therefore, Area⋅of ⋅Circle = 4⋅P⋅⎛⎝E⎞⎠Once again, we generate 100 random real numbers between 0 and 1.

0.283616066 0.4612243176 0.5414480067 0

0.1300481558 0.7779903412 0.788784821 0

0.9437412024 -0.2904186249 0.9874160393 0

0.4931080341 0.02450764179 0.4937166777 0

0.3988467455 -0.2919644117 0.4942893324 0

0.5472661257 -0.4665930271 0.7191726255 0

0.4044573307 0.4967113733 0.6405528243 0

-0.8321408033 0.05377113819 0.8338762809 0

-0.2867639065 0.8924378157 0.9373786817 0

0.9134509563 0.2656451464 0.9512938523 0

-0.1912560463 0.005419850349 0.1913328253 0

-0.5026929379 -0.2440924644 0.5588213677 0

-0.9852180481 -0.8306866884 1.288679548 1

-0.5433630943 -0.3951821327 0.6718722872 0

0.4481358528 0.04461431503 0.4503511737 0

L1 and L2 = 2*rand(100)-1 L3 = ((L2)^2 + (L1)^2)^(.5) L4= int(L3)

Why did we define List 1 and List 2 in this way? Recall that rand(100) generates 100 random realnumbers between 0 and 1.

What is the geometric interpretation of the value in List 2?

Why did we take the integer part of List 2?

What is the geometric representation of all the points whose value is 0 in List 3?

What is the geometric representation of all the points whose value is 1 in List 3?

Make a Histogram of your data?

Which of the values represent the experimental geometric probability of landing in the circle.

Take the average experimental geometric probability of the class. What is this value? Estimatethe area of the circle?

Repeat this experiment with a circle of radius 3 inscribed in a square? What is the ratio of thisarea to the area of the unit circle?

Do you see a pattern?Make a conjecture.

HISTORY OF MATHEMATICSMATHEMATICAL TOPIC V

ARCHIMEDES ON CIRCLES AND SPHERES

PAUL L. BAILEY

Abstract. Disclaimer: some sections of this document were lifted fromthe internet, but I no longer remember which ones.

1. Precursors of Archimedes

1.1. Pythagorean Irrational Numbers. The Pythagoreans (ca. 500 B.C.)proved the existence of irrational numbers in the form of “incommensurablequantities”. This tore at the fabric of their world view, based on the supremacyof whole numbers, and it is legend that the demonstrator of irrational numberswas thrown overboard at sea.

1.2. Zeno’s Paradoxes. Zeno (ca. 450 B.C.) developed his famous “paradoxesof motion”.

1.2.1. The Dichotomy. The first paradox asserts the non-existence of motion onthe grounds that which is in locomotion must arrive at the half-way stage beforeit arrives at the goal.

1.2.2. Achilles and the Tortoise. The second paradox asserts that it is impossiblefor Achilles to overtake the tortoise when pursuing it, for he must first reach apoint where the tortoise had been, but the tortoise had in the meantime movedforward.

1.2.3. The Arrow. The third paradox is that the flying arrow is at rest, whichresult follows from the assumption that time is composed of moments.

1.2.4. The Stadium. The fourth paradox concerns bodies which move alongsidebodies in the stadium from opposite directions, from which it follows, accordingto Zeno, that half the time is equal to its double.

1.3. Eudoxus Method of Exhaustion. Eudoxus (ca. 370 B.C.) is remem-bered for two major mathematical contributions: the Theory of Proportion,which filled the gaps in the Pythagorean theories created by the existence ofincommensurable quantities, and the Method of Exhaustion, which dealt withZeno’s Paradoxes. This method is based on the proposition: If from any magni-tude there be subtracted a part not less than its half, from the remainder anotherpart not less than its half, and so on, there will at length remain a magnitudeless than any preassigned magnitude of the same kind.

Date: June 5, 2007.

1

2

Archimedes credits Eudoxus with applying this method to find that the vol-ume of “any cone is on third part of the cylinder which has the same base withthe cone and equal height.”

1.4. Euclid’s Elements. Euclid of Alexandria (ca. 300 B.C.) wrote The Ele-ments, which may be the second most published book in history (after the Bible).The work consists of thirteen books, summarizing much of the basic mathematicsof the time, spanning plane and solid geometry, number theory, and irrationalnumbers.

2. Results from Euclid

Result 1. The circumferences of two circles are to each other as their diameters.

Using modern notation, this says that if we are given two circles with diame-ters D1 and D2, and circumferences C1 and C2, then

C1

C2=D1

D2, whence

C1

D1=C2

D2.

From this, one may conclude that for any given circle, the ratio between thecircumference and the diameter is a constant:

C

D= p, so C = pD.

We shall call p the circumference constant.

Result 2. The areas of two circles are to each other as the squares of theirdiameters.

That is, if A1 and A2 represent the area of circles with diameters D1 and D2,then

A1

A2=D2

1

D22

, whenceA1

D21

=A2

D22

,

which says that there is an area constant for any circle:A

D2= k, so A = kD2.

However, Euclid doesn’t mention, and possibly doesn’t realize, that p and k arerelated.

Result 3. The volumes of two spheres are to each other as the cubes of theirdiameters.

Thus if V1 and V2 are the volumes of spheres of diameter D1 and D2, then

V1

V2=D3

1

D32

, whenceV1

D31

=V2

D32

;

again, one sees that, for again given sphere, there is a volume constant m suchthat

V

D3= m, so V = mD3.

Note that in each of these three cases (circumference, area, volume), theoriginal statements by Euclid compare like units (e.g. length is to length as areais to area), whereas the modern tendency is to compare aspects of the sameobject (e.g. area is to length squared).

3

3. Measurement of a Circle

Proposition 1. The area of any circle is equal to a right-angled triangle inwhich one of the sides about the right angle is equal to the radius, and the otherto the circumference, of the circle.

Let be C be the circumference, r the radius, and A the area of the circle. LetT be the area of a right triangle with legs of length r and C. Then T = 1

2rC.Archimedes claims that A = T , so A = 1

2rC.

Lemma 1. Let h be the apothem and let Q be the perimeter of a regular polygon.Then the area of the polygon is

P =12hQ.

Proof. Suppose the polygon has n sides, each of length b. Clearly Q = nb. Thenthe area is subdivided into n triangles of base b and height h, so

P = n(12hb) =

12hQ.

�Lemma 2. Consider a circle of area A and let ε > 0. Then there exists aninscribed polygon with area P1 and a circumscribed polygon with area P2 suchthat

A− ε < P1 < A < P2 < A+ ε.

Proof. Archimedes simply says: “Inscribe a square, then bisect the arcs, thenbisect (if necessary) the halves and so on, until the sides of the inscribed polygonwhose angular points are the points of the division subtend segments whose sumis less than the excess of the area of the circle over the triangle.” �Proof of Proposition. By double reductio ad absurdum.

Suppose that A > T . Then A − T > 0, so there exists an inscribed regularpolygon with area P such that A − P < A − T . Thus P > T . If Q is theperimeter and h the apothem of the polygon, we have

P =12hQ <

12rC = T,

a contradiction.On the other hand, suppose that A < T . Then T − A > 0, so there exists

a circumscribed polygon with area P such that P − A < T − A. Thus P < T .However, if Q is the perimeter and h the apothem of the polygon, we have

P =12hQ >

12rC = T,

a contradiction.Therefore, as Archimedes writes, “since then the area of the circle is neither

greater nor less than [the area of the triangle], it is equal to it.” �

4

Proposition 2. The ratio of the circumference of any circle to its diameter isless the 31

7 but greater than 3 1071 .

Proof. Inscribe a hexagon. Compute the area:

π =C

D>Q

D=

6r2r

= 3.

Archimedes next doubles the number of vertices to obtain a regular do-decagon. The computation of its area requires accurate extraction of

√3, which

Archimedes estimates as(

1.732026 ≈)265

153<√

3 <1351780

(≈ 1.732051

),

which is impressively close. The Archimedes continues with 24, 48, and finally96 sides, at each stage extracting more sophisticated square roots.

Next circumscribe a hexagon and continue to 96 sides. �

In decimal notation, my calculator says that

31071

=22371≈ 3.14085 < π ≈ 3.14159 < 3

17

=227≈ 3.14286.

4. On the Sphere and the Cylinder

The two volume work entitled On the Sphere and the Cylinder is Archimedesundisputed masterpiece, probably regarded by Archimedes himself as the apexof his career. These two volumes are constructed in a manner similar to Euclid’sElements, in that it proceeds from basic definitions and assumptions, throughsimpler known results, onto the new discoveries of Archimedes.

Among the results in this work are the following. This first describes thesurface area of a sphere in terms of the area of a circle, thus comparing area toarea.

Proposition 3. The surface of any sphere is equal to four times the greatestcircle in it.

Technique of Proof. Double reductio ad absurdum: assumption that the area ismore leads to a contradiction, as does assumption that the area is less. One needsto understand the area of a cone to accomplish these estimates (why?). �

Let us translate this into modern notation. Let r be the radius of the sphereand let S be its surface area. Then the radius of the greatest circle in it is πr2.Thus Archimedes shows that

S = 4πr2.

The next proposition describes the volume of a sphere in terms of the volumeof a cone.

Proposition 4. Any sphere is equal to four times the cone which has its baseequal to the greatest circle in the sphere and its height equal to the radius of thesphere.

Note that again, Archimedes has expressed the volume of the sphere in termsof the volume of a known solid; this is because the Greeks did not have modernalgebraic notation. Using modern notation, we let r be the radius and let V be

5

the volume of the sphere. The volume of the cone of radius r and height r, asdetermined by Eudoxus, is 1

3πr3. Thus

V =43πr3.

6

In this way, Archimedes found the relationship between the circumferenceconstant p, the area constant k (in Measurement of a Circle), and the volumeconstant m: We have

C = pD, A = kD2, , and V = mD3,

and Archimedes has shown (in modern notation) that

C = πD (that is, p = π)

A = πr2 = π(D

2

)2

=π

4D (so k =

π

4)

V =43πr3 =

43π(D

2

)3

=π

6πD3 (so m =

π

6)

From here, Archimedes now describes an astounding discovery.Suppose we have a sphere of radius r, surface area S, and volume V . Inscribe

this sphere in a right circular cylinder, whose radius would also be r and whoseheight would be 2r. Then the surface area Acyl of the cylinder is simply theareas of the base and top circle, plus the area of the rectangle which forms thetube of the cylinder:

Acyl = 2(πr2) + (2πr)(2r) = 6πr2.

ThusAcyl : Asph = (6πr2) : (4πr2) = 3 : 2.

Moreover, the volume of the cylinder is the area of the circular base times theheight:

Vcyl = (πr2)(2r) = 2πr3.

Again, we have

Vcyl : Vsph = (2πr3) : (43πr3) = 3 : 2.

This so intrigued Archimedes that he requested that his tombstone be en-graved with a sphere inscribed in a cylinder, together with the ratio 3 : 2. Appar-ently, Marcellus, the conqueror of Syracuse, was so impressed with Archimedes,that he granted this wish.

Department of Mathematics and CSci, Southern Arkansas UniversityE-mail address: [email protected]

May 23 2007.GWB - 1/4 - Wed May 23 2007 14:57:17

May 23 2007.GWB - 2/4 - Wed May 23 2007 14:57:40

May 23 2007.GWB - 3/4 - Wed May 23 2007 14:58:06

May 23 2007.GWB - 4/4 - Wed May 23 2007 14:58:30

Surface Area

This activity is rather simple, yet informative for the student. Using nets for various polyhedra, they can formulate the surface area as the sum of the areas of the mutually disjoint polygons that make up the net. For example, below is the net for a cylinder. Find the area of each of the shapes that make the net. Cut out the shape below and make the cylinder. The surface area of a solid is the area of the surface enclosing it. What is the surface area of the cylinder? Repeat this process for the nets you are given in class. Can you devise anymore formulas? If so, write them below along with your reasons for believing this.

Surface Area of a Sphere

Author: Laura Jewell Grades: 6-12

Here is a concrete way to show students why the surface area of a sphere is derived from the formula 4 pi x r2.

Students should already understand that the surface area of an object can be represented by how much wrapping paper it would take to cover it. Ask them to then picture a sphere (a balloon or ball) and a piece of paper that is cut as wide as its diameter and as long as its circumference.

If you wrap the ball with the paper, you see that it would cover the entire sphere if it weren't for all the overlaps (which would fit into the gaps if you cut them out).

The formula for the surface area of the paper is:

length x width = circumference x diameter This is easily understood by looking at the picture above. Now substitute formulas we know:

C = 2 pi x r and d = 2 r

C x d = 4 pi x r2 = surface area

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Construction of Perpendicular Bisector (Whistler Alley Mathematics)

http://whistleralley.com/math.htm

Construct the perpendicular bisector of a line segment.

Or, construct the midpoint of a line segment.

1. Begin with line segment XY.

2. Place the compass at point X. Adjust the compass radius so that it is more than (1/2)XY. Draw two arcs as shown here.

3. Without changing the compass radius, place the compass on point Y. Draw two arcs intersecting the previously drawn arcs. Label the intersection points A and B.

4. Using the straightedge, draw line AB. Label the intersection point M. Point M is the midpoint of line segment XY, and line AB is perpendicular to line segment XY.

Employee

Text Box

For each of the following steps, explain why the procedure works and what it is accomplishing.

Given point P on line k, construct a line through P, perpendicular to k.

1. Begin with line k, containing point P.

2. Place the compass on point P. Using an arbitrary radius, draw arcs intersecting line k at two points. Label the intersection points X and Y.

3. Place the compass at point X. Adjust the compass radius so that it is more than (1/2)XY. Draw an arc as shown here.

4. Without changing the compass radius, place the compass on point Y. Draw an arc intersecting the previously drawn arc. Label the intersection point A.

5. Use the straightedge to draw line AP. Line AP is perpendicular to line k.

Given point R, not on line k, construct a line through R, perpendicular to k.

1. Begin with point line k and point R, not on the line.

2. Place the compass on point R. Using an arbitrary radius, draw arcs intersecting line k at two points. Label the intersection points X and Y.

3. Place the compass at point X. Adjust the compass radius so that it is more than (1/2)XY. Draw an arc as shown here.

4. Without changing the compass radius, place the compass on point Y. Draw an arc intersecting the previously drawn arc. Label the intersection point B.

5. Use the straightedge to draw line RB. Line RB is perpendicular to line k.

Construct the bisector of an angle.

1. Let point P be the vertex of the angle. Place the compass on point P and draw an arc across both sides of the angle. Label the intersection points Q and R.

2. Place the compass on point Q and draw an arc across the interior of the angle.

3. Without changing the radius of the compass, place it on point R and draw an arc intersecting the one drawn in the previous step. Label the intersection point W.

4. Using the straightedge, draw ray PW. This is the bisector of ∠QPR.

Given a line and a point, construct a line through the point, parallel to the given line.

1. Begin with point P and line k.

2. Draw an arbitrary line through point P, intersecting line k. Call the intersection point Q. Now the task is to construct an angle with vertex P, congruent to the angle of intersection.

3. Center the compass at point Q and draw an arc intersecting both lines. Without changing the radius of the compass, center it at point P and draw another arc.

4. Set the compass radius to the distance between the two intersection points of the first arc. Now center the compass at the point where the second arc intersects line PQ. Mark the arc intersection point R.

5. Line PR is parallel to line k.

10-15 Minute Interactive Geometry Lessons Using Cabri Jr on the TI-84

3T International Conference Chicago, IL

March 9-11, 2007

Presenters:

Leigh Heilemeier Dakota Jr/Sr High School

Dakota, IL [email protected]

Catherine Stevens Le-Win High School

Lena, IL [email protected]

Geometry: Parallel Lines and Angles

Using your graphing calculator, you will discover some of the properties of angles that are created by parallel lines that are cut by a transversal. Open the file “prllines” in Cabri Jr. Your screen should resemble the one below. Using the picture label the angles as alternate interior, same-side interior, corresponding and alternate exterior appropriately.

Alternate Interior Angles: Same-Side Interior Angles: Corresponding Angles:

Alternate Exterior Angles:

1. Use the measure tool, and measure two pairs of alternate interior angles. What conjecture can you make about alternate interior angles? 2. Use the measure tool, and measure two pairs of same-side interior angles. What conjecture can you make about alternate interior angles? 3. Use the measure tool, and measure two pairs of corresponding angles. What conjecture can you make about corresponding angles? 4. Use the measure tool, and measure two pairs of alternate exterior angles. What conjecture can you make about alternate exterior angles?

Geometry: Sum of Interior/Exterior Angles of Polygons Partner 1: You will be using the graphing calculator and the CabriJr application to complete this activity. Follow the directions below.

1. Begin by pressing the APPS key, and then select CabriJr. 2. You will be opening several different files, all listed in the table below, and using the

measure tool measure all interior angles and exterior angles of the polygon. 3. Record the sum of both the interior and exterior angles in the table below.

Shape File Name Number of

Sides Sum of Interior

Angles Sum of Exterior

Angles

Triangle Triangle 3

Quadrilateral Quad1

Pentagon Pent1

Hexagon Hex1

Heptagon Hept1

Octagon Oct1

Based on your findings, what conjecture can you make about the sum of the exterior angles of any polygon? Based on your findings, what conjecture can you make about the sum of the interior angles of any polygon? Collaborate with your partner. Together can you create a “formula” for finding the sum of the interior angles of any polgyon? Partner 2:

Below are five different polygons. One polygon at a time, pick one vertex and draw all possible diagonals. Knowing that a triangle’s interior angles have a sum of 180 degrees, determine the sum of the interior angles for each polygon. Quadrilateral Pentagon Sum of Interior Angles = _______ Sum of Interior Angles = _______ Hexagon Heptagon Sum of Interior Angles = _______ Sum of Interior Angles = _______ Octagon Sum of Interior Angles = _______ Based on your findings, what conjecture can you make about the sum of the interior angles of any polygon? Collaborate with your partner. Together can you create a “formula” for finding the sum of the interior angles of any polgyon?

Transformations Compositions of Reflections

Reflection of a figure over intersecting lines. Steps

1. Draw two intersecting lines. F2:Line. 2. Label the point of intersection O. F5:Alph-Num. 3. Measure ∠O. F5: Measure →Angle. 4. Draw a triangle above the two lines. F2:Triangle. 5. Reflect the triangle over the first line. F4:Reflect 6. Reflect the image of the triangle over the second line. F4:Reflect.

Make a sketch of your screen Q1. Is the third triangle a reflection of the first? _______________________________ If not, what transformation would map the first triangle onto the third? ____________________________________ Q2: What is the center of rotation? ________________________________________ Investigation What is the degree of rotation that will map the first triangle onto the third? Using the calculator draw segments (F2:Segments) from O to a vertex of the first triangle and from O to the corresponding image point of the third triangle. Measure the angle form. Q3: How does the angle of the rotated figures compare to the measure of the angle formed by the intersecting lines. ____________________________________

Reflection of a figure over Parallel lines. Due of the complexity in constructing the parallel lines and finding the distance between them. This activity may be best done as a demo. Steps

1. Draw two Parallel lines. F2:Line 2. Measure distance between parallel lines. F5: Measure →D & Length 3. Draw a triangle on the left side of the first line. F2:Triangle 4. Reflect the triangle over the first line. F4:Reflection 5. Reflect the image of the triangle over the second line. F4:Reflection

Make a sketch of your screen. Q1. Is the third triangle a reflection of the first? _______________________________ If not, what transformation would map the first triangle onto the third? ____________________________________ Investigation Using the calculator draw a segment (F2:Segment) from a vertex of the first triangle to the corresponding image point of the third triangle.

Measure the length of the segment. F5:Measure →D & Length Q3: How does the length of the segment from first triangle to the third triangle compare to the distance between the parallel lines? ________________________________

Centers of a Triangle

Medians of a triangle Q1: What is the definition of a median in a triangle?

________________________________________________________________

Using the TI-84 and Cabri App, construct a triangle and its medians, use the steps below. Steps

1. Start with a new sketch F1: New.

2. Make a triangle, F2: Triangle.

3. Find the midpoint of each side of the triangle, F3: Midpoint.

4. Draw a segment from a vertex to the midpoint of the opposite side,

creating the median, F2: Segment.

5. Draw in the other two medians.

6. Put a point at the intersection of the medians, F2: Point, Intersection.

This point of concurrency is called the centroid.

7. Make a sketch of your screen.

Q3: Does the centroid ever go outside of the triangle? _________________________

Investigation

1. Measure the length from the vertex to the centroid, F5: Measure, D. & Length

_________

2. Measure the length of the median. F5: Measure, D. & Length _________

Q4: Make a conjecture about the length from the vertex to the centroid

and the length of the entire median.

________________________________________________________________

3. Use F5: Calculate to check your conjecture.

Centers of a Triangle (part 2)

Perpendicular Bisectors of a triangle

Q1: What is the definition of a perpendicular bisector of a triangle?

_______________________________________________________________

Using the TI-84, Cabri Apps, use the steps below to construct the perpendicular bisectors of a triangle. Steps

1. Start with a new sketch F1: New.

2. Make a triangle, F2: Triangle.

3. Construct the perpendicular bisector of one side of the triangle. F3: Perp. Bis.

4. Repeat this step for the other two sides.

5. Put a point at the intersection of the perpendicular bisectors, F2: Point,

Intersection. This point of concurrency is called the circumcenter.

6. Make a sketch of the screen.

Q2: Does the circumcenter ever go outside of the triangle? ___________________

If so, when? ______________________________________________________

Q3: When is the circumcenter on the triangle? ______________________________ Investigation Draw a circle with center, C, and one of the vertices of the triangle. F2: Circle. Q4: What do you observe about the circle and the triangle?

_______________________________________________________________

Q5: What can you say about the circumcenter and its relationship to the vertices of the

triangle? _____________________________________________________

Central & Inscribed Angles

Using a graphing calculator you will investigate two types of angles within a circle. Using your books definitions, identify the central angle and the inscribed angle in the circle below. The central angle is The inscribed angle is Using Cabri Jr, create a drawing of a circle similar to the one above. Use the CIRCLE command, and then use POINT ON to place points A, C and D on the circle. By selecting the SEGEMENT tool, create an inscribed and central angle within your circle using points A, C and D. Measure each angle and record the measurements here: Central Angle __________ Inscribed Angle __________ Drag the sides of both angles to create a new central and inscribed angles. Record the new measurements here: Central Angle __________ Inscribed Angle __________ Create on more new set of angles as you did above, and record the new measurements here: Central Angle __________ Inscribed Angle __________ By looking at your angle measurements, what conjecture can you make about an Inscribed angle?

Midsegments of Triangles The midsegment of a triangle is a segment that joins two sides of a triangle through the midpoints. See the diagram to the left. Complete this investigation by following the steps below: * Using Cabri Jr, create a new file and draw a TRIANGLE from the construct menu. * Select F3, to construct the midpoint on two sides of the triangle. Then select the SEGMENT tool and construct a segment between the two midpoints. * From the measure menu, F5, select D&LENGTH from the MEASURE submenu and measure the length of the midsegment. Using the same tools, measure the length of the base of the triangle. * Select the measure menu again, this time use SLOPE from the MEASURE submenu. Determine the slope of both the midsegment and the base of the triangle. Do you see any relationships between the lengths and between the slopes? Grab a vertex of one of the base angles, and drag the triangle around. How are the measures changing? Do they still follow the same relationship? Use what you’ve discovered to write a conjecture about the midsegment of a triangle.

Midsegment

Activity 1. Use Geometry Sketchpad to construct a square of area 2. Explain your construction and give a procedure. 2. Use Geometry sketchpad to construct a square of area 3. . Explain your construction and give a procedure.

3. Use Geometry Sketchpad to construct a line of length 2

51+ . . Explain your

construction and give a procedure. \

Multiple Choiceandand

Constructed Response

Writing an EOC comparableWriting an EOC comparableQuestion or Problem.

Multiple ChoiceMultiple ChoiceAligned with Arkansas FrameworksM i l tMeasures single conceptClear and conciseG d i t lGrade appropriate languageWell-defined answerPlausible Distractors in numerical or alphabeticalPlausible Distractors in numerical or alphabetical orderCorrect (grammatically and mathematically)Correct (grammatically and mathematically)Avoids negatives, determiners, All/None of aboveRepeat words from stem

Open ResponseOpen Response

Is aligned with the Arkansas FrameworksIs aligned with the Arkansas FrameworksRequires higher level cognitive skillsIs written clearly and conciselyIs written clearly and conciselyIs appropriate difficulty levelH t th t l t t h thHas parts that relate to each otherIs relevant to the life of a typical student*Is complex enough to support a 4-scoring rubric

Example Multiple ChoiceExample Multiple ChoiceB C

DA

Which of the following could be used to justify ?∠DAB + ∠ ABC = 180 °

a. Alternate Exterior Angles Theorem

b. Alternate Interior Angles Theorem

c. Consecutive Interior Angle Theorem

d. Corresponding Angles Theorem

Joe drew three noncollinear points on a ppiece of paper. He would like to construct the circle through these points. Which of g pthe following should he construct to find the center of the circle?the center of the circle?

A. The Angle Bisectors

B. The Altitudes

C The MediansC.The Medians

D.The Perpendicular Bisectors

xº

BC

AD

B bb M d i l d H th id

E F

Bobby sees a Merry-go-round in a playground. He measures the sides of the Merry-go-round and finds that it is a regular hexagon and sees that it spins on the intersection of all its’ diagonals. There are seats at each of the six vertices that are labeled as above.

1. At what angle would Bobby have to spin the Merry-go-round to ensure seat A inhabits the location previously occupied by seat B?

2 Assume that all the diagonals bisect each other Use this information2. Assume that all the diagonals bisect each other. Use this information to find x.

3. Prove the triangles in the figure above are all congruent.

Suppose Plane A is traveling from Pheonix represented by P on the figure below, to Lima, Peru, Represented by L. The radius of the earth is 6377km.

1. Let C be the center of the Earth. Find the shortestdistance the plane can fly to travel from P to T.

2. Plane A wants to continue to Hawaii, represented by H. Fuel costs $3.00 a gallon and it the plane can travel three miles to every gallon. The plane had 300 gallons of fuel in its’ tank went it landed in Lima. How much fuel does the plane need to make it to Hawaii, if the central angle fromplane need to make it to Hawaii, if the central angle from Lima to Hawaii is 83º.

P

56ºr

H

LL

PART III Released Open-Response Items — Geometry

32

RELEASED MATERIALS. MAY BE DUPLICATED.

GEOMETRY OPEN-RESPONSE ITEM E

E. Carol is a school crossing guard standing in front of the school. At 9:00 A.M., the school casts a40-foot shadow. Carol is 5.5 feet tall and her shadow at 9:00 A.M. is 8.8 feet long.

1. Determine the height of the school. Show or explain all of your work even if you use mentalmath or a calculator. Include units in your answer.

2. On some days, Donna helps Carol. Donna is 5.25 feet tall and at 9:00 A.M. she stands completelyin the school’s shadow. Determine how far away from the school Donna can stand and still becompletely in the school’s shadow. Show or explain all of your work even if you use mentalmath or a calculator. Include units in your answer.

BE SURE TO LABEL YOUR RESPONSES (1) AND (2).

0303--01161



Summer Workshop


1. Properties of Quadrilaterals PPT (R.4.G.1)

2. Polygons Interior and Exterior Angles(R.4.G.2)

3. Tesselate(R.4.G.3)

4. Circles and Arcs(R.4.G.5) a. Tangents, Secants and Theorems b. Arcs and Arc Length

5. Orthographic Drawing Activity(R.4.G.7)

a. Construction with Blocks b. Drawing using Parallel Lines

4. What is Euclidean Geometry?(R.4.G.8) a. What is Non-Euclidean Geometry? b. Why is it Non-Euclidean Geometry necessary? c. Lobachevsky’s negation of the fifth postulate. 5. Exploring non-Euclidean Geometry a. Distance geodesics (Lines) are great circles b. Stereographic Projection and Infinity . c. Metrics on Spheres. 6. Hyperbolic and Elliptic Geometry

Q d il t lQuadrilaterals

Classification

T idTrapezoid

D fi i i A Definition: A trapezoid is a quadrilateral with quadrilateral with one set of parallel lines

P ll lParallelogram

D fi i i A Definition: A parallelogram is a quadrilateral quadrilateral whose opposite sides are parallel.

PARALLELOGRAMSPARALLELOGRAMS

iProperties:Opposite Angles are Congruentare Congruent


iProperties:Opposite Angles are Congruentare Congruent

• Corresponding Angle Thm• Alternate Interior Angle Thm


P iProperties:Consecutive Angles are Angles are Supplementary

A corollary to Consecutive I t i A lInterior Angle


P iProperties:Opposite Sides are CongruentCongruent

Rh bRhombus

Definition:Definition:A rhombus is a parallelogram

whose sides are all congruentg

Properties:

Diagonals are perpendicular

Rh bRhombus

P i Properties: Diagonals Bisect

AnglesAngles

KITESKITES

D fi i A ki i Defintion: A kite is a quadrilateral whose adjacent adjacent (neighboring) sides are congruent

Properties: Diagonals are perpendicularare perpendicular

R t lRectangle

D fi i iDefinition:A rectangle is a parallelogram with parallelogram with four right angles

Properties: Diagonals are

t congruent

SSquare

D fi i iDefinition:A square is a rectangle with four rectangle with four congruent sides

Cl ifi ti Q d il t lClassification QuadrilateralsQuadrilaterals Trapezoid Isosceles

TrapezoidKites

Parallelograms

Trapezoid

RectanglesRhombusesRhombuses

Square


P iProperties:Opposite Angles are Congruentare CongruentConsecutive Angles are gSupplementaryOpposite Sides are C tCongruentDiagonals Bisect each othereach other

Geometry Proof

Show if the opposite angles are supplementary then ABCD is rectangle.

REMEMBER: You are only given that ABCD is a parallelogram.

Statements Reasons

0. m⁄ 1 + m ⁄ 4 = 180 º 0.

1. m ⁄ 1 + m ⁄ 2 = 180 º 1.

2. m⁄ 4 + m ⁄ 2 = 180 º 2.

3. m ⁄ 1 = m ⁄ 4 3.

4. m. ⁄ 1 = 90 = m ⁄ 4 4.

5. m ⁄ 3 + m ⁄ 1 = 180 º 5.

6. m ⁄ 1 + m ⁄ 2 = 180º 6.

7. m ⁄ 3 = m ⁄ 2 7.

8. m. ⁄ 3 = 90 = m ⁄ 2 8.

6. Let M be the midpoint of AD and N be the midpoint of BC. Show that MN bisects the

diagonals, AC and BD.

1 2 3 4

A B

CD

NM

Polygon: A closed figure made up by a number of line segments which intersect only at the endpoints.

EXAMPLES:

NON‐EXAMPLES:

Convex: A polygon is convex if any two points in the interior of the polygon and the line segment between lies completely within the interior.

Which of the examples of polygons are convex?

Try it. Pick any two points inside the polygon, and imagine the line segment between them, does it stay inside the polygon, or venture out?

Regular Polygons and Angles

Number of Sides

Name Sum of Interior Angles

Measure of Interior Angle

Measure of Exterior Angle

3

4

5

6

7

8

9

10

11

12

N

Geometry

Practice Test Name:

Problem 1. Circle the answer which completes the sentence best.(a) The sum of the angles of an pentagon is { 5(180) , 3(180), 3(360) }.(b) A regular triangle is also called { isosceles, right, equilateral } triangle.(c) The { median , altitude, base } of a trapezoid goes from the midpoint

of one leg to the midpoint of the other.(d) The exterior angles of a polygon is { always, never, sometimes } 360o.(e) The { base, leg, altitude } is one of the parallel sides of the trapezoid.

Problem 2. Show that if the opposite angles of a parallelogram are supplemen-tary, then the parallelogram is a rectangle.Write a paragraph proof.

Problem 3. Which of the following is not a polygon?

Problem 4. How many sides does a regular convex polygon have if it’s exteriorangle measures 60o.

1

Employee

Rectangle

Employee

Oval

Employee

Polygonal Line

Employee

Polygon

Employee

Text Box

a)

Employee

Text Box

b)

Employee

Text Box

c)

Employee

Text Box

d)

2

Problem 5. Suppose an irregular hexagon has four angles which measure 115o,112o, 130o and 125o. If two sides adjacent to the missing angles are congruent.What are the measurements of the last two sides.

Problem 6. What is the measure of the interior angle of a regular convex 12-gon.

Tesselation Activity

A tessellation is a pattern in which shapes fit together to cover a plane without overlaps, or gaps. Sometimes it is called a tiling. There are conditions which these shapes must satisfy in order to fit into such pattern. We will investigate what these conditions are and which shapes satisfy these conditions. You are given numerous polygons. For each of the polygons below, test whether or not the figure tessellates. Record it into the table below. Then fill out the remaining information. Do you see a pattern? Make a conjecture.

Name of Polygon

Number of Sides

Tessellate(Y/N)

Measure of Interior Angle

Measure of Exterior Angle

Triangle

3

Y

60° 120°

Isosceles Triangle

Equilateral Triangle

Trapezoid

Square

Rectangle

Parallelogram

Pentagon

Hexagon

Octagon

Conjecture:

What kind of transformation is a tessellation? Is it a translation? A reflection? Or a rotation? Tessellation by translation In order for this tessellation to work, the measure of angle ے a and ے b must be supplementary. Since this is a translation, it is necessary that horizontally consecutive angles be supplementary. Tesselation by rotation

As you can see, the interior angles of this polygon must be able to rotate about a vertex and create one entire revolution. Therefore, it is necessary for the interior angles to divide 360°. Tesselation by reflection

For a polygon to tessellate by reflection, the vertically consecutive interior angles must be supplementary.

a b

Circles

Vocabulary A circle is a set of point equidistant to a given point, the center. A chord is a segment whose endpoints are on the circle. A diameter is a chord which passes through the center A radius is a segment from the radius to the center. A tangent line to a circle is a line which intersects the circle at only one point A secant line of a circle is a line which intersects the circle at two points.

r

Theorem 1: A line is tangent to a circle if and only if it is perpendicular to the radius at the point of tangency (the intersection of the tangent line and the circle).

Let D be the point of tangency.Let A be the center of thecircle.Let B be the point on the linewhose distance to A isminimal. Then AB is perpendicular toBD and AB ≤ r = AD.But B is on the exterior of thecircle because the tangent lineintersects the circle only once.Therefore, AB ≥ AD = r. Hence AB = r, and B = D.

A

BD

Central Arcs

A central angle is an angle whose vertex is the center of the circle. A minor arc is an arc subtended by a central angle of less than 180º. A major arc is an arc subtended by a central angle of greater than 180º.

The measure of an Arc

The measure of an arc is the measure of the central angle subtending it. We know the circumference of a circle of radius, r, is C = 2Är.

a

D

CA

B

Theorem: If an angle is inscribed in a circle, then it’s measure is half the measure of the intercepted arc. The following is not a proof, but an idea. Where does it fail?

r

mCD = m∠CAD

CD = 2πr(m∠CAD

360°)

On the other hand, CD is also anarc on the large circle with radiusBC = 2r.

CD = 2π(2r)(m∠CBD

360°)FEB A

D

C

Then if angle a is 2Ä/k in radians. Then the arc subtended by angle a is a 1/kth portion of the circle, and therefore, the length of the arc is 1/k(2Är).

Proof: Since the red triangle has two sides which are the radius of the circle, then it must be isosceles. This is why it has two angles with measure y. Since we also get a right triangle ACD, then

( 90 ) 90 1802 180 1802 02

y y xy xy xy x

+ + − + =+ − =− ==

y

90 - x

D090x180 - x

y

A

C

Orthographic Construction

Class: Geometry Title of Lesson: Construction of Orthographic Drawings Purpose of Lesson: Students will be able to construct top, front and side view of a three dimensional object. Timed Activity: 15 minutes 1. Split the class into pairs. 2. Each pair gets 12 blocks and two pieces of large graph paper. 3. Have one person in the group construct a three dimensional block structure. 4. The other person in the group draws the top, front and side view. 5. When you believe you are done, raise your hand. One point will be awarded for each correct viewpoint, and one if the structure is creative. 6. Have group members switch roles and try again. SCORE: Block 1 Block 2 Block 3 Block 4 Block 5 TOTAL

Orthographic Construction

Class: Geometry Title of Lesson: Construction given Orthographic Drawings Purpose of Lesson: Students will be able to construct a three dimensional object given top, front and side view. Timed Activity: 15 - 20 minutes 1. Split the class into pairs. 2. Each pair gets 12 blocks. 3. Look at the attached sheet of paper 4. Build a three dimensional block structure that has the specified top, front and side view. Your team member must be at desk level making sure that the structure satisfies the given views. 5. When you have finished with one structure, call your instructor to verify your construction has the specified views. (1 point is granted per correct view) 6. Once your structure has been graded, move on to the next one.

TOP FRONT RIGHT SIDE POINTS

EUCLIDEAN AND NON-EUCLIDEAN


Euclidean Geometry refers to the geometry imposed on planes, spacesand generalizations of such. Note, that to define a point on the realplane, we need only an ordered pair of real numbers. Real space, then,is the set of all ordered triples of real numbers, and n-dimensional realspace is the set of all ordered n-tuples of real numbers. However, thismethod of constructing Euclidean Space came from DesCartes in the late1500’s, and is called Cartesian Space. A distance function was defined bya generalization of the pythagorean theorem,

d((x1, x2, . . . , xn), (y1, y2, . . . , yn)) =√

(x1 − y1)2 + (x2 − y2)2 + . . . (xn − yn)2

Euclid did not have the real numbers or algebra at his disposal. Forexample, in the cartesian plane, we can define parallel lines to be distinctlines with the same slope because we have so much algebraic structureimposed on the plane by the cartesian coordinate system. In EuclideanGeometry, we define parallel lines to be those which do not intersect. Thealgebraic definition of parallel lines was constructed to be equivalent tothe geometric definition.

He still, however, attempted to impose structure on the accumulationof geometric knowledge by creating a hierarchy of assumptions and truthscalled an axiomatic approach to mathematics. We can categorize all ofhis mathematical statements into two categories.

(A) assumptions: definitions, axioms and postulates(B) truths logically deduced by the assumptions: Corollaries, Propo-

sitions, Lemmas and Theorems

Date: June 4, 2007.

1

2

Euclidean Geometry, which is defined by 5 postulates, was the only ge-ometry which mathematicians acknowledged for several centuries. Thesepostulates are:

(1) A straight line can be drawn from any point to any point.(2) A finite straight line can be produced continuously in a straight

line.(3) A circle may be described with any point as center and any dis-

tance as a radius.(4) All right angles are equal to one another.(5) If a transversal falls on two line in such a way that the interior

angles on one side of the transversal are less than two right angles,then the lines meet on the side on which the angles are less thantwo right angles.

The contrapositive of the last postulate is the following statement:

If a transversal falls on two lines and the lines do not meet on one side,then the interior angles on that side of the transversal are greater than orequal to two right angles.

Later, Playfair, found an equivalent statement to the parallel postulate:

Given a line, and a noncolinear point, there exists exactly one linethrough that point parallel to the given line.

Therefore, if one can construct a geometric space in which these pos-tulates do not hold, then this space would be a Non-Euclidean Geometry.

We will study many different modifications of the parallel postulate,each of which yield a different Non-Euclidean Geometry. We will startwith spherical geometry, which is the most elementary, easy to visualize,and whose distance metric can be computed with some basic arithmetic.


The Geometry of a Sphere Although, we believe Euclidean Space is a universal model for all surfaces we need to simulate. Locally, it is a useful and relatively accurate model, but globally, the surface of the earth cannot be modeled by a plane. We can see that the earths’ surface is two-dimensional, and, hence needs to be modeled by a two dimensional surface. It would be very useful to impose a geometry on the earths’ surface so that we can find an accurate model for the distance between two points. If a plane is traveling over the surface of the earth, it must stay within a spherical shell around the earth. In the interest of economics and time efficiency, it would be practical to find the shortest route required for the plane to travel from point A to point B. In Euclidean space, this would be the line segment starting at point A and ending at point B. However, the line segment would go through the sphere, and if it were a plane, it might crash into the earth. Not a good idea. We need a path on the sphere itself.

Stereographic Projection Let the figure below represent the unit sphere in three dimensional space. We can construct a one-to-one continuous function from the punctured sphere to the xy-plane, called stereographic projection. Let Lp be the line through the point p on the xy-plane, and the point (0, 0, 1). This line will intersect the sphere at exactly one other point. This is a well-defined one-to-one continuous function from the xy-plane to the unit sphere punctured at the “North Pole”.

An algebraic equation for this function is attached to the end of this document. The question is, why this is a one-to-one function to the punctured unit sphere? Note there is no point which yields a line whose intersection with the sphere only occurs at the point (0, 0, 1). If we add the point at infinity, however, that would correspond to the North Pole. The South Pole corresponds to the origin. Another way to view this transformation is to fold the plane so that points that correspond to infinity around the plane actually are identified as one point.

(0, 0, 1)

p

q

Geodesics However, we are mainly interested in the path which yields the shortest distance between two points on the sphere. This would be the analogy to a line on the sphere. Therefore, it makes sense to find where lines on the plane get mapped onto the sphere. We will start with lines through the origin.

In general, a line is called a geodesic. By distance geodesic, one means, a path with minimal length on the surface. Clearly, in Euclidean Space, the distance geodesic between points A and B is the line segment connecting them. One can see that line segments through the origin will get mapped to great circles through the North Pole (∞) and South Pole (origin). If this is not clear, recall that to each point on the plane, we correspond a line through infinity. The union of all these lines is a plane containing the line segment and through the point ∞. The intersection of this plane with the sphere is the image of the line segment, and is therefore a geodesic between the image of the endpoints of the line segment.

Distance Metric on the Sphere Computing the distance metric on the sphere is rather simple since all the lines are great circles of radius equal to that of the sphere. For example, if we are given the angle between the points on the sphere, we can compute the arclength of the arc of the great circle between them. Suppose the sphere below is of radius r. Let A and B be points on the sphere. Suppose the central angle between the points is given in radians, Ө. Then the distance between A and B is

D(A,B) = r Ө

Ө

A

B r

How This Fails the Parallel Postulate Recall there are two equivalent versions of the parallel postulate.

Given a line, and a noncolinear point, there exists exactly one line through that point parallel to the given line. Below is a picture of a sphere and a “line” on the sphere. Note q is a point not on the line.

As you can see, there is no “line” (great circle) through q that does not intersect ( is parallel to ) the line l. Also, rather than there being only one “line” between two points, there are infinitely many. Clearly, each of the longitude lines on the globe are great circles (lines) between the North and South poles. Also, note that the illustration below is an example of a 270˚ triangle. And there are two transversals who are perpendicular to the same line, and they intersect.

q

l

Variations Of the

Fifth Postulate

Given a line, and a non-collinear point, there exists one line through the point parallel to the given line.

Given a line, and a non-collinear point, there does not exist a line through the point parallel to the given line.

Principal Curvatures have same signBoth are concave up or concave down

Given a line, and a non-collinear point, there exist infinitely many lines through the point parallel to the given line.

Below is a crocheted example of a hyperbolic surface with and illustration of a violation of the original fifth postulate.

Principal Curvatures are opposite signsConcave down – negativeConcave down negativeConcave up - positive

Curvature?

L. Marizza A. Bailey

Arkansas School of Mathematics, Sciences and the Arts

Hot Springs, AR

[email protected]

Summer Workshop

Outline

Content Standard 5

1. Linear Equations

a. Finding slopes of lines.(CG.5.2.G.2) b. Writing equations in slope intercept

form.(CGT.5.G.3) c. Writing equations in standard form.(CGT.5.G.4)

2. Using TI-interactive to visualize

translation.(CGT.5.G.5)

Slopes of Lines

Suppose you and your friend are riding a bicycle up a hill. You are faster than your friend, so you get to the other side of the hill far before he even reaches the hill. He calls you on your cell phone and asks you how steep the hill is. You respond, “100 feet hight.” What is wrong with your answer?

Both of the hills above are 100 feet high, but one is steeper than the other. Which one would you consider steeper?

If you answered the one on the left, ask yourself why you think this is so.

On the first hill, you only travel 40 feet horizontally, but your elevation changes by 100 feet!

On the second hill, you travel 150 feet horizontally before you reach the top at 100 feet.

Therefore, in order to express steepness of a hill, you need to discuss the change in horizontal distance AND the change in height.

The slope of the first hill is 100 feet over 40 feet, or

100 5 2.540 2

= =

The slope of the second hill is 100 feet over 150 feet, or

100 2 .667150 3

=

The slope is, therefore, the ratio of vertical change over horizontal change.

100

40 ft 150 ft

How do we relate this to lines on the plane?

If we want to find the slope of this line, we need to find the ratio of the vertical change over the horizontal change. First I need to pick two points.

Note that the points that are clearly on the line are (‐1, 1), (0, 4), (1, 7).

Find the ratio of the vertical change over the horizontal change for (‐1,1) and (0,4).

The vertical change is 3 units, and is derived by 4 ‐1 = 3

The horizontal change is 1 unit, and is derived by 0 ‐ ‐1 = 1.

Therefore, the slope is

4 1 30 1−

=− −

If we pick two other points, (‐1,1) and (1,7) we get

7 1 6 31 1 2−

= =− −

Regardless of which two points I choose, we always get the same slope.

(0,4) , (1,7) will give slope

7 4 31 0−

=−

Why is this?

If we draw the right triangles generated by these points, drawn like the one above, we will see that they are all similar. If two polygons are similar, the ratios of their corresponding sides are equal. Since the slope is the ratio of the length of the legs of the right triangles, they will always be the same. Therefore, the slope is independent of your choice of points.

Find the slope of each line.

Solid line:

Dashed line:

Dotted line:

What is the relationship between their slopes?

Is it expected.

In general, if 1 1 2 2, ) and ( , )( y yx x are on a line, then the formula for the slope, m, is

2 1

2 1

y yx

mx

−−

=

Practice:

1) Find the slope of each line :

a) b)

c) d)

2) Find the slope of the line through these points

a) (2,‐3) and (1,5) b) (‐1,4) and (‐5,‐2) c) (‐ ½ , 2) and ( ¾ , ‐5) d) (.53, 6) and (1.37, 6.7)

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