chapter 9 the pythagorean theorem - high school...

CHAPTER

9

But serving up an action, suggesting

the dynamic in the static, has become a

hobby of mine . . . . The “flowing” on that

motionless plane holds my attention to such a

degree that my preference is to try and make it into

a cycle.

M. C. ESCHER

Waterfall, M. C. Escher, 1961 ©2002 Cordon Art B. V.–Baarn–Holland.All rights reserved.

The PythagoreanTheorem

O B J E C T I V E S

In this chapter you will

● discover the PythagoreanTheorem, one of the mostimportant concepts inmathematics

● use the PythagoreanTheorem to calculate thedistance between any two points

● use conjectures related tothe Pythagorean Theoremto solve problems

CHAPTER 9 The Pythagorean Theorem 461

Escher has cleverly used right angles to form hisartwork known as Waterfall. The picture containsthree uses of the impossible tribar created byBritish mathematician Roger Penrose (b 1931) in1954. In 1934 Swedish artist Oscar Reutersvard(b 1915), “father of impossible figures,” had createdan impossible tribar that consisted of a triangulararrangement of cubes.

The shapes topping the towers in Escher’s work are,on the left, a compound of three cubes and, on theright, a stellation of the rhombic dodecahedron.

[Ask] “What impossible things do you see?”[Water seems to be traveling up an incline, yet it isrunning a mill wheel.] “Which surfaces appear tobe horizontal? Vertical? Sloped? There are threeimpossible tribars in the picture; where are they?”[They all have flowing water along two sides; twiceone of the bars is replaced by the waterfall, andonce one bar is replaced by a group of fourcolumns.]

C H A P T E R 9O B J E C T I V E S

● Understand thePythagorean Theoremmore deeply

● Discover the Converse ofthe Pythagorean Theorem

● Practice working withradical expressions

● Discover relationshipsamong the lengths of thesides of a 45°-45°-90°triangle and among thelengths of the sides of a30°-60°-90° triangle

● Apply the PythagoreanTheorem and itsconverse

● Discover and apply thePythagorean relationshipon a coordinate plane(the distance formula)

● Derive the equation of acircle from the distanceformula

● Practice using geometrytools

● Develop readingcomprehension,problem-solving skills,and cooperative behavior

● Learn new vocabulary

Penrose tribar

L E S S O N

9.1

462 CHAPTER 9 The Pythagorean Theorem

The puzzle in this investigation is intended to help yourecall the Pythagorean Theorem. It uses a dissection,which means you will cut apart one or more geometric figures and make the pieces fit intoanother figure.

Step 1 Construct a scalene right triangle in the middle of your paper. Label thehypotenuse c and the legs a and b.Construct a square on each side ofthe triangle.

Step 2 To locate the center of the square on thelonger leg, draw its diagonals. Label the center O.

Step 3 Through point O, construct line jperpendicular to the hypotenuse and line kperpendicular to line j. Line k is parallel to the hypotenuse. Lines j and k dividethe square on the longer leg into four parts.

Step 4 Cut out the square on the shorter leg and the four parts of the square on thelonger leg. Arrange them to exactly cover the square on the hypotenuse.

The Theorem ofPythagorasIn a right triangle, the side opposite theright angle is called the hypotenuse.The other two sides are called legs. Inthe figure at right, a and b represent thelengths of the legs, and c represents thelength of the hypotenuse.

There is a special relationship between the lengths of the legs and the length of thehypotenuse. This relationship is known today as the Pythagorean Theorem.

I am not young enough to

know everything.

OSCAR WILDE

L E S S O N

9.1

InvestigationThe Three Sides of a Right Triangle

You will need

● scissors

● a compass

● a straightedge

● patty paper O

j

k

ab

c

FUNKY WINKERBEAN by Batiuk. Reprinted with special permission of North America Syndicate.

c a

b

In a right triangle, the side opposite the right angle is called the hypotenuse, here with length c.

The other two sides are legs, here with lengths a and b.

trying a special case first—here, an isosceles righttriangle. As needed, point out that good pieces mightbe formed if they draw lines through the smallersquares parallel to edges of the largest square.

[Language] A dissection is the result of separatingsomething into pieces.

Step 1 Using the Dissection of Squares worksheets orthe Sketchpad demonstration will speed the investiga-tion, but the use of many different triangles drawn bythe students strengthens the inductive conclusion.The constructions are quicker with patty paper than

with compass and straightedge. It is also easy tocreate several examples using geometry software.

Step 2 As needed, remind students that the legs arethe sides other than the hypotenuse, so the “longerleg” is not the hypotenuse. Suggest that studentsminimize clutter by making these diagonals verylight or by drawing only the portion near the centerof the square.

Step 4 Ask students to take care in drawing andcutting out pieces so they will fit together well.Students may want to tape the pieces together.

PLANNING

LESSON OUTLINE

One day:15 min Investigation

5 min Sharing

10 min Examples

15 min Closing and Exercises

MATERIALS

� construction tools � scissors� Pythagorean Theorem (W) for One step� Dissection of Squares (W), optional� Sketchpad demonstration Three

Triangles, optional

TEACHING

Many students may already knowthe Pythagorean Theorem as a2 � b2 � c2. In this lesson theyreview what the letters stand forand discover proofs showing whythe relationship holds for allright triangles.

INTRODUCTION

Direct students’ attention toImproving Your Visual ThinkingSkills on page 454. Ask what theycan conclude about right trian-gles, and help them state thePythagorean Theorem usingareas of squares and the termshypotenuse and legs.

Guiding the Investigation

One step Hand out a copy of thePythagorean Theorem worksheetto each group. Challenge studentsto cut up one or both of thesmaller squares and assemble thepieces on top of the largestsquare. As you circulate, youmight remind students of theproblem-solving technique of

History

Pythagoras of Samos (ca. 569–475 B.C.E.),depicted in this statue, is often described as“the first pure mathematician.” Samos was aprincipal commercial center of Greece and islocated on the island of Samos in the AegeanSea. The ancient town of Samos now lies inruins, as shown in the photo at right.

Mysteriously, none of Pythagoras’s writings stillexist, and we know very little about his life. Hefounded a mathematical society in Croton, inwhat is now Italy, whose members discovered irrational numbers and the fiveregular solids. They proved what is now called the Pythagorean Theorem,although it was discovered and used 1000 years earlier by the Chinese andBabylonians. Some math historians believe that the ancient Egyptians alsoused a special case of this property to construct right angles.

A theorem is a conjecture that has been proved. Demonstrations like the one in the investigation are the first step toward proving the Pythagorean Theorem.

Believe it or not, there are more than 200 proofs of the Pythagorean Theorem.Elisha Scott Loomis’s Pythagorean Proposition, first published in 1927, containsoriginal proofs by Pythagoras, Euclid, and even Leonardo da Vinci and U. S.President James Garfield. One well-known proof of the Pythagorean Theorem is included below. You will complete another proof as an exercise.

Paragraph Proof: The Pythagorean Theorem

You need to show that a2 � b2 equals c2 for the right triangles in the figure at left.The area of the entire square is �a � b�2 or a2 � 2ab � b2. The area of any triangleis ��

12��ab, so the sum of the areas of the four triangles is 2ab. The area of the

quadrilateral in the center is �a2 � 2ab � b2� � 2ab, or a2 � b2.

If the quadrilateral in the center is a square then its area also equals c2. You nowneed to show that it is a square. You know that all the sides have length c, but you also need to show that the angles are right angles. The two acute angles in the right triangle, along with any angle of the quadrilateral, add up to 180°. The acute angles in a right triangle add up to 90°. Therefore the quadrilateral anglemeasures 90° and the quadrilateral is a square. If it is a square with side length c,then its area is c2. So, a2 � b2 � c2, which proves the Pythagorean Theorem. �

b a

b

a

ba

b

ac

c

cc

The Pythagorean Theorem

In a right triangle, the sum of the squares of the lengths of the legs equals thesquare of the length of the hypotenuse. If a and b are the lengths of the legs,and c is the length of the hypotenuse, then �? . a2 � b2 � c2

C-82

Step 5 State the Pythagorean Theorem. SHARING IDEAS

You might make a transparencyof the Dissection of Squaresworksheets for students to use in presenting their ideas.

Ask about symmetry in thedissected square on the hypot-enuse. The method of the Inves-tigation gives 4-fold rotationalsymmetry.

[Ask] “What if the triangle isn’t a right triangle? Do you thinkthere’s still a relationship amongthe lengths of the sides?” You neednot answer this question now;it’s addressed later. [Link] ThePythagorean Theorem is a specialcase of the Law of Cosines

�c2 � a2 � b2 � 2ab cos C,where �C is the angle oppositeside c�; when m�C � 90°, wehave cos C � 0.

[Ask] “What is the longest side ofa right triangle?” “Is it the same asthe longest leg?” [The hypotenuse,not the longest leg, is the longestside.] If you ask why the longestside is always the hypotenuse andwhat can be said about the longerof the two legs, you can reviewthe Triangle Inequality Conjectureand the Side-Angle InequalityConjecture.

[Ask] “What is a theorem?” [It’s aconjecture that has been proveddeductively within a deductivesystem.] So far in this course noaxiom system has been devel-oped, so there are no real theo-rems; this conjecture is called atheorem in this book because ithas been proved within an axiomsystem and it’s so well known bythat name.

If students are not very familiarwith the Pythagorean Theorem,ask how this theorem about areasof squares might be used tocalculate lengths. Direct students’attention to the two examples.

LESSON 9.1 The Theorem of Pythagoras 463

LESSON OBJECTIVES

� Understand the Pythagorean Theorem more deeply� Practice using geometry tools� Learn new vocabulary

NCTM STANDARDS

CONTENT PROCESS

Number � Problem Solving

� Algebra � Reasoning

� Geometry � Communication

� Measurement � Connections

Data/Probability � Representation

The Pythagorean Theorem works for right triangles, but does it work for alltriangles? A quick check demonstrates that it doesn’t hold for other triangles.

Let’s look at a few examples to see how you can use the Pythagorean Theorem tofind the distance between two points.

How high up on the wall will a 20-foot ladder touch if the foot of the ladder is placed 5 feet from the wall?

The ladder is the hypotenuse of a right triangle, so a2 � b2 � c2.

�5�2 � �h�2 � �20�2 Substitute.

25 � h2 � 400 Multiply.

h2 � 375 Subtract 25 from both sides.

h � �375� � 19.4 Take the square root of each side.

The top of the ladder will touch the wall about 19.4 feet up from the ground.

Notice that the exact answer in Example A is �375�. However, this is a practicalapplication, so you need to calculate the approximate answer.

Find the area of the rectangular rug if the widthis 12 feet and the diagonal measures 20 feet.

Use the Pythagorean Theorem to find the length.

a2 � b2 � c2

�12�2 � �L�2 � �20�2

144 � L2 � 400

L2 � 256

L � �256�L � 16

The length is 16 feet. The area of the rectangle is12 � 16, or 192 square feet.

EXAMPLE A

� Solution

EXAMPLE B

� Solution

h

5 ft

20 ft

12 ft

L 20 ft

a 2 � b 2 � 45c 2 � 25

a 2 � b 2 � c 2 a 2 � b 2 � c 2 a 2 � b 2 � c 2

a 2 � b 2 � 45c 2 � 45

a 2 � b 2 � 45 c 2 � 57.2

62 � 72 � 82

7

8

6

Acute triangle

172 � 252 � 382

17 25

38

Obtuse triangle

For an interactive version of this sketch, visit .www.keymath.com/DG

� EXAMPLE A

After working through theexample, ask students whatsquare roots are and how to findthem using their calculators.Remind them that many of thesquare roots they’ll find withtheir calculators are approxima-tions. [Ask] “What are somenumbers whose square roots arewhole numbers?” [4, 9, 16, 25, 36,and so on] Students will begin torecognize more examples ofperfect squares as they workthrough the chapter.

The equation h2 � 375 is not aperfect model for this problembecause it has two solutions, onepositive and one negative. Thenegative solution is ignoredbecause all distances in geometryare positive.

� EXAMPLE B

In Example A, the book mentionsthat 19.4 is an approximation of�375�. Point out that in ExampleB the calculation is exact. Seewhether students recognize 256as a perfect square. Again, thenegative square root is beingignored.

Process text (the reason for eachalgebraic step) is included witheach step in Example A. InExample B, it has been left up tostudents to figure out what alge-braic steps are being carried out.[Ask] “Why does [this step]follow from the previous step?”

Assessing ProgressYou can assess students’ under-standing of right triangle, square,diagonal, and perpendicular.Also watch for their ability tofollow instructions and to worktogether in groups. See how wellthey realize when exact answersare appropriate and when theyneed to use approximations.

Closing the Lesson

The Pythagorean Theorem is a claim about areas of squares built on the sides of a right triangle:The area of the square on the hypotenuse is thesum of the areas of the squares on the legs. Itsprimary applications are in finding the length ofone side of a right triangle in which the lengths of the other two sides are known. The theorem can be proved by dissection.


EXERCISESIn Exercises 1–11, find each missing length. All measurements are in centimeters. Giveapproximate answers accurate to the nearest tenth of a centimeter.

1. a � �? 2. c � �? 3. a � �?

4. d � �? 5. s � �? 6. c � �?

7. b � �? 8. x � �? 9. The base is a circle.x � �?

10. s � �? 11. r � �?

12. A baseball infield is a square, each side measuring 90 feet. To thenearest foot, what is the distance from home plate to second base?

13. The diagonal of a square measures 32 meters. What is the area ofthe square?

14. What is the length of the diagonal of a square whose area is 64 cm2?

15. The lengths of the three sides of a right triangle are consecutiveintegers. Find them.

16. A rectangular garden 6 meters wide has a diagonal measuring 10 meters. Find the perimeter of the garden. 28 m

3, 4, 5

512 m2

5

s

13 cm 3.5 cm

9

x 41x

1.51.5

3.97

8

b

25

40 cm3.6 cm24 cm

6cs

s

24

10d

8

6

8.5 cm26 cm10 cm

a6

8

12

c

15513

a

5.3 cm19.2 cm12 cm

r

(0, 0)

(5, 12)

y

x

�

Second base

Home plate

127 ft

11.3 cm


BUILDINGUNDERSTANDING

After students have solved someexercises, you may want to haveseveral groups report on howthey solved them.

ASSIGNING HOMEWORK

Essential 1–16

Portfolio 17, 18

Journal 17, 18

Group 18

Review 19–22

|

� Helping with the Exercises

[Alert] As needed, remindstudents that the hypotenuse isalways the longest side and isopposite the right angle.

Exercise 4 Students might wonderwhy the upper triangle is a righttriangle. [Ask] “What kind offigure is the quadrilateral? Doyou see congruent triangles?”

Exercise 7 [Alert] Some studentsmay want to use the measure 8,which is not needed in the calculation.

Exercise 10 If students are havingdifficulty, ask what shape thequadrilateral is.

Exercise 11 This is good prepara-tion for work with the unit circlein trigonometry.

Exercises 12–16 Students may findit helpful to draw and labelpictures.

Exercise 12 As an extension, youmight have students measure thelengths of the distances betweenthe bases on a local baseball field.

17. One very famous proof of the Pythagorean Theorem is by theHindu mathematician Bhaskara. It is often called the “Behold” proofbecause, as the story goes, Bhaskara drew the diagram at right andoffered no verbal argument other than to exclaim, “Behold.” Usealgebra to fill in the steps, explaining why this diagram proves thePythagorean Theorem.

18. Is �ABC � �XYZ? Explain your reasoning.

Review

19. The two quadrilaterals are squares. 20. Give the vertex arrangement for the Find x. 2-uniform tessellation.

21. Explain why m � n � 120°. 22. Calculate each lettered angle, measure,or arc. EF� is a diameter; �1 and �2 aretangents.

120°m

n

x

225 cm2

36 cm2

36/32.4.3.4x � 21 cm

�

History

Bhaskara (1114–1185, India) was one of the first mathematicians to gain athorough understanding of number systems and how to solve equations,several centuries before European mathematicians. He wrote six books onmathematics and astronomy, and led the astronomical observatory at Ujjain.

4 cm

5 cmA B

C

4 cm

5 cmX Y

Z

�2

�1

E

F

c

h

b

arg

n

e

u

v

t

s fd

106°

58°

Sample answer: Yes, �ABC � �XYZ bySSS. Both triangles are right triangles, soyou can use the Pythagorean Theorem tofind that CB � ZY � 3 cm.

17. The area of the large squareis 4 � area of triangle � area ofsmall square.

c2 � 4 � �12� � ab � (b � a)2

c2 � 2ab � b2 � 2ab � a2

c2 � a2 � b2

Exercise 18 This problem illustratesa special case in which SSA is acongruence shortcut. That is, whenthe non-included angle is a rightangle, there is only one trianglethat can be formed by SSA. Thisshortcut is sometimes called HL(Hypotenuse-Leg). Students canprove HL using the thinkingprocess used in the solution to thisexercise. Given two right triangleswith congruent, correspondinghypotenuses with length c andcorresponding legs with length a,use the Pythagorean Theorem toshow that the other two corre-sponding legs are congruent; thetriangles are congruent by SSS.Students will prove the HLTheorem in Lesson 13.7

21. Mark the unnamed angles as shown in the figure below.By the Linear Pair Conjecture,p � 120° � 180°. � p � 60°. ByAIA, m � q. By the TriangleSum Conjecture, q � p � n �180°. Substitute m � q and p �60° to get m � 60° � n � 180°.� m � n � 120°.

Or use the Exterior AngleConjecture q � n � 120°. ByAIA q � m. Substituting,m � n � 120°.

p

n

m

q

120°

c

b

b �a

a


22. a � 122°, b � 74°, c � 106°, d � 16°, e � 90°,f � 74°, g � 74°, h � 74°, n � 74°, r � 32°,s � 74°, t � 74°, u � 32°, v � 74°

8.1

4.1

7.4

6.3

CREATING A GEOMETRY FLIP BOOK

Have you ever fanned the pages of a flip book andwatched the pictures seem to move? Each page shows a picture slightly different from the previous one. Flipbooks are basic to animation technique. For moreinformation about flip books, see .

Here are two dissections that you can animate to demonstrate the Pythagorean Theorem.(You used another dissection in the Investigation The Three Sides of a Right Triangle.)

You could also animate these drawings to demonstrate area formulas.

Choose one of the animations mentioned above and create a flip book that demonstratesit. Be ready to explain how your flip book demonstrates the formula you chose.

Here are some practical tips.� Draw your figures in the same position on each page so they don’t jump around when

the pages are flipped. Use graph paper or tracing paper to help.� The smaller the change from picture to picture, and the more pictures there are, the

smoother the motion will be.� Label each picture so that it’s clear how the process works.

ba

a

www.keymath.com/DG

These five frames start off the photo series titled The Horse in Motion, by photographer,innovator, and motion picture pioneer Eadweard Muybridge (1830–1904).

[Context] Eadweard Muybridge,the man who created The Horsein Motion, was born in Englandbut moved to the United Statesas a boy. As part of scientificresearch to improve techniquesof horseracing, he devised aningenious method of setting up a dozen or more cameras to gooff in rapid sequence. He was a leader in the early days ofphotography, improving onexisting methods, including the art of film developing. Heinvented a precursor to themovie projector based on theidea of a zoetrope, a slitted drumwith pictures on the inside that is spun to show the illusion ofmotion when the pictures arewatched through the slits.

EXTENSIONS

A. Have students research andtry one of numerous otherdissections that demonstrate thePythagorean Theorem.

B. Use Take Another Lookactivity 1, 2, or 3 onpages 501–502.


Suggest that students use a small graph papertablet to help keep the nonmoving features inthe same position from page to page. They canglue tracing papers of the flip book onto cardsfor a firmer flip.

OUTCOMES

� Movement in the flip book is smooth.� The student can explain the dissection used.� The student adds other animation, for

example, drawing a hand flipping the pages ofa flip book, thus creating a flip book of a flipbook!

� An animation is created using geometrysoftware.

Supporting the

L E S S O N

9.2


Any time you see someone

more successful than you are,

they are doing something

you aren’t.

MALCOLM X

L E S S O N

9.2The Converse of thePythagorean TheoremIn Lesson 9.1, you saw that if a triangle is a righttriangle, then the square of the lengthof its hypotenuse is equal to the sumof the squares of the lengths ofthe two legs. What about theconverse? If x, y, and zare the lengths of the three sides of a triangleand they satisfy thePythagorean equation,a2 � b2 � c2, must thetriangle be a right triangle?Let’s find out.

InvestigationIs the Converse True?

You will need

● string

● a ruler

● paper clips

● a piece of patty paper

Three positive integers that work in the Pythagorean equation are calledPythagorean triples. For example, 8-15-17 is a Pythagorean triple because 82 � 152 � 172. Here are nine sets of Pythagorean triples.

3-4-5 5-12-13 7-24-25 8-15-17

6-8-10 10-24-26 16-30-34

9-12-15

12-16-20

Step 1 Select one set of Pythagorean triples from the list above. Mark off four points,A, B, C, and D, on a string to create three consecutive lengths from your set of triples.

Step 2 Loop three paper clips onto the string. Tie the ends together so that points A and D meet.

Step 3 Three group members should each pull a paper clip at point A, B, or C to stretch the string tight.

17 cm

01 2 3 4

56 7 8 9

1011 12 13 14

1516 17

15 cm8 cmA B C D

LESSON OBJECTIVES

� Discover the Converse of the Pythagorean Theorem� Learn new vocabulary� Develop reading comprehension, problem-solving skills, and

cooperative behavior

You might let the class do this investigation outside,with the students themselves acting as “ropestretchers” of long ropes.

Step 1 Suggest that leaving excess string at bothends will make it easier to tie the ends together inStep 2.

PLANNING

LESSON OUTLINE


10 min Sharing

5 min Closing

10 min Exercises

MATERIALS

� rulers� string (two 1-meter strings per group)� paper clips (six per group)� protractors� patty paper

TEACHING

The Converse of the PythagoreanTheorem can also be proved.

Guiding the Investigation

One step Pose this problem: “Whatcan you say about triangles inwhich the side lengths satisfy theequation a2 � b2 � c2?” As youcirculate, encourage groups to tryvarious triples from the list in thestudent book or to make up theirown lengths, perhaps not integers.

As needed, remind students thatthe converse of a true statementmight be false.

The Pythagorean triples arearranged in families. [Ask] “Howare the triples in a column related?” [They are multiples ofthe first triple in that column.]“What would be the next triplein the second column?” [15, 36,39] “Why is 3-4-5 or 5-12-13called a primitive Pythagoreantriple?” [The numbers in thetriple have no common factor.]

Step 4 With your paper, check the largest angle. What type of triangle is formed?

Step 5 Select another set of triples from the list. Repeat Steps 1–4 with your new lengths.

Step 6 Compare results in your group. State your results as your next conjecture.

This ancient Babyloniantablet, called Plimpton 322,dates sometime between1900 and 1600 B.C.E. Itsuggests several advancedPythagorean triples, such as1679-2400-2929.

Converse of the Pythagorean Theorem

If the lengths of the three sides of a triangle satisfy the Pythagorean equation,then the triangle �? . is a right triangle

C-83

History

Some historians believe Egyptian “rope stretchers” used the Converse of thePythagorean Theorem to help reestablish land boundaries after the yearlyflooding of the Nile and to help construct the pyramids. Some ancient tombsshow workers carrying ropes tied with equally spaced knots. For example,13 equally spaced knots would divide the rope into 12 equal lengths. If oneperson held knots 1 and 13 together, and two others held the rope at knots4 and 8 and stretched it tight, they could have created a 3-4-5 right triangle.

Step 4 right triangle Step 4 Students can also use acorner of a piece of paper (or of some other object such as acarpenter’s square) to verify aright angle.

SHARING IDEAS

Have students show their workwith a variety of lengths. Let theclass discuss measurement errors.Elicit the idea that a slight errorin the measurement of a lengthcan result in a measurablydifferent angle. Keep askingstudents whether they think theconverse is always true. Agree on a statement of the conjecturewithout resolving the question of truth.

[Ask] “Will the conjecture hold ifyou use different measurementunits so that the triples change,possibly to non-integers?”

[Ask] “Suppose one triangle has sides whose lengths are aPythagorean triple and anothertriangle has sides whose lengthsare a multiple of that Pythagoreantriple. How are the trianglesrelated?” [They are similar.]Students can try drawing suchtriangles and looking for patterns,but you need not answer thequestion yet. It foreshadows theideas of similarity in Chapter 11.

Also ask whether students whobelieve the conjecture is true canprove it deductively. Then havethe class read the proof outlinein the student book.

MAKING THE CONNECTION

Pythagoras himself is believed tohave studied in Egypt, and hemay have learned the trianglerelationship there. Althoughsome historians discount therope stretchers tale, there’s nodoubt that Egyptian mathemati-cians knew the relationship.

LESSON 9.2 The Converse of the Pythagorean Theorem 469

NCTM STANDARDS

CONTENT PROCESS

� Number � Problem Solving





EXERCISESIn Exercises 1–6, use the Converse of the Pythagorean Theorem to determine whethereach triangle is a right triangle.

1. 2. 3.

4. 5. 6.

In Exercises 7 and 8, use the Converse of the PythagoreanTheorem to solve each problem.

7. Is a triangle with sides measuring 9 feet, 12 feet, and 18 feet aright triangle?

8. A window frame that seems rectangular has height 408 cm,length 306 cm, and one diagonal with length 525 cm. Is thewindow frame really rectangular? Explain.

no

1.73 1.41

2.23

10 20

24

2212

18

12 36

35

130

120

508

15

17

�

The proof of the Converse of the Pythagorean Theorem is very interesting becauseit is one of the few instances where the original theorem is used to prove theconverse. Let’s take a look. One proof is started for you below. You will finish it asan exercise.

Proof: Converse of the Pythagorean Theorem

Conjecture: If the lengths of the three sides of a trianglework in the Pythagorean equation, then the triangle isa right triangle.

Given: a, b, c are the lengths of the sides of �ABC and a2 � b2 � c2

Show: �ABC is a right triangle

Plan: Begin by constructing a second triangle, righttriangle DEF (with �F a right angle), with legs oflengths a and b and hypotenuse of length x. The plan is to show that x � c, so that the triangles arecongruent. Then show that �C and �F are congruent.Once you show that �C is a right angle, then �ABC isa right triangle and the proof is complete.

B A

C

a b

c

E D

F

a b

x

yes yes no

no no no

No, the given lengths are not a Pythagorean triple.

Exercise 7 Students’ justifications might citePythagorean multiples. The numbers 9, 12, and 18have a common factor of 3, so a primitive would be3-4-6. This is not a Pythagorean triple; the familiar3-4-5 triple says that if the two legs have lengths 3and 4, the hypotenuse must have length 5 to give aright triangle.

Exercise 8 After students decide that the angle isn’tright, you might ask whether students could haveknown the window frame was rectangular if theangle had turned out to be right. Having one rightangle is not a sufficient condition for a quadrilateralto be a rectangle.


ProofAsk students to critique thisoutline in order to deepen theirunderstanding of it. As theydiscuss it, monitor their facialexpressions and try to includestudents who seem to have ideasbut aren’t speaking up. You maywant to let the discussion lead tofilling in details as a class andthen writing up a good modelproof. Or, if your students arefairly comfortable with proof,you can challenge them to writeup the details as homework.

Assessing ProgressYou can check how well studentscan measure lengths and angles,experiment systematically, andfollow a deductive proof.

Closing the Lesson

The Converse of the PythagoreanTheorem is true and can be usedto determine right angles. Acommon proof actually uses thePythagorean Theorem itself.


These exercises help studentspractice both the PythagoreanTheorem and its converse.

ASSIGNING HOMEWORK

Essential 1–7, 9–15, 19, 20

Performanceassessment 8

Portfolio 16

Journal 20

Group 17, 18

Review 21–25

|


Exercise 2 This is a good place topoint out the power of recog-nizing Pythagorean multiples.The triple 50-120-130 is basedon the Pythagorean primitive 5-12-13, so sides of those lengthsform a right triangle.

In Exercises 9–11, find y.

9. Both quadrilaterals 10. 11.are squares.

12. The lengths of the three sides of a right 13. Find the area of a right triangle with triangle are consecutive even integers. hypotenuse length 17 cm and one leg Find them. length 15 cm.

14. How high on a building will a 15-foot ladder 15. The congruent sides of an isosceles triangle touch if the foot of the ladder is 5 feet from measure 6 cm, and the base measures 8 cm.the building? Find the area.

16. Find the amount of fencing in linear feet needed for the perimeter of a rectangular lot with a diagonal length 39 m and a side length 36 m.

17. A rectangular piece of cardboard fits snugly on a diagonal in this box.

a. What is the area of the cardboard rectangle?

b. What is the length of the diagonal of the cardboard rectangle?

18. Look back at the start of the proof of the Converse of the Pythagorean Theorem. Copy the conjecture, the given, the show, the plan, and the two diagrams. Use the plan to complete the proof.

19. What’s wrong with this picture? 20. Explain why �ABC is a right triangle.

Review

21. Identify the point of concurrency from theconstruction marks. centroid

6 m

3 m 9 m

B

A D C

3.75 cm 2 cm

4.25 cm

74.8 cm

1442 cm2

102 m

17.9 cm214.1 ft

60 cm26, 8, 10

15 cm

y 25 cm2

y � 25 cmy � 17.3 m

40 cm60 cm

20 cm

�

25

(–7, y)

y

x

18 m

25 m y

y � 24 units

Sample answer: Thenumbers given satisfythe PythagoreanTheorem, so the trian-gle is a right triangle;but the right angleshould be inscribed inan arc of 180°. Thusthe triangle is not aright triangle.

Sample answer:BD2 � 62 � 32 � 27;BC2 � BD2 � 92 � 108;then AB2 � BC2 �AC2 (36 � 108 � 144),so �ABC is a right tri-angle by the Converseof the PythagoreanTheorem.

Exercise 9 Here is another situa-tion in which Pythagorean triplescan be applied. For the trianglewith y as the hypotenuse, the legshave lengths 15 cm and 20 cm.Each length has a commonfactor of 5 cm. Because 3-4-5 is acommon Pythagorean primitive,the hypotenuse must have length5(5 cm) � 25 cm.

Exercise 10 This exercise providesgroundwork for Chapter 12 work with the unit circle intrigonometry.

Exercises 12–16 Encouragestudents to draw pictures.

Exercise 12 Students might missthe condition that the integersare even. If appropriate, [Ask] “Ifx is the first even integer, howwould the second integer bedescribed?” [x � 2] “Is thereonly one triple of consecutiveeven integers?” [Half of any suchtriple is a triple of consecutiveintegers, say, x � 1, x, and x � 1.The algebraic equation (x � 1)2 �x2 � (x � 1)2 has only two solu-tions: x � 0 and x � 4.]

Exercise 15 [Ask] “In an isoscelestriangle, what does an altitudefrom the vertex angle to the basedo to the base?” [It is the perpen-dicular bisector of the base; it isthe same as the median.]

Exercise 16 [Language] Linear feetrefers to the length of the fence,whereas square feet measures area.

Exercise 17 [Alert] Some studentswill have difficulty visualizingthe box. They might create athree-dimensional model.Students are being graduallyintroduced to the PythagoreanTheorem in three dimensions.As in Exercises 11 and 13–16,exactness of answers will vary.The answers given assume thatmeasurements given in the exer-cises are exact and thus answersto the nearest square cm ornearest tenth of a cm are reason-able. In real life the exactness ofan answer depends on how itwill be used and further knowl-edge of given measurements.

18. Because �DEF is a right triangle, a2 � b2 � x2.By substitution, c2 � x2 and c � x. Therefore,�EFD � �BCA by SSS and �C � �F by CPCTC.Hence, �C is a right angle and �BCA is a righttriangle.

LESSON 9.2 The Converse of the Pythagorean Theorem 471

3.8

22. Line CF is tangent to circle D at C. 23. What is the probability of randomly selecting The arc measure of CE� is a. Explain three points that form an isosceles triangle why x � ��

12��a. from the 10 points in this isometric grid?

24. If the pattern of blocks continues, what will be the surface area of the 50th solid inthe pattern?

25. Sketch the solid shown, but with the two blue cubes removed and the red cube moved to cover the visible face of the green cube.

790 square units

D

E

C

a

x

F

�130�

IMPROVING YOUR ALGEBRA SKILLS

Algebraic Sequences III

Find the next three terms of this algebraic sequence.

x9, 9x8y, 36x7y2, 84x6y3, 126x5y4, 126x4y5, 84x3y6, �? , �? , �?

The outlines of stackedcubes create a visualimpact in this untitledmodule unit sculpture by conceptual artist Sol Lewitt.

Exercise 22 This exercise can beapproached through finding thatthe complement of x is 90 � �

a2�,

through using the properties ofan isosceles right triangle, or byconsidering the limit as a secantline through C rotates to becomethe tangent line (and the inter-cepted arc becomes the arc withmeasure a).

Exercise 23 If students are havingdifficulty, ask how they mightcount triangles systematically.One approach is to considerwhere the vertex angle might go.Symmetry helps.

Exercise 24 As needed, focusstudents’ attention on thenumber of square faces beingadded at each step.

EXTENSIONS

A. Pose this problem: If the sumof the squares of the lengths ofthe two shorter sides of a triangleis less than the square of thelength of the longest side, whatcan you conjecture about theangle opposite the longest side? If the sum of the squares of thelengths of the two shorter sides of a triangle is greater than thesquare of the length of the longestside, what can you conjectureabout the angle opposite thelongest side? [If the sum is less,the angle is obtuse. If the sum isgreater, the angle is acute.]

B. Have students use geometrysoftware to draw triangles. Havethem label and measure anglesand sides, calculate squares of thelengths of the sides or constructsquares on the sides and measuretheir area, and drag vertices untilthe sum of the squares of thelengths of the two smallest sidesequals the square of the length ofthe largest side. Students shouldfind a right angle.


Students might think of number combinations,Pascal’s triangle, or symmetry. Or they mightsee this pattern: After the first term, each coeffi-cient can be determined by multiplying theprevious coefficient by the exponent on x in theprevious term and then dividing by the numberof the term (counting from zero). For example,

IMPROVING ALGEBRA SKILLS

the term that includes x6 is the third term, so its

coefficient is �7(3

36)� � 84. The next three terms

in the sequence are 36x2y7, 9xy8, and y9.

6.3

2.3

1.8

See page 775 for answers toExercises 22 and 25.

USING YOUR ALGEBRA SKILLS 8 Radical Expressions 473

Radical ExpressionsWhen you work with the Pythagorean Theorem, you often get radical expressions,such as �50�. Until now you may have left these expressions as radicals, or you mayhave found a decimal approximation using a calculator. Some radical expressionscan be simplified. To simplify a square root means to take the square root of anyperfect-square factors of the number under the radical sign. Let’s look at anexample.

Simplify �50�.

One way to simplify a square root is to look for perfect-square factors.

Another approach is to factor the number as far as possible with prime factors.

You might argue that 5�2� doesn’t look any simpler than �50�. However, in thedays before calculators with square root buttons, mathematicians used paper-and-pencil algorithms to find approximate values of square roots. Working with thesmallest possible number under the radical made the algorithms easier to use.

Giving an exact answer to a problem involving a square root is important in anumber of situations. Some patterns are easier to discover with simplified squareroots than with decimal approximations. Standardized tests often express answersin simplified form. And when you multiply radical expressions, you often have tosimplify the answer.

Squaring and taking the square root are inverse operations—they undo each other.So, equals 5. 52

Write 50 as a setof prime factors.

Look for any square factors (factors that appear twice).

�� 50 5 � 5 � 2 52 � 2 52� � � 22 5

The largest perfect-squarefactor of 50 is 25.

25 is a perfect square, soyou can take its square root.

� 50 25 � 2 � 25 � � 22 5

LGEBRA SKILLS 1 ● USING YOUR ALGEBRA SKILLS 1 ● USING YOUR ALGEBRA SKILLS 8 ● USING YOUSING YOUR ALGEBRA SKILLS 8

EXAMPLE A

� Solution

LESSON OBJECTIVES

� Learn to simplify square roots� Learn to multiply radical expressions

NCTM STANDARDS

CONTENT PROCESS

� Number Problem Solving

� Algebra Reasoning

Geometry Communication

Measurement Connections


USING YOURALGEBRA SKILLS 8

PLANNING

LESSON OUTLINE

One day or partial day:15 min Examples

20 min Exercises

MATERIALS

none

TEACHING

Being able to work with radicalexpressions will help studentslater see patterns in special righttriangles.

One step Draw a right trianglewith legs labeled with lengths 4 and 8. Ask what multiple of4 the length of the hypotenuse is. Students might divide thehypotenuse into four pieces andlook at squares of each piece.Or they might rewrite �80� as a product of 4 and a square root. During Sharing, ask aboutworking backward, as inExample B.

INTRODUCTION

[Language] The symbol for thenonnegative square root, ��, isan example of a radical. Roots to other powers are also calledradicals.

� EXAMPLE A

You might give a definition suchas “Prime factors are factors thatcan’t be broken down intosmaller factors.”

Emphasize that 5�2� equals �50�exactly; it is not an approximation.

ALGEBRA SKILLS 8 ● USING YOUR ALGEBRA SKILLS 8 ● USING YOUR ALGEBRA SKILLS 8 ● USING Y

Multiply 3�6� by 5�2�.

To multiply radical expressions, associate and multiply the quantities outside theradical sign, and associate and multiply the quantities inside the radical sign.

�3�6��5�2�� 3 � 5 � �6 � 2� � 15 � �12� � 15 � �4 � 3� � 15 � 2�3� � 30�3�

EXAMPLE B

� Solution

EXERCISESIn Exercises 1–5, express each product in its simplest form.

1. ��3��2�� 2. ��5��2 3. �3�6��2�3�� 4. �7�3��2 5. �2�2��2

In Exercises 6–20, express each square root in its simplest form.

6. �18� 7. �40� 8. �75� 9. �85� 10. �96�

11. �576� 12. �720� 13. �722� 14. �784� 15. �828�

16. �2952� 17. �5248� 18. �8200� 19. �11808� 20. �16072�

21. What is the next term in the pattern? �2952�, �5248�, �8200�, �11808�, �16072�, . . .

14�82�12�82�10�82�8�82�6�82�

6�23�2819�2�12�5�24

4�6��85�5�3�2�10�3�2�

81475�6�18�2�

�

IMPROVING YOUR VISUAL THINKING SKILLS

Folding Cubes II

Each cube has designs on three faces. When unfolded, which figure at right could it become?

1. A. B. C. D.

2. A. B. C. D.

�20992�� 16�82�


IMPROVING VISUAL THINKING SKILLS

If students are having difficulty, point outthat a good problem-solving technique is toeliminate as many choices as possible. Forexample, 1A can be eliminated because thebase of the green face is touching the redface, unlike its position in the original cube.

1. D 2. C

|


Exercise 2 Squaring and taking the square root areopposite operations.

Exercise 9 It is possible to factor 85 as 5 � 17, butneither factor is a perfect square.

Exercise 21 If students are mystified, refer them toExercises 16–20.

� EXAMPLE B

You may want to mention thatthe commutative property ofmultiplication allows you torewrite 3 � �6� � 5 � �2� as 3 � 5 � �6� � �2�.

SHARING IDEAS

You might ask the class to critiquethis shortcut for Example A as ashortcut for all cases: ��5 � 5 ��2�immediately becomes 5�2�because “you can move anyfactor that appears twice insidethe radical to one appearanceoutside the radical.” Keep askingfor justification, using the generalrules about square roots of prod-ucts and of squares.

Ask whether the relationship �a � a � b�� a�b� can be repre-sented by lengths, rememberingthat square roots are often repre-sented by sides of squares. Thisquestion foreshadows work witharea ratios in Lesson 11.5.

Assessing ProgressYou can assess students’ previousunderstanding of radical expres-sions as well as their ability tosee and generalize patterns.

Closing the Lesson

Two rules are useful in rewritingsquare roots and in multiplyingradical expressions: The (posi-tive) square root of a product isthe product of the square roots,and the square root of the squareof a number is the (absolutevalue of) the number.


Though students are taught tosimplify radical expressions, theyare not taught to rationalize thedenominators. You might decideto teach that also.

ASSIGNING HOMEWORK

Essential 1–21 odds

Group 2–20 evens

LESSON 9.3 Two Special Right Triangles 475

L E S S O N

9.3In an isosceles triangle,

the sum of the square roots

of the two equal sides is

equal to the square root of

the third side.

THE SCARECROW IN THE 1939FILM THE WIZARD OF OZ

L E S S O N

9.3Two Special Right TrianglesIn this lesson you will use the Pythagorean Theorem to discover some relationshipsbetween the sides of two special right triangles.

One of these special triangles is an isosceles right triangle, also called a 45°-45°-90°triangle. Each isosceles right triangle is half a square, so they show up often inmathematics and engineering. In the next investigation, you will look for a shortcutfor finding the length of an unknown side in a 45°-45°-90° triangle.

Investigation 1Isosceles Right Triangles

Step 1 Sketch an isosceles right triangle. Label the legs l and the hypotenuse h.

Step 2 Pick any integer for l, the length of the legs.Use the Pythagorean Theorem to find h.Simplify the square root.

Step 3 Repeat Step 2 with several differentvalues for l. Share results withyour group. Do you see anypattern in the relationshipbetween l and h?

Step 4 State your nextconjecture in terms oflength l.

Isosceles Right Triangle Conjecture

In an isosceles right triangle, if the legs have length l, then the hypotenuse has length �? . l�2�

C-84

45°

45°

An isosceles right triangle

l

l

h

11

? 22

?

33

?

10 10

?

LESSON OBJECTIVES

� Practice simplifying square roots� Discover relationships among the lengths of the sides of a

45°-45°-90° triangle and a 30°-60°-90° triangle� Develop problem-solving skills and cooperative behavior

NCTM STANDARDS

CONTENT PROCESS





Data/Probability Representation

PLANNING

LESSON OUTLINE


10 min Sharing

5 min Closing

5 min Exercises

MATERIALS

� square and isometric dot paper

TEACHING

Two special right triangles occurso often in real life (and oncollege entrance exams andachievement tests) that it’s goodto know the ratios of their sidelengths.

One step Ask students to draw ondot paper right triangles inwhich the two legs are congruentand right triangles in which thehypotenuse has length twice thatof one of the legs. They shouldlook for the length of the thirdsides of these triangles andmeasure the angles.

Guiding Investigation 1

Step 3 If students are havingdifficulty seeing a pattern,suggest that they systematicallytry consecutive integers andmake a table of the results.

Investigation 230º-60º-90º TrianglesLet’s start by using a little deductive thinking to find therelationships in 30°-60°-90° triangles. Triangle ABC isequilateral, and CD� is an altitude.

Step 1 What are m�A and m�B? What are m�ACD andm�BCD? What are m�ADC and m�BDC?

Step 2 Is �ADC � �BDC? Why?

Step 3 Is AD�� BD�? Why? How do AC and AD compare? In a 30°-60°-90° triangle, will this relationship between the hypotenuse and theshorter leg always hold true? Explain.

Step 4 Sketch a 30°-60°-90° triangle. Choose anyinteger for the length of the shorter leg.Use the relationship from Step 3 and thePythagorean Theorem to find the lengthof the other leg. Simplify the square root.

Step 5 Repeat Step 4 with several different valuesfor the length of the shorter leg. Shareresults with your group. What is therelationship between the lengths of thetwo legs? You should notice a pattern in your answers.

Step 6 State your next conjecture in terms of the length of the shorter leg, a.

30°-60°-90° Triangle Conjecture

In a 30°-60°-90° triangle, if the shorter leg has length a, then the longer leghas length �? and the hypotenuse has length �? . 2a

C-85

You can also demonstrate this propertyon a geoboard or graph paper, asshown at right.

The other special right triangle is a 30°-60°-90° triangle. If you fold anequilateral triangle along one of itsaltitudes, the triangles you get are 30°-60°-90° triangles. A 30°-60°-90° triangle ishalf an equilateral triangle, so it also shows up often in mathematics andengineering. Let’s see if there is a shortcut for finding the lengths of its sides.

2

1

12

22

2

3

323

30°

60°

A 30°-60°-90° triangle

C

A D B

1

2? 2

4?

7

14

?

Step 1 60°; 30°; 90°

Step 2 yes, SAS, ASA,or SAA

Step 3 Yes, CPCTC.AC � 2AD; yes all 30°-60°-90° triangles are similar.

a�3�


Steps 1–3 Students can use pattypaper to answer some of thequestions in Steps 2 and 3without doing the measurementsin Step 1. They might also recallthat every altitude is a median.

SHARING IDEAS

After the class reaches consensusabout what conjectures to recordin their notebooks, ask how torestate the conjectures usingratios. [For a 45°-45°-90° triangle,the side lengths have the ratio1:1:�2�; for a 30°-60°-90° triangle,they have the ratio 1:�3�:2.]

Ask whether these ratios can be represented geometrically,remembering that square rootsare often shown as sides ofsquares. Students can draw the triangles on dot paper (orisometric dot paper) and calculatethe areas of appropriate squares.

[Ask] “How do you know whichangle is the 30° angle in a 30°-60°-90° triangle?” [By theSide-Angle Inequality Conjec-ture, it’s the angle opposite theshortest side.]

[Ask] “How can these specialtriangles be constructed withstraightedge and compass?”[Students can construct perpen-dicular lines and then lay outequal segments from the inter-section point to get a 45°-45°-90°triangle. Now that they know the30°-60°-90° Triangle Conjecture,they can construct one leg and ahypotenuse to determine thethird side.]

Wonder aloud whether the 30°-60°-90° Triangle Conjecturecan be proved for all triangles,even if none of the lengths areintegers. After students havemade suggestions, direct theirattention to the proof in thestudent book, asking them tocritique and rewrite the reasoningin order to understand it better.

Assessing ProgressAssess students’ ability to generate isosceles righttriangles and equilateral triangles, to simplifysquare roots, to measure angles, to find patterns,and to follow a deductive proof. Also check theirunderstanding of altitudes of isosceles triangles,SAS, and the Pythagorean Theorem.

Closing the Lesson

In a 45°-45°-90° triangle, if the legs have length l,then the hypotenuse has length l�2�. In a 30°-60°-90° triangle, if the leg opposite the 30° angle haslength a, then the hypotenuse has length 2a and theother leg has length a�3�.


You can use algebra to verify that the conjecture will hold true for any 30°-60°-90° triangle.

Proof: 30°-60°-90° Triangle Conjecture

�2a�2 � a2 � b2 Start with the Pythagorean Theorem.

4a2 � a2 � b2 Square 2a.

3a2 � b2 Subtract a2 from both sides.

a�3� � b Take the square root of both sides.

Although you investigated only integer values, the proof shows that any number,even a non-integer, can be used for a. You can also demonstrate this property forinteger values on isometric dot paper.

1

2

2

3

264

3

3

3

3

3

3

3

3 3

2a30°

a

b

EXERCISESIn Exercises 1–8, use your new conjectures to find the unknown lengths.All measurements are in centimeters.

1. a � �? 2. b � �? 3. a � �? , b � �?

4. c � �? , d � �? 5. e � �? , f � �? 6. What is the perimeter ofsquare SQRE?E

S

R

Q

18 217 3

f

e30°

20

c

60°

d

72 cm

60°5

b

a

b

13 2a

72

13 cm72�2� cm

�

You will need

Construction toolsfor Exercises 19 and 20

10 cm, 5�3� cm

10�3� cm, 10 cm 34 cm, 17 cm



The exercises give practice withthe two special right triangles.

ASSIGNING HOMEWORK

Essential 1–11


Portfolio 17

Group 12–16

Review 19–23

|


Exercise 3 If students aren’t surewhat to do, suggest that in a 30°-60°-90° triangle it often helpsto locate the shortest side first.[Ask] “Where is the 30-degreeangle?”

7. The solid is a cube. 8. g � �? , h � �? 9. What is the area ofd � �? the triangle?

10. Find the coordinates of P. 11. What’s wrong with this picture?

12. Sketch and label a figure to demonstrate that �27� is equivalent to 3�3�.(Use isometric dot paper to aid your sketch.)

13. Sketch and label a figure to demonstrate that �32� is equivalent to 4�2�.(Use square dot paper or graph paper.)

14. In equilateral triangle ABC, AE�, BF�, and CD� are all angle bisectors,medians, and altitudes simultaneously. These three segments divide the equilateral triangle into six overlapping 30°-60°-90° triangles and six smaller, non-overlapping 30°-60°-90° triangles.

a. One of the overlapping triangles is �CDB. Name the other five triangles that are congruent to it.

b. One of the non-overlapping triangles is �MDA. Name the other five triangles congruent to it.

15. Use algebra and deductive reasoning to show that the Isosceles Right Triangle Conjecture holds true for any isosceles right triangle.Use the figure at right.

16. Find the area of an equilateral triangle whose sides measure 26 meters.

17. An equilateral triangle has an altitude that measures 26 meters.Find the area of the triangle to the nearest square meter.

18. Sketch the largest 45°-45°-90° triangle that fits in a 30°-60°-90° triangle. What is theratio of the area of the 30°-60°-90° triangle to the area of the 45°-45°-90° triangle?

390 m2

169�3� m2

�MDB, �MEB, �MEC, �MFC, �MFA

�CDA, �AEC, �AEB, �BFA, �BFC

17

158

30°60°

(?, ?)

(0, –1)

(1, 0)

P

45°

y

x

8

g

130

12030°h

12 cm

dE

A

D

G

C

B

H

F

16 cm212�3� cm

F

C

M

E

BDA

x

cx

50 cm, 100 cm

A 30°-60°-90° triangle must havesides whose lengths are multiples of1, 2, and �3�. The triangle showndoes not reflect this rule.

� , �1��2�

1��2�

Exercise 7 This is another exampleof using the PythagoreanTheorem in three dimensions.[Ask] “How would you generalizethe Pythagorean Theorem tothree dimensions?” [In a rightrectangular prism, the space diag-onal (d) can be found from thethree dimensions of the prism (a, b, c): d2 � a2 � b2 � c2.

Exercise 10 In situations in whichyou want to find the coordinatesof a point, it’s often useful to drawsegments whose lengths are thosecoordinates. [Link] Students willwork with the unit circle intrigonometry.

12. possible answer:

13. possible answer:

Exercise 16 Ask students how to find a special triangle in apicture of this situation.

Exercise 18 Students using geom-etry software might discover thatletting the hypotenuses (not theright angles) coincide produces a slightly larger triangle.

18.

30°

45°

45°

15°

x

2x

x

x

2

x 3

4

424

333

3

1

2

6

2

21

1

3

3

�3��1 15. c2 � x2 � x2 Start with the Pythagorean Theorem.

c2 � 2x2 Combine like terms.

c � x�2� Take the square root of both sides.


�


Mudville Monsters

The 11 starting members of the Mudville Monsters football team and their coach,Osgood Gipper, have been invited to compete in the Smallville Punt, Pass, and KickCompetition. To get there, they must cross the deep Smallville River. The only way acrossis with a small boat owned by two very small Smallville football players. The boat holdsjust one Monster visitor or the two Smallville players. The Smallville players agree to helpthe Mudville players across if the visitors agree to pay $5 each time the boat crosses the

river. If the Monsters have a total of $100 among them, do they have enough moneyto get all players and the coach to the other side of the river?

Review

Construction In Exercises 19 and 20, choose either patty paper or a compass andstraightedge and perform the constructions.

19. Given the segment with length a below, construct segments with lengths a�2�, a�3�,and a�5�.

20. Mini-Investigation Draw a right triangle with sides of lengths 6 cm, 8 cm, and 10 cm.Locate the midpoint of each side. Construct a semicircle on each side with themidpoints of the sides as centers. Find the area of each semicircle. What relationship do you notice among the three areas?

21. The Jiuzhang suanshu is an ancient Chinese mathematicstext of 246 problems. Some solutions use the gou gu, theChinese name for what we call the PythagoreanTheorem. The gou gu reads �gou�2 � �gu�2 � (xian)2.Here is a gou gu problem translated from the ninthchapter of Jiuzhang.

A rope hangs from the top of a pole with three chih of itlying on the ground. When it is tightly stretched so thatits end just touches the ground, it is eight chih from thebase of the pole. How long is the rope?

22. Explain why m�1 � m�2 � 90°. 23. The lateral surface area of the cone below is unwrapped into a sector. What is the angle at the vertex of the sector?

1

2

80°

�763� � 12.2 chih

a

l � 27 cm, r � 6 cm

?

r

ll

19. Construct an isosceles righttriangle with legs of length a,construct a 30°-60°-90° trianglewith legs of lengths a and a�3�,and construct a right trianglewith legs of lengths a�2� anda�3�.

Exercise 20 This mini-investiga-tion foreshadows area ratios inLesson 11.5.

20. Areas: 4.5� cm2, 8� cm2,12.5� cm2. 4.5� � 8� � 12.5�,that is, the sum of the areas ofthe semicircles on the two legsis equal to the area of the semi-circle on the hypotenuse.

22. Extend the rays that form the right angle.m�4 � m�5 � 180° by theLinear Pair Conjecture, and it’s given that m�5 � 90°.� m�4 � 90°.m�2 � m�3 � m�4 �m�2 � m�3 � 90° � 180°.� m�2 � m�3 � 90°.m�3 � m�1 by AIA.� m�1 � m�2 � 90°.

EXTENSION

Use Take Another Lookactivity 4 or 5 on page 502.

1

34

2

5

2a

2a3a

a 2a 3a5a

a

a


Each Monster visitor must go across alone. Itmakes no sense for one to return, so beforeeach Monster crosses, the two Smallville playersmust cross and leave one on the other side tobring the boat back on the next trip. This willrequire 23 round trips for the 11 players and the coach. The team has enough money for


20 one-way trips. Even without seeing how toarrange the Smallville players to return the boatat least 11 times, the Monsters don’t haveenough money.

8.5

9.1

2.6 8.7


A Pythagorean FractalIf you wanted to draw a picture to state thePythagorean Theorem without words, you’dprobably draw a right triangle with squares oneach of the three sides. This is the way you firstexplored the Pythagorean Theorem in Lesson 9.1.

Another picture of the theorem is even simpler:a right triangle divided into two right triangles.Here, a right triangle with hypotenuse c is dividedinto two smaller triangles, the smaller withhypotenuse a and the larger with hypotenuse b. Clearly, their areas add up to thearea of the whole triangle. What’s surprising is that all three triangles have the sameangle measures. Why? Though different in size, the three triangles all have the sameshape. Figures that have the same shape but not necessarily the same size are calledsimilar figures. You’ll use these similar triangles to prove the PythagoreanTheorem in a later chapter.

A beautifully complex fractal combinesboth of these pictorial representationsof the Pythagorean Theorem. Thefractal starts with a right trianglewith squares on each side. Thensimilar triangles are built onto thesquares. Then squares are builtonto the new triangles, and so on.In this exploration, you’ll createthis fractal.

b

aa b

c

LESSON OBJECTIVES

� Explore a right triangle fractal� Look for and describe patterns in the fractal

NCTM STANDARDS

CONTENT PROCESS

Number Problem Solving

Algebra � Reasoning


� Measurement Connections


EXPLORATION

PLANNING

LESSON OUTLINE

One day:45 min Activity

MATERIALS

� The Geometer’s Sketchpad� The Right Triangle Fractal (W),

optional� Sketchpad demonstration A Right

Triangle Fractal, optional

TEACHING

Students construct a fractal basedon the Pythagorean Theorem.

The student book includes a very brief mention of similarity.You need not elaborate; the topicwill be formally addressed inChapter 11. Also, a proof of thePythagorean Theorem based onsimilar triangles is an exercise inLesson 13.7.

uiding the Activity

Sketchpad provides many tools for constructions like this. Forexample, half of this fractal can be made easily by constructing asquare, then constructing a righttriangle on one side, and thenusing the iteration tool. The fullfractal can be made with customtools. If students are not proficientenough using Sketchpad to createthe fractal on their own, use The Right Triangle Fractal worksheetwith detailed steps for creating thefractal in Sketchpad or use theSketchpad demonstration.

Step 2 Changing the ratio of thelengths of the legs of the originaltriangle affects the shape of thefractal. See page 56 for a quiltdesign that uses isosceles righttriangles.

G

ActivityThe Right Triangle Fractal

You will need

● the worksheet The Right TriangleFractal (optional)

The Geometer’s Sketchpad software uses customtools to save the steps of repeated constructions.They are very helpful for fractals like this one.

Step 1 Use The Geometer’s Sketchpad to create thefractal on page 480. Follow the Procedure Note.

Notice that each square has two congruenttriangles on two opposite sides. Use a reflectionto guarantee that the triangles are congruent.

After you successfully make the Pythagoreanfractal, you’re ready to investigate its fascinatingpatterns.

Step 2 First, try dragging a vertex of the original triangle.

Step 3 Does the Pythagorean Theorem still apply to the branches of this figure? That is,does the sum of the areas of the branches on the legs equal the area of thebranch on the hypotenuse? See if you can answer without actually measuring allthe areas.

Step 4 Consider your original sketch to be a single right triangle with a square built oneach side. Call this sketch Stage 0 of your fractal. Explore these questions.

a. At Stage 1, you add three triangles and six squares to your construction. On a piece of paper, draw a rough sketch of Stage 1. How much area do you addto this fractal between Stage 0 and Stage 1? (Don’t measure any areas toanswer this.)

b. Draw a rough sketch of Stage 2. How much area do you add between Stage 1and Stage 2?

c. How much area is added at any new stage?

d. A true fractal exists only after an infinite number of stages. If you could builda true fractal based on the construction in this activity, what would be its total area?

Step 5 Give the same color and shade to sets of squares that are congruent. What doyou notice about these sets of squares other than their equal area? Describe anypatterns you find in sets of congruent squares.

Step 6 Describe any other patterns you can find in the Pythagorean fractal.

yes

1. Use the diameter of a circleand an inscribed angle tomake a triangle that alwaysremains a right triangle.

2. It is important to constructthe altitude to thehypotenuse in each trianglein order to divide it intosimilar triangles.

3. Create custom tools to makesquares and similar trianglesrepeatedly.

Step 4c At every new stage,the area added equals the area of the Stage 0 figure plus the area of the original triangle.

Step 4d Its area would be infinite.

Step 5 Sample observa-tions: Each square on the hypotenuse branch has a mirror image on one of the leg branches. The squares of equal area on the leg branches all face the same direction; that is, corre-sponding sides in all the squares are parallel. The number of squares of a particular area on the leg branches is one more than the number ofsquares of the next largest area on those branches.

Step 3 At each stage, the sum ofthe areas of the squares addedto each branch equals the areaof the original square on thatbranch.

Step 4a

At Stage 1, the area of thetriangle on the hypotenusebranch equals the area of theoriginal triangle, and the sum ofthe areas of the triangles on theother two branches also equalsthe area of the original triangle.The sum of the two addedsquares equals the area of oneof the original squares by thePythagorean Theorem. Thus the total added area is twice the area of the original triangleplus the areas of all the originalsquares, or the area of theStage 0 figure plus the area of the original triangle.

Step 4b

The total added area is, again,the area of the Stage 0 figureplus the area of the originaltriangle.

Step 4 At each stage, triangles whose total area istwice that of the original are added, and squareswhose total area is twice that of the square on theoriginal hypotenuse are added. If the originaltriangle has the usual lengths a, b, and c, then anarea equal to ab � 2c2 is added at each stage. Thisnumber becomes larger without bound as thefractal design expands.

Step 5 Students can see many patterns. For example,there is reflectional symmetry over a midsegment ofthe square on the original hypotenuse.

Step 6 Students’ responses will vary. Check them forvalidity and justification.

EXPLORATION A Pythagorean Fractal 481

L E S S O N

9.4


Story ProblemsYou have learned that drawing a diagram will help you to solve difficult problems.By now you know to look for many special relationships in your diagrams, such ascongruent polygons, parallel lines, and right triangles.

What is the longest stick that will fit inside a 24-by-30-by-18-inch box?

Draw a diagram.

You can lay a stick with length d diagonally at thebottom of the box. But you can position an evenlonger stick with length x along the diagonal ofthe box, as shown. How long is this stick?

Both d and x are the hypotenuses of right triangles, butfinding d 2 will help you find x.

302 � 242 � d2 d2 � 182 � x2

900 � 576 � d2 1476 � 182 � x2

d2 � 1476 1476 � 324 � x2

1800 � x2

x � 42.4

The longest possible stick is about 42.4 in.

EXAMPLE

� Solution

You may be disappointed if

you fail, but you are doomed

if you don’t try.

BEVERLY SILLS

L E S S O N

9.4

EXERCISES1. A giant California redwood tree 36 meters tall cracked

in a violent storm and fell as if hinged. The tip of theonce beautiful tree hit the ground 24 meters from thebase. Researcher Red Woods wishes to investigate thecrack. How many meters up from the base of the treedoes he have to climb?

2. Amir’s sister is away at college, and he wants to mail her a 34 in. baseball bat. Thepacking service sells only one kind of box, which measures 24 in. by 2 in. by 18 in.Will the box be big enough? No. The space diagonal of the box is 30.1 in.

10 m

�

FUNKY WINKERBEAN by Batiuk. Reprinted with special permission of North America Syndicate.

x18 in.

24 in. 30 in.d

24 m

x

LESSON OBJECTIVES

� Apply the Pythagorean Theorem and its converse� Develop reading comprehension and problem-solving skills

NCTM STANDARDS

CONTENT PROCESS



� Geometry Communication



PLANNING

LESSON OUTLINE

One day:25 min Example and Exercises

15 min Sharing

5 min Closing

MATERIALS

� calculators

TEACHING

The Pythagorean Theorem hasmany applications. This lessonconsists primarily of exercises.

One step Pose the problem fromthe Example and have studentsdiscuss it without opening theirbooks.

� EXAMPLE

The distance x is called the spacediagonal. To find its length, youneed d2, but you don’t need tofind d.

Assessing ProgressThrough their work on the exercises, you can assessstudents’ understanding ofthe Pythagorean Theorem.

SHARING IDEAS

Have several groups preparesome of their solutions to theexercises on transparencies andshare them with the class. Keepasking whether the results seemreasonable.

Closing the Lesson

Emphasize that a good first stepin solving application problems isto draw a picture. Then studentsshould examine the picture forcommon geometric shapes, suchas right triangles, and apply whatthey know about those shapes.

3. Meteorologist Paul Windward and geologist Rhaina Stone are rushing to a paleontology conference in Pecos Gulch.Paul lifts off in his balloon at noon from Lost Wages,heading east for Pecos Gulch Conference Center. With the wind blowing west to east, he averages a land speed of30 km/hr. This will allow him to arrive in 4 hours, just as the conference begins. Meanwhile, Rhaina is 160 km north of Lost Wages. At the moment of Paul’s lift off, Rhaina hops into an off-roading vehicle and heads directly for the conference center. At what average speed must she travel to arrive at the same time Paul does?

4. A 25-foot ladder is placed against a building. The bottom ofthe ladder is 7 feet from the building. If the top of theladder slips down 4 feet, how many feet will the bottomslide out? (It is not 4 feet.)

5. The front and back walls of an A-frame cabin are isoscelestriangles, each with a base measuring 10 m and legsmeasuring 13 m. The entire front wall is made of glass 1 cm thick that cost $120/m2. What did the glass for thefront wall cost?

6. A regular hexagonal prism fitsperfectly inside a cylindrical box withdiameter 6 cm and height 10 cm. Whatis the surface area of the prism? Whatis the surface area of the cylinder?

7. Find the perimeter of an equilateral triangle whose medianmeasures 6 cm.

8. APPLICATION According to the Americans with DisabilitiesAct, the slope of a wheelchair ramp must be no greater than�112�. What is the length of ramp needed to gain a height of

4 feet? Read the Science Connection on the top of page 484and then figure out how much force is required to go upthe ramp if a person and a wheelchair together weigh200 pounds. 48.2 ft; 16.6 lb

area: 60 m2; cost: $7200

8 ft

50 km/hr

Career

Meteorologists study the weather and the atmosphere. They also look at airquality, oceanic influence on weather, changes in climate over time, and evenother planetary climates. They make forecasts using satellite photographs,weather balloons, contour maps, and mathematics to calculate wind speed orthe arrival of a storm.

surface area of prism � �27�3� � 180� cm2 �226.8 cm2; surface area of cylinder � 78� cm2 � 245.0 cm2

cm36��3�

LESSON 9.4 Story Problems 483


You may want to divide up theexercises and have different groupswork on different exercises.

ASSIGNING HOMEWORK

Essential 1–5


Portfolio 9

Group 7–10

Review 11–20

|


Exercise 1 If students are havingdifficulty, [Ask] “If the tree wasoriginally 36 meters tall and it cracked with x meters leftstanding, how much has fallen?”[36 � x] You might also usespecific numbers until studentssee the general case.

Exercise 5 As needed, suggest thatstudents draw altitudes to createright triangles.

Exercise 6 [Alert] Some studentsmay have forgotten that a hexagoninscribed in a circle has sides thesame length as the circle’s radius.

Exercise 7 After students havedrawn pictures, you might[Ask] “Which side of the 30°-60°-90° triangle do we know?” [the longer leg]

For Exercises 9 and 10, refer to the above Science Connection about inclined planes.

9. Compare what it would take to lift an object these three different ways.

a. How much work, in foot-pounds, is necessary to lift 80 pounds straight up 2 feet?

b. If a ramp 4 feet long is used to raise the 80 pounds up 2 feet, how much force, inpounds, will it take?

c. If a ramp 8 feet long is used to raise the 80 pounds up 2 feet, how much force, inpounds, will it take?

10. If you can exert only 70 pounds of force and you need to lift a 160-pound steeldrum up 2 feet, what is the minimum length of ramp you should set up?

Review

11. If the area of the red square piece is 4 cm2, what are the dimensions of the other six pieces?

Horse with RiderCat

4.6 ft

20 lb

40 lb

160 ft-lb

Swan

Rabbit

Recreation

This set of enameled porcelain qi qiao bowls can be arranged to form a 37-by-37 cm square (as shown) or other shapes, or used separately. Each bowlis 10 cm deep. Dishes of this type are usually used to serve candies, nuts, driedfruits, and other snacks on special occasions.

The qi qiao, or tangram puzzle, originated in China and consists of sevenpieces—five isosceles right triangles, a square, and a parallelogram. The puzzleinvolves rearranging the pieces into a square, or hundreds of other shapes (a few are shown below).

�

Science

It takes less effort to roll objects up an inclined plane, or ramp, than to lift them straight up. Work is a measure of force applied over distance, and you calculate it as a product of force and distance. For example, a force of100 pounds is required to hold up a 100-pound object. The work required to lift it 2 feet is 200 foot-pounds. But if you use a 4-foot-long ramp to roll it up, you’ll do the 200 foot-pounds of work over a 4-foot distance.So you need to apply only 50 pounds of force at any given moment.

Private collection, Berkeley, California.Photo by Cheryl Fenton.

11.

Making the Connection[Language] Qi qiao ispronounced [che– che–au. ].

22

2222

22 22

24

24

24

2

2

2

2


9.1


Fold, Punch, and Snip

A square sheet of paper is folded vertically, a hole is punched out of the center, and then one of the corners is snipped off.When the paper is unfolded it will look like the figure at right.

Sketch what a square sheet of paper will look like when it is unfolded after the following sequence of folds, punches,and snips.

Fold once. Fold twice. Snip double-fold corner. Punch opposite corner.

12. Make a set of your own seven tangram pieces and create the Cat, Rabbit, Swan, andHorse with Rider as shown on page 484.

13. Find the radius of circle Q. 14. Find the length of AC�. 15. The two rays are tangent to the circle. What’s wrong withthis picture?

16. In the figure below, point A� 17. Which congruence shortcut 18. Identify the point ofis the image of point A after can you use to show that concurrency in �QUOa reflection over OT��. What �ABP � �DCP? from the construction are the coordinates of A�? marks.

19. In parallelogram QUID, m�Q � 2x � 5° and m�I � 4x � 55°. What is m�U?

20. In �PRO, m�P � 70° and m�R � 45°. Which side of the triangle is the shortest? PO�

115°

Q

U

O

B D

A C

P

30°(8, 0)

A

T

O

A�

y

x

SAA

54°

226°

B

A

DC

45° 30°

36 cm

A B

C

y

r

Qx

150°(?, 6)

orthocenter

12 units 18�2� cm

�4, 4�3��

12.

15. Draw radii CD�� and CB�.�ABC � �ADC � 90°. Forquadrilateral ABCD 54° �90° � m�C � 90° � 360°, som�C � 126°. BD� � 126° but126° � 226° � 360°.

Exercises 19, 20, 18 [Language] Thefigure names in these exercisesform a legal phrase: quid pro quo.Quid pro quo means “somethingfor something.” It’s used to meanthe consideration for a contract,that is, what each party gets outof it (such as money or advan-tage). A similar colloquialexpression is “tit for tat.”

EXTENSION

Have students show algebraicallythat 3-4-5 is the only Pythagoreantriple of consecutive positiveintegers.

LESSON 9.4 Story Problems 485


9.3

7.1

5.5

4.3

9.1

4.5

6.2

3.7

L E S S O N

9.5


We talk too much; we should

talk less and draw more.

JOHANN WOLFGANGVON GOETHE

L E S S O N

9.5Distance in CoordinateGeometryViki is standing on the corner of Seventh Street and 8th Avenue, and her brotherScott is on the corner of Second Street and 3rd Avenue. To find her shortest sidewalkroute to Scott, Viki can simply count blocks. But if Viki wants to know her diagonaldistance to Scott, she would need the Pythagorean Theorem to measure across blocks.

You can think of a coordinate plane as a grid of streets with two sets of parallellines running perpendicular to each other. Every segment in the plane that is not in the x- or y-direction is the hypotenuse of a right triangle whose legs are in the x- and y-directions. So you can use the Pythagorean Theorem to find the distance between any two points on a coordinate plane.

y

x

7th Ave

8th Ave

6th Ave

5th Ave

4th Ave

3rd Ave

2nd Ave

1st Ave

Firs

t St

Seco

nd

St

Fou

rth

St

Th

ird

St

Fift

h S

t

Sixt

h S

t

Seve

nth

St

Eig

hth

St

V

S

Investigation 1The Distance Formula

You will need

● graph paper

In Steps 1 and 2, find the length of each segment by using the segment as thehypotenuse of a right triangle. Simply count the squares on the horizontal andvertical legs, then use the Pythagorean Theorem to find the length of thehypotenuse.

Step 1 Copy graphs a–d from the next page onto your own graph paper. Use eachsegment as the hypotenuse of a right triangle. Draw the legs along the grid lines.Find the length of each segment.

LESSON OBJECTIVES

� Discover the Pythagorean relationship on a coordinate plane(the distance formula)

� Derive the equation of a circle from the distance formula� Use the distance formula to solve problems� Develop problem-solving skills and cooperative behavior

NCTM STANDARDS

CONTENT PROCESS






PLANNING

LESSON OUTLINE

One day:25 min Investigation and Examples

5 min Sharing

5 min Closing

10 min Exercises

MATERIALS

� graph paper� The Distance Formula (T), optional

TEACHING

If your curriculum or yourstudents’ background requiresthat you emphasize both thedistance formula and the equa-tion of a circle, you may want toplan two days for this lesson.

One step Pose this problem: “Vikiis standing on a street cornerand is trying to talk with Scottby walkie-talkie. He says he’s also at a corner, but static keeps Viki from understanding whichcorner. If each block is one-tenthmile long, what are the possiblestraight-line distances Scottcould be from Vicki and still bewithin a half mile?” As needed,encourage students to draw right triangles to find diagonaldistances. During Sharing, askfor an equation describing Scott’spossible locations if he’s exactly

�12� mile away but not necessarily

at a corner.

a. b.

c. d.

Step 2 Graph each pair of points, then find the distances between them.

a. (�1, �2), (11, �7) b. (�9,�6), (3, 10)

What if the points are so far apart that it’s notpractical to plot them? For example, what is thedistance between the points A(15, 34) andB(42, 70)? A formula that uses the coordinates ofthe given points would be helpful. To find thisformula, you first need to find the lengths of thelegs in terms of the x- and y-coordinates. Fromyour work with slope triangles, you know how tocalculate horizontal and vertical distances.

Step 3 Write an expression for the length of the horizontal leg using the x-coordinates.

Step 4 Write a similar expression for the length of the vertical leg using they-coordinates.

Step 5 Use your expressions from Steps 3 and 4, and the Pythagorean Theorem, to findthe distance between points A(15, 34) and B(42, 70).

Step 6 Generalize what you have learned about the distance between two points in acoordinate plane. Copy and complete the conjecture below.

2013

5

5

y

x

5

5

y

x5

5

y

x

B (42, 70)

A (15, 34)?

?

Distance Formula

The distance between points A�x1, y1� and B�x2, y2� is given by

�AB�2 � ��? �2 � ��? �2 or AB � ��? �2 ��? �2�

C-86

Let’s look at an example to see how you can apply the distance formula.

�20� � 2�5� � 4.5

�68� � 2�17� � 8.2

�40� � 2�10� � 6.3

�x2 � x1�2 � � y2 � y1�2 ��x2 � x�1�2 � ��y2 � y1��2�

–5

y

x

–4

�100� � 10


Step 1 Students used triangleslike this earlier in finding slopesof lines. You might provide TheDistance Formula worksheet sostudents aren’t tempted to writein their books. You might usepair share within groups orjigsaw between groups to save time.

Step 2 As needed, encouragestudents to use right triangles sothey can see the distance formulaas a special application of thePythagorean Theorem. Pair shareworks well for this step also.

Step 3 Students may want todiscuss how to handle negativedistances. Some students maywant to take the absolute valueso that all distances are positive.Others may be comfortable withnegative values as a measure of directed distance. Still otherstudents may advocate alwayssubtracting the smaller x-coordinate from the larger so the distance is always positive.

Step 3 Subtract one x-coordinatefrom the other.

Step 4 Subtract one y-coordinatefrom the other.

Step 5AB2 � (42 � 15)2 � (70 � 34)2;AB � �729 �� 1296� � 45

Step 6 Students may rememberthe distance formula from algebraI, perhaps using d instead of AB.

LESSON 9.5 Distance in Coordinate Geometry 487

EXAMPLE A

� Solution

EXAMPLE B

� Solution

Find the distance between A(8, 15) and B(�7, 23).

�AB�2 � �x2 � x1�2 � � y2 � y1�2 The distance formula.

� ��7 � 8�2 � �23 � 15�2 Substitute 8 for x1, 15 for y1, �7 for x2, and 23 for y2.

� ��15�2 � �8�2 Subtract.

�AB�2 � 289 Square �15 and 8 and add.

AB � 17 Take the square root of both sides.

The distance formula is also used to write the equation of a circle.

Write an equation for the circle with center (5, 4) and radius 7 units.

Let (x, y) represent any point on the circle. The distance from (x, y) to the circle’s center, (5, 4), is 7. Substitute this information in the distance formula.

So, the equation in standard form is �x � 5�2 � �y � 4�2 � 72.

(x � 5)2 � (y � 4)2 � 72

Substitute(x , y) for (x2, y2).

Substitute(5, 4) for (x1, y1).

Substitute 7as the distance.

Investigation 2 The Equation of a Circle

You will need

● graph paper

Find equations for a few more circles and then generalize the equation for anycircle with radius r and center (h, k).

Step 1 Given its center and radius, graph each circle on graph paper.

a. Center � (1, �2), r � 8 b. Center � (0, 2), r � 6

c. Center � (�3, �4), r � 10

Step 2 Select any point on each circle; label it (x, y). Use the distance formula to write an equation expressing the distance between the center of each circle and (x, y).

Step 3 Copy and complete the conjecture for the equation of a circle.

Equation of a Circle

The equation of a circle with radius r and center (h, k) is

�x � �? �2 � �y � �? �2 � ��? �2

C-87

(5, 4)

(x, y)

y

x

(h, k)

(x, y)

y

x

r

Step 2a(x � 1)2 � (y � 2)2 � 64

Step 2bx2 � (y � 2)2 � 36

Step 2c(x � 3)2 � (y � 4)2 � 100

h k r

� EXAMPLE A

Because the distance formula is derived directly from the Pythagorean Theorem,the book begins with the a2 � b2 � c2 version of thedistance formula. You could alsoshow that you can start with

AB ��x2 � x�1�2 � ��y2 � y1��2�.[Alert] If students are usingcalculators, they can easily getincorrect answers if parenthesesare omitted or not used correctly.

� EXAMPLE B

Students may benefit from beingreminded that an equation for ageometric figure is satisfied by thecoordinates of a point if and onlyif the point lies on the figure.


Step 1 [Alert] Students may inat-tentively think of diameterinstead of radius.

Step 1a

Step 1b

Step 1c

Step 2 As needed, encouragestudents to use the model inExample B.

After doing this investigation,you may want to work withanother example. [Ask] “What is the equation for a circle withcenter (2, �3) and radius 4?”[(x � 2)2 � ( y � 3)2 � 16]

y

x

(–3, –4)

(x, y)

y

x

(0, 2)

(x, y)

y

x

(x, y)

(1, –2)

SHARING IDEAS

Students may have come up with a variety of distanceformulas. For example, they may have either x2 � x1or y2 � y1 first, or they might have x1 � x2 in place of x2 � x1 or y1 � y2 in place of y2 � y1. Throughstudent discussion, elicit the idea that these are all the same formula after squaring and adding.

[Ask] “In Example A, how would you know tosubstitute the coordinates (8, 15) for �x1, y1� insteadof for �x2, y2�?” Student experimentation can leadto the realization that it doesn’t matter.

When discussing the equation of a circle, askwhether you could get an equivalent formula bytaking the square root, as with the distance formula.Students may say r � �(x � h�)2 � (�y � k)�2�.[Ask] “Should the equation include negative values of r, r � ��(x �h)�2 � ( y� �k)2�? [No; r is a radius,so it cannot be negative.] The equation with r2 iscalled the standard form.

Ask how students might use the equation to grapha circle with center at (1, �2) and radius 8. Theyneed to solve the equation for y in terms of x.They’ll get ( y � 2)2 � 64 � (x � 1)2, from


EXAMPLE C

� Solution

Let’s look at an example that uses the equation of a circle in reverse.

Find the center and radius of the circle �x � 2�2 � � y � 5�2 � 36.

Rewrite the equation of the circle in the standard form.

Identify the values of h, k, and r. The center is (�2, 5) and the radius is 6.

�x � h�2 � �y � k�2 � r 2

�x � (�2)�2 � �y � 5�2 � 62

EXERCISESIn Exercises 1–3, find the distance between each pair of points.

1. (10, 20), (13, 16) 2. (15, 37), (42, 73) 3. (�19, �16), (�3, 14)

4. Look back at the diagram of Viki’s and Scott’s locations on page 486. Assume eachblock is approximately 50 meters long. What is the shortest distance from Viki toScott to the nearest meter?

5. Find the perimeter of �ABC with vertices A(2, 4), B(8, 12), and C(24, 0).

6. Determine whether �DEF with vertices D(6, �6), E(39, �12), and F(24, 18) isscalene, isosceles, or equilateral.

For Exercises 7 and 8, find the equation of the circle.

7. Center � (0, 0), r � 4 8. Center � (2, 0), r � 5

For Exercises 9 and 10, find the radius and center of the circle.

9. �x � 2�2 � � y � 5�2 � 62 10. x2 � � y � 1�2 � 81

11. The center of a circle is (3, �1). One point on the circle is (6, 2). Find the equationof the circle.

12. Mini-Investigation How would you find thedistance between two points in a three-dimensional coordinate system?Investigate and make a conjecture.

a. What is the distance from the origin (0, 0, 0)to (2, �1, 3)?

b. What is the distance between P(1, 2, 3) andQ(5, 6, 15)?

c. Complete this conjecture:If A�x1, y1, z1� and B�x2, y2, z2� are twopoints in a three-dimensional coordinatesystem, then the distance AB is

��? �2 ��? �2�� ?��2�.

�176� � 4�11� units

�14� units

(x � 3)2 � (y � 1)2 � 18

center is (0, 1), r � 9center is (2, �5), r � 6

(x � 2)2 � y2 � 25x2 � y2 � 16

isosceles

52.4 units

354 m

45 units5 units

�

y

z

x

5

5

4

–5

–5

–4

This point is the graph of the ordered triple(2, �1, 3).

34 units

Sharing Ideas (continued)which they may want to take square roots to get y � 2 � �64 � (�x � 1)�2� and hence y � �2 � �64 � (�x � 1)�2�. When they graph this function on a calculator, they’ll get only a semi-circle, because in taking the square root they elimi-nated negative values of y � 2. To graph the otherhalf of the circle, they’ll need to graph the second function y � �2 � �64 � (�x � 1)�2�.

Their graphs still might not look very circular.Suggest that they use a friendly window.

11c. AB � ��x1 ��x2�2 �� y1 �� y2�2 ��z1 �� z2�2�Exercise 12 The radius is determined to be �18� or3�2�. The equation is then (x � 3)2 � ( y �1)2 �

�3�2��2 or (x �3)2 � ( y �1)2 � ��18��2, orsimply (x �3)2 � ( y � 1)2 � 18. This is a goodexample of a case in which the “unsimplified” form�18� is more useful than the “simplified” form,because it’s easier to square.

LESSON 9.5 Distance in Coordinate Geometry 489

� EXAMPLE C

Point out that (x � 2)2 can berewritten as (x � (�2))2.

Assessing ProgressYou can assess students’ famil-iarity with coordinates of points,circles and their radii, righttriangles and their hypotenuses,and the Pythagorean Theorem.Also check students’ ability towork separately with the hori-zontal and vertical distances.

Closing the Lesson

The distance formula,AB � ��x2 ��x1�2 �� y2 �� y1�2�,and the related equation of acircle, (x � h)2 � ( y � k)2 � r2,are both derived from thePythagorean Theorem.


These exercises motivate theneed for a formula to find thedistance between points whenplotting the points is impractical.

ASSIGNING HOMEWORK

Essential 1–10


Portfolio 6

Journal 17

Group 11

Review 13–17

|


Exercise 3 [Alert] Students maynot be careful enough in substi-tuting the negative numbers intothe distance formula.

Exercise 11 As needed, encouragestudents to break apart the prob-lem and consider a diagonal ofa horizontal or vertical rectangleas an intermediate step. Or askwhether the distance could be aspace diagonal of an imaginarybox.

Review

13. Find the coordinates of A. 14. k � �? , m � �? 15. The large triangle is equilateral. Find x and y.

16. Antonio is a biologist studying life in a pond.He needs to know how deep the water is.He notices a water lily sticking straight up from the water, whose blossom is 8 cm above the water’s surface. Antonio pulls the lily to oneside, keeping the stem straight, until the blossomtouches the water at a spot 40 cm from where the stem first broke the water’s surface. How isAntonio able to calculate the depth of the water?What is the depth?

17. C�U�R�T� is the image of CURT under a rotation transformation. Copy the polygon and its image onto patty paper. Find the center of rotation and the measure of the angle of rotation. Explain your method.

96 cm

12

yx

km

60°

3

A (?, ?)

(0, –1)

(1, 0)

150°

y

x

�


The Spider and the Fly (attributed to the British puzzlist Henry E. Dudeney, 1857–1930)

In a rectangular room, measuring 30 by 12 by 12 feet, a spider is at point A on the middle of one of the end walls, 1 foot fromthe ceiling. A fly is at point B on the center of the oppositewall, 1 foot from the floor. What is the shortest distance that the spider must crawl to reach the fly, which remains

stationary? The spider never drops or uses its web, butcrawls fairly.

12 ft

12 ft

30 ft

k � �2�, m � �6��23��, �

12�� x � , y � 12

��3�

6��3�

The angle of rotation is approximately 77°. Connect twopairs of corresponding points. Construct the perpendi-cular bisector of each segment. The point where theperpendicular bisectors meet is the center of rotation.

U'

C'

R'

T' U

C

T

R

Exercise 15 As needed, askwhether the triangle is a specialkind. If a student rationalizes thedenominator, the solutions are x � 2�3� and y � 4�3�.

Exercise 17 If students are havingdifficulty, ask if they know anyproperties of the center of a rotation.

The Spider and the FlyHere is another path shorterthan 42 feet. It is not theshortest.



[Context] Henry E. Dudeney (1857–1930,England) was a puzzle inventor whosepuzzles continue to challenge people today.His spider-and-fly problem first appearedin an English newspaper.

If students are stuck, wonder aloud whetherit would help to think of the problem in

two dimensions. They might draw severaldifferent nets of the room and drawstraight lines between the points on thenet. [Ask] “Which net will give the shortestdistance? What does that mean for theoriginal room?”

The shortest path measures 40 feet.

32

2440Side

wall A

BSidewall

Ceiling

Floor

Backwall

1 ft 1 ft30 ft

30 ft

40 ft

6 ft

12 ft

6 ft

9.3

7.1

9.1 9.1

37

1740.7Side

wall

Sidewall

Ceiling

Back wall

6 ft1 ft 30 ft

6 ft

11 ft

EXPLORATION Ladder Climb 491

You will need

● a graphing calculator

Ladder ClimbSuppose a house painter rests a20-foot ladder against a building,then decides the ladder needs torest 1 foot higher against thebuilding. Will moving the ladder1 foot toward the building do thejob? If it needs to be 2 feet lower,will moving the ladder 2 feet awayfrom the building do the trick?Let’s investigate.

ActivityClimbing the WallSketch a ladder leaning against avertical wall, with the foot of theladder resting on horizontal ground. Label the sketch using y for height reached bythe ladder and x for the distance from the base of the wall to the foot of the ladder.

Step 1 Write an equation relating x, y, and the length of the ladder and solve it for y. Younow have a function for the height reached by the ladder in terms of the distancefrom the wall to the foot of the ladder. Enter this equation into your calculator.

Step 2 Before you graph the equation, think about the settings you’ll want for the graphwindow. What are the greatest and least values possible for x and y? Enterreasonable settings, then graph the equation.

Step 3 Describe the shape of the graph.

Step 4 Trace along the graph, starting at x � 0. Record values (rounded to the nearest0.1 unit) for the height reached by the ladder when x � 3, 6, 9, and 12. If youmove the foot of the ladder away from the wall 3 feet at a time, will each moveresult in the same change in the height reached by the ladder? Explain.

Step 5 Find the value for x that gives a y-value approximately equal to x. How is thisvalue related to the length of the ladder? Sketch the ladder in this position. Whatangle does the ladder make with the ground?

Step 6 Should you lean a ladder against a wall in such a way that x is greater than y?Explain. How does your graph support your explanation?

Step 1y � �400 �� x2�

Step 2 0 � x � 20;0 � y � 20

Step 3 a quarter-circle

Step 4 (3, 19.8); (6, 19.1);(9, 17.9); (12, 16)No. At first the height reached by the ladderdecreases slowly, but the farther the ladder is pulled out, the faster the height reached by it decreases.

Step 5 When x � y �at x � 14, the length of theladder divided by �2��, theladder forms a 45°angle with the floor andthe wall.

LESSON OBJECTIVES

� Create an algebraic model for the ladder problem� Review graphing an equation� Apply the Pythagorean Theorem to understand how rates of

change vary

NCTM STANDARDS

CONTENT PROCESS

Number Problem Solving





EXPLORATION

PLANNING

LESSON OUTLINE

One day:25 min Activity

20 min Sharing and Closing

MATERIALS

� rulers� graphing calculators

TEACHING

[ESL] Do the trick means “accom-plish the task.”

uiding the Activity

Step 2 Ask students to conjecturewhether the amount of changewill always be the same and,if not, what it depends on. Asneeded, help students see wherethe maximum value of each variable occurs (on an axis).

Step 3 Unless students happenedto choose a friendly window inStep 2, they may say that thegraph is part of a parabola rather than an arc of a circle.

Step 5 Suggest that studentsstudy a table of these data.

SHARING IDEAS

As students present their ideasabout Steps 4–6, [Ask] “At whatpoint will a one-foot change in xresult in a one-foot change in y?”[at the instant y � x]

Closing the Lesson

Rates of change of a functionwill vary if the graph of thefunction is curved.

G

See page 775 for answer to Step 6.

L E S S O N

9.6


Circles and thePythagorean TheoremIn Chapter 6, you discovered a number of properties that involved right angles inand around circles. In this lesson you will use the conjectures you made, along withthe Pythagorean Theorem, to solve some challenging problems. Let’s review two ofthe most useful conjectures.

Tangent Conjecture: A tangent to a circle isperpendicular to the radius drawn to the point of tangency.

Angles Inscribed in a Semicircle Conjecture: Anglesinscribed in a semicircle are right angles.

Here are two examples that use these conjectures alongwith the Pythagorean Theorem.

TA�� is tangent to circle N at A. TATA � 12�3� cm. Find the area of the shaded region.

TA�� is tangent at A, so �TAN is a right angle and �TAN is a 30°-60°-90°triangle. The longer leg is 12�3� cm, so the shorter leg (also the radius of thecircle) is 12 cm. The area of the entire circle is 144� cm2. The area of the shadedregion is �360

36�

060

�, or �56�, of the area of the circle. Therefore the shaded area is

�56�(144�), or 120� cm2.

AB � 6 cm and BC � 8 cm. Find the area of the circle.

Inscribed angle ABC is a right angle, so ABC� is a semicircle and AC� is adiameter. By the Pythagorean Theorem, if AB � 6 cm and BC � 8 cm, then AC � 10 cm. Therefore the radius of the circle is 5 cm and the area of the circleis 25� cm2.

C

B A

T

N

A

G

30°

EXAMPLE A

� Solution

EXAMPLE B

� Solution

You must do things you think

you cannot do.

ELEANOR ROOSEVELT

L E S S O N

9.6

from the center of the smaller wheel. What lengthbelt should you locate?” As needed, encouragestudents to review earlier conjectures about circlesand special right triangles.

Assessing ProgressAs students discuss the examples and work throughthe exercises, you can assess their understanding oftangent segments, arc lengths, the PythagoreanTheorem, and special right triangles.

Closing the Lesson

Much of the power of geometry comes fromcombining different conjectures. For example, usingwhat we know about tangent segments, arc lengths,the Pythagorean Theorem, and special right trian-gles can help us solve a variety of problems.

PLANNING

LESSON OUTLINE

One day:10 min Examples

30 min Exercises

5 min Closing

MATERIALS

� graph paper� Exercise 8 (T) for One step

TEACHING

The Pythagorean Theorem canbe powerful when combined withearlier conjectures about circles.You may either start with theone-step investigation; or, formore structure, begin with theexamples. The lesson consistsprimarily of exercises.

� EXAMPLE A

Some students might miss thefact that the angle at vertex Nhas measure 60°. They may alsobe confused by the proportionalreasoning.

� EXAMPLE B

Ask students how they know that AC� is a diameter. [By theInscribed Angle Conjecture, aright angle inscribed in a circleintercepts a semicircle.]

One step Pose this problem, asyou display the transparency forExercise 8: “In repairing an oldmachine, you must find a beltthat will make one wheel turnanother. The wheels’ diametershave lengths 36 cm and 24 cm,and their centers are 60 cmapart. Because the wheels turn inopposite directions, the belt hasto cross itself; marks indicatethat the crossing point is 24 cm

EXERCISESIn Exercises 1–4, find the area of the shaded region in each figure. Assumelines that appear tangent are tangent at the labeled points.

1. OD � 24 cm 2. HT � 8�3� cm 3. HA � 8�3� cm 4. HO � 8�3� cm

5. A 3-meter-wide circular track is shown at right. The radius of the inner circleis 12 meters. What is the longest straight path that stays on the track? (Inother words, find AB.)

6. An annulus has a 36 cm chord of the outer circle that is also tangent to theinner concentric circle. Find the area of the annulus.

7. In her latest expedition, Ertha Diggs has uncovereda portion of circular, terra-cotta pipe that shebelieves is part of an early water drainage system.To find the diameter of the original pipe, she lays ameterstick across the portion and measures thelength of the chord at 48 cm. The depth of theportion from the midpoint of the chord is 6 cm.What was the pipe’s original diameter?

8. APPLICATION A machinery belt needs to be replaced.The belt runs around two wheels, crossing betweenthem so that the larger wheel turns the smallerwheel in the opposite direction. The diameter of thelarger wheel is 36 cm, and the diameter of thesmaller is 24 cm. The distance between the centersof the two wheels is 60 cm. The belt crosses 24 cmfrom the center of the smaller wheel. What is thelength of the belt?

9. A circle of radius 6 has chord AB� of length 6. If point C is selected randomly on thecircle, what is the probability that �ABC is obtuse? �

56�

230 cm

102 cm

324� cm2

18 m

O H120°R

A

T

H

120°

T

R H60°O

D

B

Y

105°

�

You will need

Construction toolsfor Exercise 20

A

B

O

T

36 cm 24 cm

60 cm

24

456� cm2 �32� � 32�3�� cm2

��64

3�� 16�3�� cm2

�64�3� � �64

3�� cm2

LESSON 9.6 Circles and the Pythagorean Theorem 493

LESSON OBJECTIVE

� Apply the Pythagorean relationship to problems involvingcircles

NCTM STANDARDS

CONTENT PROCESS







You might have several groupspresent their solutions to a few of the exercises. Distribute transparencies and pens to help students prepare theirpresentations.

ASSIGNING HOMEWORK

Essential 1–9


Portfolio 14

Group 10–17

Review 18–23

|


Exercise 3 If students are havingdifficulty, wonder aloud whetherthere’s a way to draw auxiliarylines to create right triangles,perhaps special right triangles.

3. �64�3� � �64

3�� cm2

Exercise 4 As needed, remindstudents that the area of asegment is the differencebetween the area of a sector and the area of a triangle.

Exercise 7 Some students maystop after finding the radiusinstead of finding the diameter.

Exercise 8 The answer in terms of� is �40� � 60�3�� cm.

Exercise 9 As needed, [Ask] “Doesa chord that’s congruent to aradius remind you of anything?”[Mark the radius six times arounda circle to form a hexagon.] “Whathappens if point C falls in the arcsintercepted by the various sides ofthe inscribed regular hexagon?”[The arc on the hexagon sideopposite side AB would be theonly one where the third vertexwould create an acute triangle.]

In Exercises 10 and 11, each triangle is equilateral. Find the area of the inscribed circleand the area of the circumscribed circle. How many times greater is the area of thecircumscribed circle than the area of the inscribed circle?

10. AB � 6 cm 11. DE � 2�3� cm

12. The Gothic arch is based on the equilateral 13. Each of three circles of radius 6 cm is tangent triangle. If the base of the arch measures to the other two, and they are inscribed in a 80 cm, what is the area of the shaded region? rectangle, as shown. What is the height of

the rectangle?

14. Sector ARC has a radius of 9 cm and an angle 15. APPLICATION Will plans to use a circular that measures 80°. When sector ARC is cut cross section of wood to make a square table.out and AR� and RC� are taped together, they The cross section has a circumference ofform a cone. The length of AC� becomes the 336 cm. To the nearest centimeter, what is the circumference of the base of the cone. What is side length of the largest square that he can the height of the cone? cut from it?

16. Find the coordinates of point M. 17. Find the coordinates of point K.

K

(0, –1)

(1, 0)

210°

y

x

M

(0, –1)

(1, 0)

135°x

y

h

A

C

R

9 cm

80°

76 cm�77� cm

6

680 cm

80 cm

�12 � 6�3�� cm

D E

F

A B

C

x

Inscribed circle: 3� cm2.Circumscribed circle:12� cm2. The area of thecircumscribed circle isfour times as great as thearea of the inscribed circle.

Inscribed circle: � cm2.Circumscribed circle:4� cm2. The area of the circumscribed circle isfour times as great as thearea of the inscribed circle.

3931 cm2

�� , �1��2�

1��2� ��2

3��, � �12��

Exercise 12 As in Exercise 8, thisexercise represents a physicalsituation, so an answer in termsof �, ��

64300�� 1600�3�� cm2,

may not be as appropriate as arounded answer.

Exercise 13 If students are stuck,ask where the marked lengthstransfer to on the sides of therectangle and what they knowabout the rest of the height.

Exercise 14 You might encouragestudents to make a model forthis exercise.

Exercise 16 Remind students asneeded that they haven’t workedmuch with obtuse angles, so theymight benefit from drawing anacute triangle with point M atone vertex.

Exercises 16, 17 [Link] This kind ofreasoning is used in trigonometry.


Review

18. Find the equation of a circle with center (3, 3) and radius 6.

19. Find the radius and center of a circle with the equation x2 � y2 � 2x � 1 � 100.

20. Construction Construct a circle and a chord in a circle. With compass andstraightedge, construct a second chord parallel and congruent to the first chord.Explain your method.

21. Explain why the opposite sides of a regular hexagon are parallel.

22. Find the rule for this number pattern:

1 � 3 � 3 � 4 � 0

2 � 4 � 3 � 5 � 1

3 � 5 � 3 � 6 � 2

4 � 6 � 3 � 7 � 3

5 � 7 � 3 � 8 � 4

n � ��? � � ��? � � ��? � � ��? �

23. APPLICATION Felice wants to determine thediameter of a large heating duct. She places a carpenter’s square up to the surface of thecylinder, and the length of each tangent segment is 10 inches.

a. What is the diameter? Explain your reasoning.

b. Describe another way she can find thediameter of the duct.

…

center � (1, 0), r � 10

(x � 3)2 � (y � 3)2 � 36

�

IMPROVING YOUR REASONING SKILLS

Reasonable ’rithmetic I

Each letter in these problems represents a different digit.

1. What is the value of B? 2. What is the value of J?

3 7 2 E F 6

3 8 4 � D 7

� 9 B 4 D D F D

C 7 C A J E D

H G E D

n � (n � 2) � 3 � (n � 3) � (n � 1)

Possible answer: Measure the circumferencewith string and divide by �.

Exercise 19 This exercise requiresthat students be able to recognizethe square of the binomial (x � 1)2.

20. The diameter is thetransversal, and the chords areparallel by the Converse of theParallel Lines Conjecture. Thechords are congruent becausethey are the same distance fromthe center. Sample construction:

21. Any long diagonal of aregular hexagon divides it intotwo congruent quadrilaterals.Each angle of a regular hexagon is �

(6 � 2

6

) � 180°� � 120°, and

the diagonal divides two of the120° angles into 60° angles.Look at the diagonal as atransversal.

The alternate interior angles are congruent, thus the oppositesides of a regular hexagon areparallel.

23a. Because a carpenter’ssquare has a right angle andboth radii are perpendicular tothe tangents, a square is formed.The radius is 10 in., thereforethe diameter is 20 in.

10 in.

10 in.

10 in.

10 in.

d � 20 in.

120°

120°

120°

120°

60°

60°60°

60°

LESSON 9.6 Circles and the Pythagorean Theorem 495

IMPROVING REASONING SKILLS

If students are having difficulty, suggest thatthey make a table of the letters, listing whatthey can say about each letter.

1. A � 0 and C � 1, so B � 5.

2. D � 2; 7F � 4 ends in F, so F is 1 or 6.Trying each possibility leads to J � 6.

EXTENSION

Ask students to show that the shaded area in thisfigure is equal to the area of the triangle. SeeLeonardo’s Dessert—No Pi by Herbert Wills forfurther study of problems like this.

a b

9.5

9.5

6.3

2.6

2.3

6.2


C H A P T E R R E V I E W

9 ● CHAPTER 11 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTCHAPTER

9R E V I E W

If 50 years from now you’ve forgotten everything else you learnedin geometry, you’ll probably still remember the PythagoreanTheorem. (Though let’s hope you don’t really forget everythingelse!) That’s because it has practical applications in themathematics and science that you encounter throughout youreducation.

It’s one thing to remember the equation a2 � b2 � c2. It’s anotherto know what it means and to be able to apply it. Review yourwork from this chapter to be sure you understand how to usespecial triangle shortcuts and how to find the distance between two points in a coordinate plane.

EXERCISESFor Exercises 1–4, measurements are given in centimeters.

1. x � �? 2. AB � �? 3. Is �ABC an acute, obtuse,or right triangle?

4. The solid is a rectangular 5. Find the coordinates of 6. Find the coordinates ofprism. AB � �? point U. point V.

7. What is the area of the 8. The area of this square is 9. What is the area oftriangle? 144 cm2. Find d. trapezoid ABCD?

12 cm 20 cm 15 cm

D

A

C

B

d40 cm

30°

246 cm2d � 12�2� cm200�3� cm2

U (?, ?)

(0, 1)

(1, 0)30°

y

x6

8

24

A B

26 cm

260

24070

A B

C

13

12

A B

C

1525

x

10 cm20 cm

V (?, ?)

(0, 1)

(1, 0)

225°

y

x

obtuse

��

23��, �

12��

�1

2�� , ��

�1

2��

�

You will need

Construction toolsfor Exercise 30

PLANNING

LESSON OUTLINE

First day:10 min Reviewing

35 min Exercises

Second day:30 min Exercises

15 min Student self-assessment

REVIEWING

Remind students of the equation ofa circle, (x � h)2 � ( y � k)2 � r2,and ask what that means. Try toelicit the idea that it’s the set of allpoints (x, y) whose coordinatessatisfy the equation. Ask whythose points form a circle. Bringout the idea of the distanceformula. Finally, ask where thedistance formula comes from. Seethat students understand it comesfrom the Pythagorean Theorem.Wonder aloud how the distanceformula can be derived from the Pythagorean Theorem giventhat there are no a’s, b’s, or c’s in the distance formula but thePythagorean Theorem is a2 � b2 � c2. Elicit the idea thatthe Pythagorean Theorem includesa condition: a2 � b2 � c2 if aand b are the lengths of the legsof a right triangle and c is thelength of its hypotenuse. Theletters a, b, and c are variablesand can be replaced by any otherrepresentations of those lengths.

ASSIGNING HOMEWORK

It’s probably best to assign every student to work on everyproblem from Exercises 1–29.The odds 15–29 could be donein groups. Use the mixed reviewto prepare for a unit exam.

10. The arc is a semicircle. 11. Rays TA and TB are tangent to 12. The quadrilateral is a What is the area of the circle O at A and B respectively, square, and QE � 2�2� cm.shaded region? and BT � 6�3� cm. What is the What is the area of the

area of the shaded region? shaded region?

13. The area of circle Q is 350 cm2. Find the area of square ABCD to thenearest 0.1 cm2.

14. Determine whether �ABC with vertices A(3, 5), B(11, 3), and C(8, 8) is an equilateral, isosceles, or isosceles right triangle.

15. Sagebrush Sally leaves camp on her dirt bike traveling east at 60 km/hr with a full tank of gas. After 2 hours,she stops and does a little prospecting—with no luck.So she heads north for 2 hours at 45 km/hr. She stops again, and this time hits pay dirt. Sally knows that she can travel at most 350 km on one tank of gas. Does she have enough fuel to get back to camp? If not,how close can she get?

16. A parallelogram has sides measuring 8.5 cm and 12 cm, and a diagonal measuring 15 cm. Is the parallelogram a rectangle? If not, is the 15 cm diagonal the longer or shorter diagonal?

17. After an argument, Peter and Paul walk away from each other on separate paths at a right angle to each other. Peter is walking 2 km/hr, and Paul is walking 3 km/hr. After 20 min, Paul sits down to think. After 30 min, Peter stops.Both decide to apologize. How far apart are they? How long will it take them to reach each other if they both start running straight toward each other at 5 km/hr?

18. Flora is away at camp and wants to mail her flute back home.The flute is 24 inches long. Will it fit diagonally within a boxwhose inside dimensions are 12 by 16 by 14 inches?

19. To the nearest foot, find the original height of a fallen flagpole that cracked and fell as if hinged, forming an angleof 45 degrees with the ground. The tip of the pole hit theground 12 feet from its base. 29 ft

yes

1.4 km; 8�12� min

222.8 cm2

E

S Q

RB

OA

T

120°7 in.

25 in.

24� cm272� in.2

(2� � 4) cm2

CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9

Q

B C

A D

isosceles right

No. 15 cm is the longer diagonal.

|


Exercise 11 If students are havingdifficulty, suggest that they createa right triangle and look forspecial triangles.

Exercise 14 From the way theproblem is stated, students needto do only two calculations tofind sides of two different lengths;then they can conclude that thetriangle is an isosceles righttriangle because it’s not equilat-eral. Encourage them, though, toverify the congruence andperpendicularity of two sides.Slope can help with the latter.

15. No. The closest she cancome to camp is 10 km.

Exercise 17 After finding thedistance apart to be �2� km,students might realize that thesecond question can be answeredeasily by using closing speed:Paul and Peter’s distance apart is closing at a rate of 10 milesper hour.

Exercise 18 Students have workedenough with space diagonals that they may be ready to gener-alize and find that the spacediagonal of this box has length

�122 ��162 ��142�.

CHAPTER 9 REVIEW 497

20. You are standing 12 feet from a cylindrical corn-syrup storage tank. Thedistance from you to a point of tangency on the tank is 35 feet. What is theradius of the tank?

21. APPLICATION Read the Technology Connection above. What is the maximumbroadcasting radius from a radio tower 1800 feet tall (approximately 0.34 mile)? Theradius of Earth is approximately 3960 miles, and you can assume the ground aroundthe tower is nearly flat. Round your answer to the nearest 10 miles.

22. A diver hooked to a 25-meter line issearching for the remains of a Spanishgalleon in the Caribbean Sea. The sea is20 meters deep and the bottom is flat.What is the area of circular region thatthe diver can explore?

23. What are the lengths of the two legs of a30°-60°-90° triangle if the length of thehypotenuse is 12�3�?

24. Find the side length of an equilateral triangle with an area of 36�3� m2.

25. Find the perimeter of an equilateral triangle with a height of 7�3�.

26. Al baked brownies for himself and his two sisters. He divided the square pan ofbrownies into three parts. He measured three 30° angles at one of the corners so that two pieces formed right triangles and the middle piece formed a kite. Did he divide the pan of brownies equally? Draw a sketch and explain your reasoning.

27. A circle has a central angle AOB that measures 80°. If point C is selected randomly on the circle, what is the probability that �ABC is obtuse? �

79�

42

12 m

6�3� and 18

707 m2

50 mi

� 45 ft

EW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPT

Technology

Radio and TV stations broadcast from high towers.Their signals are picked up by radios and TVs inhomes within a certain radius. Because Earth isspherical, these signals don’t get picked up beyond the point of tangency.

r

r

12 feet

35 feet

20 m 25 m

26. No. If you reflect one of theright triangles into the centerpiece, you’ll see that the area ofthe kite is almost half again aslarge as the area of each of theother triangles.

Or students might compareareas by assuming the short legof the 30°-60°-90° triangle is 1.The area of each triangle is then

and the area of the kite is3 � �3�.

Extra

30°30°

30°

�3��2


28. One of the sketches below shows the greatest area that you can enclose in a right-angled corner with a rope of length s. Which one? Explain your reasoning.

29. A wire is attached to a block of wood at point A.The wire is pulled over a pulley as shown. How farwill the block move if the wire is pulled 1.4 metersin the direction of the arrow?

MIXED REVIEW30. Construction Construct an isosceles triangle that has a base length equal to half the

length of one leg.

31. In a regular octagon inscribed in a circle, how many diagonals pass through thecenter of the circle? In a regular nonagon? a regular 20-gon? What is the general rule?

32. A bug clings to a point two inches from the center of a spinning fan blade. The bladespins around once per second. How fast does the bug travel in inches per second?

In Exercises 33–40, identify the statement as true or false. For each false statement,explain why it is false or sketch a counterexample.

33. The area of a rectangle and the area of a parallelogram are both given by the formula A � bh, where A is the area, b is the length of the base, and h is the height.

34. When a figure is reflected over a line, the line of reflection is perpendicular to everysegment joining a point on the original figure with its image.

35. In an isosceles right triangle, if the legs have length x, then the hypotenuse haslength x�3�.

36. The area of a kite or a rhombus can be found by using the formula A � (0.5)d1d2,where A is the area and d1 and d2 are the lengths of the diagonals.

37. If the coordinates of points A and B are �x1, y1� and �x2, y2�, respectively, then AB � ��x1 ��y1�2 �� x2 �� y2�2�.

38. A glide reflection is a combination of a translation and a rotation.

false; AB � ��x2 � x�1�2 � ��y2 � y1��2�

true

False. The hypotenuse is of length x�2�.

true

true

1.6 m

s

A quarter-circleA square

s1_2

s1_2

s

A triangle


�

1.4 m

1.5 m3.9 m

?

A

4�, or approximately 12.6 in./sec

False. A glide reflection is a combination of a translation and a reflection.

28. The quarter-circle gives themaximum area.

Triangle:

A � �12� � � � �

s42�

Square:

A � �12� s � �

12� s � �

s42�

Quarter-circle:

s � �14� � 2�r

r � �2�

s�

A � �14� ��

2�

s��

2

� �s�

2�

�s�

2� �

s42�

30.

Exercise 31 Encourage students tomake a table and look for apattern.

31. 4; 0; 10. The rule is �n2� if n is

even, but 0 if n is odd.

2s___�

s

s1_2

s1_2

s��2�

s��2�

s____ 2

s____ 2

s45°

45°


3.1

2.3

6.7

8.1

7.1

9.3

8.2

9.5

7.3

39. Equilateral triangles, squares, and regular octagons can be used to createmonohedral tessellations.

40. In a 30°-60°-90° triangle, if the shorter leg has length x, then the longer leg haslength x�3� and the hypotenuse has length 2x.

In Exercises 41–46, select the correct answer.

41. The hypotenuse of a right triangle is always �? .

A. opposite the smallest angle and is the shortest side.

B. opposite the largest angle and is the shortest side.

C. opposite the smallest angle and is the longest side.

D. opposite the largest angle and is the longest side.

42. The area of a triangle is given by the formula �? , where A is the area, b is the lengthof the base, and h is the height.

A. A � bh B. A � �12�bh

C. A � 2bh D. A � b2h

43. If the lengths of the three sides of a triangle satisfy the Pythagorean equation, thenthe triangle must be a(n) �? triangle.

A. right B. acute

C. obtuse D. scalene

44. The ordered pair rule (x, y) → (y, x) is a �? .

A. reflection over the x-axis B. reflection over the y-axis

C. reflection over the line y � x D. rotation 90° about the origin

45. The composition of two reflections over two intersecting lines is equivalent to �? .

A. a single reflection B. a translation

C. a rotation D. no transformation

46. The total surface area of a cone is equal to �? , where r is the radius of the circularbase and l is the slant height.

A. �r2 � 2�r B. �rl

C. �rl � 2�r D. �rl � �r2

47. Create a flowchart proof to show that the diagonal of a rectangle divides therectangle into two congruent triangles.

D

C

C

A

B

D

true


A B

D C

False. Equilateral triangles, squares, and regular hexagons canbe used to create monohedral tessellations.

47.1 ABCD is

a rectangle

Given

2 ABCD isa parallelogram

Definition ofrectangle

Same segment

6 �� AC � AC

Definition ofrectangle

4 �D � �B

Definition ofparallelogram

3

SAA CongruenceConjecture

7 �ABC � �CDA�� DA � CB

AIA Conjecture

5 �DAC � �BCA


7.4

9.3

9.1

8.2

9.2

7.2

7.3

8.7

5.7

48. Copy the ball positions onto patty paper.

a. At what point on the S cushion should a player aim so that the cue ball bounces off and strikes the 8-ball? Mark the point with the letter A.

b. At what point on the W cushion should a player aim so that the cue ball bounces off and strikes the 8-ball? Mark the point with the letter B.

49. Find the area and the perimeter of 50. Find the area of the shaded region.the trapezoid.

51. An Olympic swimming pool has length 52. The side length of a regular pentagon is 6 cm,50 meters and width 25 meters. What is and the apothem measures about 4.1 cm.the diagonal distance across the pool? What is the area of the pentagon?

53. The box below has dimensions 25 cm, 54. The cylindrical container below has an open 36 cm, and x cm. The diagonal shown top. Find the surface area of the container has length 65 cm. Find the value of x. (inside and out) to the nearest square foot.

TAKE ANOTHER LOOK1. Use geometry software to demonstrate the

Pythagorean Theorem. Does your demonstration stillwork if you use a shape other than a square—forexample, an equilateral triangle or a semicircle?

2. Find Elisha Scott Loomis’s Pythagorean Propositionand demonstrate one of the proofs of the PythagoreanTheorem from the book.

9 ft

5 ft36 cm

65 cm

25 cm

x cm

48 cm

4 cm

3 cm

120°

12 cm

5 cm

5 cm4 cm

322 ft2

34 cm2; 22 � 4�2� � 27.7 cm


�A

BC

�40

3�� cm2

N

S

W E

51. about 55.9 m2

52. about 61.5 cm2

|

� Take Another Look

1. Demonstrations should includeshapes other than a square. (Anyregular ploygon can be used. Infact, any three similar figures willwork. Students will study similarfigures in Chapter 11.)

2. Demonstrations will vary.

48.

S

EWB

A

N

8-ball

Cue ball


1.2

8.2

9.1

9.1

8.6

8.4

8.7

3. The Zhoubi Suanjing, one of the oldest sources of Chinesemathematics and astronomy, contains the diagram at rightdemonstrating the Pythagorean Theorem (called gou gu inChina). Find out how the Chinese used and proved the gou gu, and present your findings.

4. Use the SSS Congruence Conjecture to verify the converse ofthe 30°-60°-90° Triangle Conjecture. That is, show that if a triangle has sides with lengths x, x�3�, and 2x, then it is a 30°-60°-90° triangle.

5. Starting with an isosceles right triangle, use geometry software ora compass and straightedge to start a right triangle like the oneshown. Continue constructing right triangles on the hypotenuseof the previous triangle at least five more times. Calculate thelength of each hypotenuse and leave them in radical form.

Assessing What You’ve LearnedUPDATE YOUR PORTFOLIO Choose a challenging project, Take Another Lookactivity, or exercise you did in this chapter and add it to your portfolio. Explain the strategies you used.

ORGANIZE YOUR NOTEBOOK Review your notebook and your conjecture list to besure they are complete. Write a one-page chapter summary.

WRITE IN YOUR JOURNAL Why do you think the Pythagorean Theorem isconsidered one of the most important theorems in mathematics?

WRITE TEST ITEMS Work with group members to write test items for this chapter.Try to demonstrate more than one way to solve each problem.

GIVE A PRESENTATION Create a visual aid and give a presentation about thePythagorean Theorem.


1

1

1

14

3

2

FACILITATING SELF-ASSESSMENT

To help students complete the portfolio describedin Assessing What You’ve Learned, suggest that they consider for evaluation their work onLesson 9.1, Exercises 17, 18; Lesson 9.2,Exercise 16; Lesson 9.3, Exercise 17; Lesson 9.4,Exercise 9; and Lesson 9.5, Exercise 6.


3. The small square in the centerhas sides b � a, the slantedsquare has area c2, and thetriangles each have area �

a2b�. The

equation c2 � (b � a)2 � 4��a2b��

simplifies to c2 � a2 � b2.

4. One possible proof: Given�ABC with BC � x, AC �x�3�, and AB � 2x, construct30°-60°-90° right triangle DEFwith right angle F, 30° angle D,and EF � x. DF � x�3� and DE � 2x, by the 30°-60°-90°Triangle Conjecture.�ABC � �DEF by SSS.�C � �F and is a right angle,�A � �D and is a 30° angle,and �B � �E and is a 60° angle.

5.

ASSESSING

As part of a final grade, youmight ask students to presentvisual demonstrations or anima-tions of various proofs of thePythagorean Theorem. (See TakeAnother Look activities 1–3.) Ifthis presentation replaces thechapter test, you can give a unitexam after working through themixed review.

1

1

1

1

1

11 1

1

98 7

6

5

4

3

2

c

b

b

b �a

a

a

chapter 9 the pythagorean theorem - high school...

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