Solutions Manual for Geometry: A High School Course [1 ed.] 0387941819, 9780387941813, 9781461208617

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Solutions Manual for

Geometry: A High School Course by S. Lang and G. Murrow

Philip Carlson

Solutions Manual for

Geometry: A High School Course by S. Lang and G. Murrow With 100 Figures

Springer Science+Business Media, LLC

Philip Carlson General College University of Minnesota Minneapolis, MN 55455-0434 USA

Mathematics Subject Classification (1991): 02/04, 02/05 Library of Congress Cataloging-in-Publication Data Carlson, Philip. Solutions manual for geometry: a high school course by S. Lang and G. Murrow/Philip Carlson. p. cm. Includes bibliographical references. ISBN 978-0-387-94181-3 ISBN 978-1-4612-0861-7 (eBook) DOI 10.1007/978-1-4612-0861-7

1. Geometry, Plane-Problems, exercises, etc. 1994 QA459.C24 516.2-dc20

I. Title.

93-38093

Printed on acid-free paper. © 1994 Springer Science+Business Media N ew York Originally published by Springer-Verlag New York, Ine in 1994 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher Springer Seienee+Business Media, LLC , except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely byanyone.

Production managed by Francine McNeill; manufacturing supervised by Vincent Scelta. Camera-ready copy prepared by the author. 987654321 ISBN 978-0-387-94181-3

Contents

CHAPTER 1

Distance and Angles

1

CHAPTER 2

Coordinates CHAPTER 3

Area and the Pythagoras Theorem

21

27

CHAPTER 4

The Distance Formula

38

CHAPTER 5

Some Applications of Right Triangles

43

CHAPTER 6

Polygons ............. .....................................................

53

CHAPTER 7

Congruent Triangles

56

CHAPTER 8

Dilations and Similarities

64

CHAPTER 9

Volumes

..................................................................

80

CHAPTER 10

Vectors and Dot Products

85 v

vi

CONTENTS

CHAPTER 11

Transformations

98

CHAPTER 12

Isometrics

114

CHAPTER 1

Distance and Angles

§

1. Exercises

(Pages 5,6) l.

2.

a.

P-
-Q

h.

P",

Q

c.

P

Q

d.

P

»Q

a.

p h.

~,

The two rays, RpM and RPQ, form a whole line if M, P and Q are collinear and P lies between M and Q.

+ d(Q,M)

d(P,M)

3.

d(P,Q)

4.

AB is not parallel to PQ because LAB is not parallel to LPQ.

5.

Line segments connecting any three points in the plane do not necessarily form a triangle since the three points may be collinear.

=

1

2

GEOMETRY: SOLUTION MANUAL

6.

If line L were parallel to line U there would be 2 lines parallel to line U passing through the point P. This contradicts PAR 2. Therefore line L is not parallel to line U and must intersect it.

7.

By the definition of parallelism, two lines, K and L, are parallel if K = L or K is not equal to L and does not intersect L. If P is on line L and K = L, then K and L are parallel lines.

Experiment 1-1 (Pages 7,8) 1.

p----&

s X------y

R------

x 2.

y

3

CHAPTER 1. DISTANCE AND ANGLES

3.

4.

p

5.

6.

.

Q

\

The ares with radii 7 em and 5 em do not intersect sinee 5+7

< 15.

4

7.

GEOMETRY: SOLUTION MANUAL

a.

/1b.

For a triangle, the sum of the lengths of any two sides is greater than the length of the third side.

8.

You cannot construct a triangle with sides 5 cm, 10 cm and 15 cm since the segments would be collinear.

9.

If P, Q and M are points in the plane and d(P, Q) d(P,M) the points P, Q and M are collinear.

+

d(Q,M) =

r §

2. Exercises

(Page 11)

1.

a.

Both Ygleph and Zyzzx are, at most, 100 km. from the antenna.

b.

The messenger travelling from Ygleph to the antenna and then to Zyzzx would travel at most 200 km.

c.

The maximum possible distance between Y gleph and Zyzzx is 200 km.. This distance would occur if Y gleph and Zyzzx were at opposite ends of a diameter of a circle with radius 100 km. (see diagram in lb.)

CHAPTER 1. DISTANCE AND ANGLES

Proof: d(Y,Z)

~ ~ ~

d(Y,A) + d(A,Z) by the Triangle Inequality 100 + 100 since Y and Z receive the signal 200

2.

d(A,C) ~ d(A,B) + d(B,C) by the Triangle inequality d(A,C) ~ 265 + 286 = 551 by the charts Also 286 ~ d(A,C) + d(A,B) by the Triangle inequality ~ d(A,C) + 265, since d(A,B) = 265. Hence, d(A,C) ~ 286-265=21. Therefore, 21 ~ d(A,C) ~ 551

3.

(a) (b) (c) (d) (e) (f)

4.

yes yes no, 5 + 2 < 8 yes no, 1 112 + 3 112 yes

5, collinear

•

If two sides of a triangle are 12 cm and 20 cm, the third side must be larger than l cm, and smaller than 32 cm.

20 - 12

5.

=

< x < 12 + 20, where x is the length of the third side

d(P,Q) < r 1 + r2

5

6

6.

GEOMETRY: SOLUTION MANUAL

XPh Z

Y 5

d(X,Z) + d(Z,y) = d(X,Y), by SEG postulate. Hence, 1 112 + d(Z, Y) = 5 and d(Z, y) = 3 112. 7.

Check your own work.

8.

Since X and Yare contained in the disc, d(p,X) :s: rand d(p,Y) d(X, Y) :s: d(P ,X) + d(p, Y) by the Triangle inequality Therefore, d(X, Y) :s: r + r = 2r

:s:

r.

Experiment 1-2 (pages 12, 13) 1.

2.

a.

If a number is even, then it is divisible by 2. If a number is divisible by 2, then it is even.

b.

If6x = 18, then x = 3. Ifx = 3, then 6x = 18.

c.

If a car is registered in California, then it has California license plates. If a car has California license plates, then it is registered in California.

d.

If all of the angles of a triangle have equal measure, then the triangle is equilateral. • If the triangle is equilateral, then all of its angles have equal measure.

e.

If two distinct lines are parallel, then they do not intersect. If two distinct lines do not intersect, then they are parallel.

a.

False, If the square of a number is 9, then the number is not necessarily 3 because (-3? = 9 also.

b.

False, If a man lives in the U.S., then he does not necessarily live in California. He may live in any U.S. state.

c.

False, If a2 example, a

= b2,

=

then a is not necessarily equal to b. For -3 and b = 3 and a2 = b2•

CHAPTER 1 • DISTANCE AND ANGLES

3.

d.

True

e.

False, The statement If the sum of integers x. y and z is 3y. then x. y and z are consecutive integers is a false statement. For example, let x = 5, Y = 10 and z = 15. Then x + y + z = 30 = 3y; but 5, 10 and 15 are not consecutive integers.

The converse of a true "if-then" statement is not always true. Note #2 (a, b, c and d). Make up five "if-then" statements and test them to determine whether they are true or not by testing the statement for any exceptions. If there are any exceptions, the "if-then" statement is false.

§

3. Exercises

(Pages 22-25) 1.

a. h. c.

d. 2.

3.

7

a.

b. c.

LABC or LRPQ or LYOX or LBOC or

LCBA or LB LQPR LXOY LCOB

x = 180 - 40 = 140 x = 360 - 68 = 292 x = 146 - 35 = 111 0 0 0

a.

b.

c.

d.

c-\

Iq~O

------~---~-------

8

GEOMETRY: SOLUTION MANUAL

e.

4.

(1) 45° (4) 90°

5.

90/360

6.

a.

(2) 60° (5) 20°

= 114

(3) 135° (6) 225°

The length of arc is (114) (36) = 9.

= 118 The length of arc is (118)(36) = 912 180/360 = 112 The length of arc is (112)(36) = 18 60/360 = 116 The length of arc is (116)(36) = 6

45/360

b. c. d.

(x/360)(36) =

=

x 110

7.

(11360)(40,000)

8.

a. Distance to the equator

40,000/360

(43/360)(40,000) b. (47/360)(40,000)

=

=

=

1000/9 km

=

172000/360 km = 43000/9 km 47000/9 km to North Pole since 90°-43°

9.

If 1 is the latitude of your home city or town, take [(90 -1)/360](40,000) = distance to N.P.

10.

All of the angles L AXB have the same measure.

II.

a. 30°

12.

a.

?

c. 110°

b.220°

b.

~

M~l;.oo f

cQ d.

c.

M~,a~ r

I "O°

q

~

r

~

=

47°

CHAPTER 1. DISTANCE AND ANGLES

h.

13.

Converse: If points Q, P and M lie on the same line, then m( L OPM) = 0°. It is false as shown by < ______~1~8~00_________ >

P

Q

M

Experiment 1-3 (Pages 27-29) 1.

2.

For example, a. b.

m(LA) = m(LA') = 120°, m(LB) = m(LB') = 60°

c.

Measure the angles as above.

d.

Conclusion: Opposite angles have the same measure.

a.

Produce a figure similar to fig. 1.51

b.

The measure of the angle formed by the bisectots is 90°.

Exercise a. b. c.

1. m( L A) + m( L B) = 180° since they form a straight angle. For the same reason, m( LA') + m( L B) = 180°.

mel A) = mel A')

9

10

GEOMETRY: SOLUTION MANUAL

Exercise 2. a. m( L QOP) + m( L ROP) = 180° since these adjacent angles form a straight angle. b. Let m( L QOP) = xO. Then m( L ROP) = (180-x)0 m(angle between bisectors) = (1I2)m( L QOP) + (1I2)m( L ROP) = (1I2)(m( L QOP) + m( L ROP» = (1I2)(x + (180 - x» = (112)(180) = 90°. Proof of 1.

~ ~ Because L A and L B form a straight angle, we know m( L A) + m(LB) = 180. Similarly, m( LA') + m( L B) = 180. Therefore, m(LA) + m(LB) = m(LA') + m(LB). m( L A) = m( L A') subtracting m( L B) from each side. When two lines intersect, opposite angles have the same measure. Proof of 2

ll\L-_Y ~

0

R

Let Rox bisect L QOP and Roy bisect L ROP. Since L QOP and L ROP form a straight angle we know m( L QOP) = m( L ROP) = 180. If we let m( L QOP) = x and substitute in the first equation, we obtain x + m(LROP) = 180. Thus m( L ROP) = 180 -x. The angle between the bisectors, L XOY, has measure m(LXOY)= (112) m(LQOP) + (112) m(LROP) = (112) (m( L QOP) + m( L ROP» = (112) (x + (180 - x» = (112) (180) = 90°. Thus, we have that the bisectors of a pair of linear angles form an angle whose measure is 90°.

11

CHAPTER 1. DISTANCE AND ANGLES

§ 4. Exercise (Pages 33-36) 1.

A postulate is a statement accepted as true without proof. a. Acceptable answers include all postulates given in the text. The Triangle Inequality Postulate is one example. b. e.g. Given real numbers x, y and a, if x=y then x+a=y+a.

2.

a. b.

m(LNOM) = m(LPOQ) = 90 - 55 = 35°

m( L SOQ) = m( L RON), Opposite angles

3. Given m( L WPX) = 2m( L WPY) = 2x and m( L WPY) = xO. (see figure 1.57) m( L YPV) = m( L WPX) = 2xo, opposite angles x + 2x = 180, since L WPX and L WPY form a straight angle. Thus 3x = 180 and x = 60. Therefore, m(LYPV) = 120°. O

4.

Assume RoB bisects L Aoe and Roc bisects L BOD. Prove: m( L AOB) = m( L eOD)(see fig. 1.58) Let m(LAOB) = x, m(LBOC) = y and m(LeOD = z. Since RoB bisects L AOe, x = y. Since Roc bisects L BOD, Y = z. Therefore x = z and m( L AOB) = m( L COD).

5. Assume lines PV,QT and RS meet in point 0 and line L POR. Prove: Line QT bisects L SOV. see fig. 1.59)

QT bisects

m( L POQ) = m( L QOR) since Rar bisects L POR m( L POQ) = m( L TOV), opposite angles m( L QOR) = m L SOT), opposite angles Therefore, m( L TOV) = m( L SOT). Thus ROT bisects L SOY and line QT bisects L SOV. 6.

Assume: d(P,Q) = d(R,S) Prove: d(P,R) = d(Q,S) (see fig. 1.60) Since d(P,Q) = d(R,S) we can add d(Q,R) to both sides of this equation and obtain d(Q,R) = d(P,Q) + d(Q,R) by SEG postulate = d(R,S) + d(Q,R) by assumption = d(Q.S) by SEG postulate

12

7.

GEOMETRY: SOLUTION MANUAL

Assume points X and Yare contained in a disc with radius r around a point P. Prove: d(X,Y) ::;; 2r. d(X,Y) ::;; d(P,X) + d(P,Y), by the Triangle Inequality ::;; r + r = 2r because X, Yare in the disc centered at P, radius r

8.

Assume line L intersects lines K and U so that L 1 is supplementary to L 2. Prove: m( L 3) = m( L 4) (see fig. 1.61) mel 1)+m(L2) = 180, L 1 supplementary to L2 m(L1)+m(L3) = 180, L1, L3 are linear angles Therefore, m(L1)+m(L2) = m(L1)+m(L3). Thus (L 2) = m( L 3), by subtraction and m( L 2) = meL 4) since they are opposite angles. m( L 3) = m( L 4) by substitution.

9.

Assume in triangle ABC, m( L CAB) = m( L CBA) and in triangle ABD, m( L DAB) = m( L DBA). Prove: m( L CAD) = m( L CBD) (see fig. 1.62) Since L DAB and L CAD are adjacent angles, we obtain m( L DAB) + m( L CAD) = m( L CAB). m( L CAD) = m( L CAB)-m( L DAB) by subtraction and = m( L CBA)-m( L DBA), assumption and substitution = m( L CBD) L CBA and L DBA are adjacent angles. Therefore, m( L CAD) = m( L CBD)

10.

Assume m( L b) = m( L c) Prove: m( L a) = m( L d) (see fig. 1.63) L a and L b are opposite angles as are L c and L d. By theorem 1.1 m(La) = m(Lb) and m(Lc) = meld). Then m( L b) = m( L c) by assumption, and m( L a) = m( L d) by substitution.

11.

Assume points P, B, C and Q are collinear, m( L x) = m( L y), segment BK bisects L ABC and segment CK bisects LACE. Prove: m( L KBC) = m( L KCB) (see fig. 1.64) Since L PBA and L ABC are linear angles we have m(Lx)+m(LABC) = 180 m( L ABC) = 180 - m( L x). Similarly, since

13

CHAPTER 1. DISTANCE AND ANGLES

m( L ACB) = 180 - m( L y) since L y and L ACB are linear angles. Segment 13I{ bisects L ABC so m( L KCB) = 112 m( L ABC) and = 1/2(180 - m( L x». Thus = 112(180 - m( L y), by assumption. Also, segment CK bisects L ACB and m( L KCB) = 112 m( L ACB) Thus = 112(180 - m( L y) by substitution. m( L KBC) = m( L KCB) since both equal 112(180 - m( L y).

§ 5. Exercise (Pages 41-43) 1.

Since K 1. V and L 1. V we have K parallel to L by theorem 1.2. Two lines perpendicular to the same line in a plane are parallel.

2.

Given that K 1. Ll and Ll is parallel to Lz we have K 1. Lz by PERP 2. Given two parallel lines, Ll and Lz, ifK1.Ll' then K1.Lz·

3.

Assume PR 1. PT and PQ 1. PS Prove: m( L a) = m( L b) (see fig. 1.76) By assumption, PR 1. PT thus L TRP is a right angle and m(Lb)+m(Lx)=90. Since PQ1.PS we also have m(La)+m(Lx)=90. Since both sums equal 90 we obtain m(La)+m(Lx)=m(Lb)+m(Lx). By subtraction m(La)=m(Lb).

4.

For each example, the assumption should lead to a contradiction of some accepted statement (as in the example).

5.

Assume: ABCD is a parallelogram with L A a right angle 1.77) Prove: L B, L C and L D are right angles.

(see fig.

Since L A is a right angle, AB 1. AD. Since ABCD is a parallelogram An is parallel to AB. By Perp 2 we conclude that AB 1.:Be and L B is a right angle. Also, AB is parallel to CD and BC 1. AB so Perp 2 implies that ~ 1. CD and L C is a right angle. Finally, since CD 1. Be and OC is parallel to AD we obtain CD 1. AD and L D is a right angle by Perp 2. Therefore, L B, L C and L D are right angles and ABCD is a rectangle.

14

6.

GEOMETRY: SOLUTION MANUAL

a) b) c) d) e) f) g)

h)

90 90 - 50 = 40 (see fig. 1.78) yes 180 90 - 23 = 67 LiandL3 m( L 2) = 90 - 32 = 58 m(LTOB) = 58+90 = 148 Ifm(L 1) + m(L4) = 90 by (d) we know m(L2) + m(L3) = 90 and Rar is perpendicular to Ros.

7.

Using Post Perp 1 we can conclude that L, and

8.

In quadrilateral PBQC assume L PBQ and L PCQ are right angles and m( L x) = m( L y). Prove: m( L ABQ) = m( L DCQ) (see fig. 1.79)

~

are the same line.

m( L x) + m( L CBQ) = 90 given the L PBQ is a right angle. m( L y) + m( L BCQ) = 90 since L PCQ is a right angle. By addition m(Lx) + m(LCBQ) = m(Lx)+m(BCQ) since m(Lx)=m(Ly). m( L CBQ) = m( L BCQ) by subtraction. Then mel CBQ)+m(L ABQ)= 180 L CBQ and LABQ are linear angles. m(LBCQ)+m(LDCQ)=180 again by linear angles. m(LCBQ)+m(LABQ = m(LCBQ)+m(LDCQ) they equal 180. m( L ABQ) =m( L DCQ) by subtraction. 9.

Assume L, is parallel to ~ and Prove: L, is parallel to l.:J

~

is parallel to

~.

Further assume that L, is not parallel to l.:J. By PAR 1, L, meets l.:J in a point, P. By PAR 2, given line ~ and point P there is one and only one line passing through P parallel to~. However, L, and l.:J both pass through point P and are parallel to~. This contradicts PAR 2 so we conclude that the above assumption is false and L, is parallel to l.:J.

§ 4. Experiment (pages 45-46) 1.

m(La) + m(Lb) = 54 + 36 = 90

CHAPTER 1. DISTANCE AND ANGLES

2.

m(La) + m(Lb) = 38 + 52 = 90 m(La) + m(Lb) = 77 + 13 = 90

3.

No, Two right triangles with corresponding legs of equal measures have the same shape.

4.

Two such right triangles form a rectangle.

5.

The m(La)+m(Lb)+m(Lc)=180

6.

Constructing a perpendicular at a point on a line is like bisecting a straight angle.

§

15

6. Exercise

(Pages 54-60) 1.

a. c.

m( L C) = 90 - 34 = 56 b. m(L C) = 30 d. m(LC) = 45 m(LC) = 60

2.

a. b.

m(LC)=180 - (47+110) = 23 m(LC)=30 c. m(LC)=50 d. m(L C)=36

3.

a.

c.

m(Lx)=180 - 50=130 m(L x)=90 - 64=26

a. c. d.

m( L x)= 180-130=50 b. m(Lx)=180-18=162 m( L x)=360-(100+95 +60)= 105 m( L x) = 180-83 =97

4.

5.

--~~~-a.

6.

b. m(Lx)=80 d. m( L x)= 180-60= 120

m(Lx)=25+48=73

b. m(Lx)=20+30=50

Let m(LA)=x, m(LB)=2x and m(LC)=3x. Then x+2x+3x=180 and 6x=180. Thus x=30 and the three angles have measure 30, 60 and 90.

16

GEOMETRY: SOLUTION MANUAL

7.

(see figure 1.106) Since AP bisects LA let m(LCAP)=m(LPAB)=x. Similarly, let m(LCBP)=m(LPBA)=y. Then 2x+2y+70=180 and 2x+2y=110 or x+y=55. In ~B, x+y+m(LP)=180. Combining these expressions 55+m(LP)=180. Thus, m( L P)= 180-55= 125.

8.

Since a square and a rectangle can be divided up into two triangles, the sum of the angles of a square and of a rectangle is 360°.

9.

Let SWAT be any four-sided figure. (see fig. 1.107) Prove: The sum of angles of SWAT is 360°. Connect the vertices W and T with a line segment. Note that this forms two triangles, STW and TWA. L STW and L WT A form L STA and L SWT and L TWA form L SWA. For each triangle the sum of angles is 180° so the total angle measure for SWAT is 2 x 180° = 360°.

10.

Suppose triangle ABC has two obtuse angles, L A and L B. m( LA) + m( L B) > 180° contradicting the theorem that the sum of angles in a triangle equals 180°. Therefore, no triangle can contain two obtuse angles.

11.

Yes, for example, the 30°, 60°, 90° triangle contains no obtuse angle.

12.

Assume that line M is perpendicular to line L, m(Lx)=m(Ly) and L is parallel to K. Prove: m( L a)=m( L b). (see fig. 1.108) Since L is parallel to K and M is perpendicular to L, by PERP 2 we know that M is perpendicular to K. Thus mel x)+m(La)=90 and m(Ly)+m(L b)=90. Because m(Lx)=m(Ly) we obtain m(Lx)+m(Lb)=90 by substitution. Therefore, m(Lx)+m(La)=m(Lx)+m(Lb). By subtraction we conclude m(La)=m(Lb).

13.

Given figure 1.109, assume segment CL is parallel to segment AB. Prove: m( L s)=m( L p) +m( L q)

a.

L r and L BCL are alternate angles for the parallel lines AB and Thus, m(Lr)=m(LBCL). Also, m(LBCL)+m(Ls)=180 since they form a linear pair of angles. By substitution we obtain

CHAPTER 1. DISTANCE AND ANGLES

17

m(Lr)+m(Ls)=180. In the triangle ABC, m( L r) +m( L p) +m( L q) = 180. Therefore, mel r)+m(L s)=m(L r)+m(Lp)+m(L q). By subtraction we conclude mel s)=m(Lp)+m(L q).

14.

Let PMQ be a right triangle with L M the right angle and assume segment KIR is perpendicular to segment PQ. (see fig. 1.110) Prove: The 3 angles of ",PHM have the same measure as the 3 angles of t.MHQ. m(LPHM)=m(MHQ) because they are both right angles. Since L Q belongs to both triangles, PMQ and MHQ we note that m( L Q) +m( L HMQ) +90 = 180 and m(LQ)+m(LP)+90=180. Therefore, by subtraction we obtain m( L HMQ)=m( L P). Finally, since two pairs of angles of triangles PMH and MHQ have the same measure the remaining pair of angles will also have the same measure. Therefore, the three angles of t.PHM have the same measures as the three angles of illHQ.

15.

Let TALK be a parallelogram. Prove: m( L K)=m( L A) and m( L T)=m( L L)

Construct the rays RTA and RKL • Name the new angle adjacent to LA, L y and the other new angle, L x. L T and L yare parallel angles so m(LT)=m(Ly). And m(LL)=m(Ly) because they are alternate angles. Thus, m( L T)=m( L L).In a similar way m( L K)=m( L x) and m( L A)=m( L x) so m(LK)=m(LA).

16.

Theorem 1-7

17.

Let ABC be an arbitrary triangle with side AC extended in the C direction. Prove: mel l)=m(LA)+m(LB) (see fig. 1.112) Note that L 1 and .L C form a linear pair so m( L 1) + m( L C) = 180. In ",ABC we have m(LA)+m(LB)+m(L C)=180. Thus, mel l)+m(L C)=m(LA)+m(LB)+m(L C). By subtraction we obtain our conclusion since mel l)=m(LA)+m(LB).

18

18.

GEOMETRY: SOLUTION MANUAL

Assume for L A two perpendiculars are dropped from point Q (within the angle) to each of the rays intersecting the rays at points M and N. (see fig. 1.113) Prove: L A and L C are supplementary. Since QM 1. AM and 15.3. In problem 3, 10+ 10> 14.1.

m

N , ;

;

,

30

GEOMETRY: SOLUTION MANUAL

5. 6.

7. I

t-...

......

.......

:7

.......

I'

I'--

I-....

I/_~

§

2. Exercises

(pages 101-108) 1.

(a)

52

+ x2 = 36, x2 = 11, x=.y1'i

(b)

x2

+ 122 = 132,

(b)

x2 = 2(3 2) = 18, x =

J../'L

(c)

x2 = 2(4~ = 32, x =

4J'f.

(d)

x2 = 2(5~ = 50, x =

5J'I.

(e)

x2

X2

= 169-144=25, x=5.

= 2(r2), x = r-J'L

CHAPTER 3. AREA AND THE PYTHAGORAS THEOREM

3.

31

..f5.

(a)

d2 = 12 + 22 = 5, d =

(b)

d2 = 32 + 52 = 34, d = ..[34.

(c)

d2 = 42 + 72 = 65, d =

(d)

d2 = r + (2r)2 = 5r, d = r~.

(e)

d2 = (3r)2 + (5r)2 = 34r, d = r.J34.

(f)

d2 = (4r)2 + (7r)2 = 65r, d = r.,f65.

...rr;s.

4.

Since the cube has the same length sides the formula for the diagonal is (diagonal of base)2 + (height of cube)2 = (S2 + s~ + S2 which simplifies to d2 = 3s2• (a) d2 = 3(1)2 = 3, d = ..[1. (b) d2 = 3(2)2 = 12, d = 2..[1. (c) d2 = 3(3)2 = 27, d = 3..['J'. (d) d2 = 3(4)2 = 48, d = 4./3. (e) d2 = 3r, d = N3.

5.

In the drawing of the box, note x is the length of the diagonal of the base rectangle and d is the length of the diagonal of the box.

x2 = d2 = d2 = (a) (b) (c) (d) (e) 6.

a2 + x2 + a2 +

b2

d2 = d2 = d2 = d2 = d2 =

c2 Thus, b2 + c2. 32 + 42 + 52 = 9+16+25=50, d=5..fI. 1+4+16 = 21, d=v'2I 1+9+16 = 26, d=l26. a2 + b2 + c2, d=..Jai +b2 +c2• (ra)2+(rb)2+(rc)2, d=r..,fa2+b2+c2•

In figure 3.34, note that the tower and ground form a right angle and the wire is the hypotenuse of the right triangle. Therefore, w2 = 252 + 302 = 625+900 = 1525 = (25)(61) w = 5J6f m. That is approximately 39 m.

32

7.

GEOMETRY: SOLUTION MANUAL

Given that the baseball diamond is a square with sides 90 feet in length, the distance from the home plate to second base is the diagonal of the square. d2 = 902 + 902 = 2(90)2, d approximately 127.28 feet.

9o..['f feet.

This distance is

8.

For this square the area = 144 sq. cm. so S2 = 144 and s =.Jl44 = 12. Let the diagonal be of length d. d2 = 2(12)2 thus d = 12.,(2cm.

9.

For this right triangle, let the shorter leg have length x. The other leg is then of length 2x. Since the area equals 72, 'hx(2x)=72, x2 = 72, x=W2 and 2x = 12..j2. Let the hypotenuse have length h. h 2 = (6.J'I'F + (12.J2)2 = 72+288=360, h=&/IO.

10.

(a)

IPA,1 2 = 12 + 12 = 2, IPA,I =..j2.

(b)

IPA212 = (..[2)2 + 12 = 3, IP~I =..['3.

(c)

IPA3 12 = (..['J'i + 12 = 4, IPA3 1 =..f4 = 2.

(d)

IPA"l 2 =

11.

12.

(JDY

+ 12 =n+1, IPA"l = .J(n+b.

The distance of the ship from its starting point is the length of the diagonal of a rectangle with sides 10 km and 5 km. See figure 3.36. (a)

d2 = 102 + 52 = 125, d = s..,.t)km.

(b)

Actual distance from the North Pole is 10 km. The eastward part does not take the ship further from the North Pole.

(a)

h 2 = 32 + 42 = 25, h = 5.

(b)

h 2 = 62 + 82 = 100, h = 10.

(c)

h

= 15.

(d)

h

= 5c.

(f)

(i) a = (100)(3), b = (100)(4) so c= (100)(5) = 500 (ii) 30 (iii) 45

CHAPTER 3. AREA AND THE PYTHAGORAS THEOREM

13.

The authors suggest Furthermore, they y=(2t)/(l +f). Does

33

letting x=alc and y=b/c so that x2 + y2 = 1. suggest the formulas x=(l-t~/(l +t~ and 2 x +y2 = 11

Choose t=2: x=(-3)/5, y=415, a=3, b=4, c=5 Choose t=7: x=(-48)/SO, y= 14150 so a= 14, b=48 and c=50 (Better to divide by 2) a=7, b=24, c=25. 14.

(a)

Divide 15 and 36 by 3 revealing 5, 12, 13. Thus x =3(13) = 39.

(b)

(3, 4, 5) so x = 5 m.

(c)

Divide 108 and 144 by 36 obtaining 3, 4, 5. Thus x = 36(5) = 180.

15.

Area of the right triangle equals 27 and its legs could have lengths 2x and 3x and be in the ratio 2:3. Thus, Ih(2x)(3x)=27, 3x2=27. x2=9 so x = 3. °The legs have lengths 6 and 9. h2 = 62 + 92 = 36+81=117. h =.JTI'1 = J.JTI.

16.

Let this right triangle have legs with lengths 4x and 5x. The smaller leg is 415 of the larger leg. Since the area is 320, Ih(4x)(5x)=320, 10x2 = 320. Then x2 = 32 and x~=4.J'1. The legs will have lengths 1&/2 and 20..[2.

17.

w'- = 182 + 302 + 202 = 324+900+400=1624. Thus, w = v'1624 = 2..[406 m. This is slightly more than 40 m. (See figure 3.38)

18.

(See figure 3.39) Let the larger square have sides oflength 2x. Thus its area is 4x2. The side of the smaller square is the hypotenuse of a right triangle with both legs of length x. Thus, S2 = 2X2 and s = x.J2. The area of the smaller square is A = (x.J2)2 = 2x2. This is Ih of the area of the larger square.

19.

In this diagram, it is clear that the fielder's throw is along the diagonal of a rectangle with sides 120 ft. and 90 ft.

.

- -

-t;

or

I... II I

T

I"

- r-

h-

Since 90 and 120 are multiples of 3 and 4, we obtain t = 30(5) = 150 so the throw would be 150 ft.

34

GEOMETRY: SOLUTION MANUAL

20.

Let the side of the square have length x. Then, x2 = 169 so X= 13. The diagonal has length d such that d2 = 2(13)2 so d = BIz.

21.

d =

22.

d = 8 so 64 = 2S2 and S2 = 32 = Area of the square.

23.

(a)

Let the sides of the screen have lengths 3x and 4x. Then, d2 = 252 = 625 = (3X)2 + (4X)2 = 25x2. x2 = 25 and x=5. The sides are 15 and 20 so the viewing area of the screen is (15)(20)=300 sq. cm.

(b)

d2 = 502 = 2500 = 25x2. x2 = 100 and x= 10. The sides are 30 and 40 so the viewing area is (30)(40) = 1200 which is four times the viewing area in part (a).

24.

&J2 so 128

= S2 + S2 = 2S2. Thus, s2=64 = area of the square.

In figure 3.40, construct a line through S and parallel to side PQ intersecting side RQ at point T. IST I = IPQ I' = ~-t-= -~r

t-r1'": _' tr -"-,+--If Lg

t

+"-

-=

l-+- -

202 = 42 + ISTI2 so ISTI2 = 400 - 16 = 384 and ISTI = ..j384 = 'W6. Thus, IPQI = 'W6 units. 25.

S=1

--+

IQ

Given two right triangles, aABC and AXYZ, such that I AB I = IXY I and the hypotenuse AC of AABC and the hypotenuse XZ of AXYZ have the same length. Prove: IBC I = IYZ I IABI2 + IBCI 2 = IACI 2 and IXYI 2+ IYZI 2 = IXZI2. Since I AB I = IXY I and IAC I = IXZ I by substitution we obtain IABI2 + IYZI 2 = IACI 2. Thus, IABI2 + IBCI 2 = IABI2 + IYZI 2. By subtraction IBC 12 = IYZ 12 Therefore, IBC I = IYZ I

26.

(a)

Given rectangle ABCD with right triangle AQD inscribed in it such that IABI =a, IBQI =b and IQCi =c. (See fig. 3.41) Prove IAD I = .Jb2+ 2a2 + b2.

IAQI2 = a2+b2 and IQDI2 = a2+c2 • Therefore, IADI2 = IAQI2+IQDI2 = a2+b2 +a2+c2 = b2+2a2+c2. Thus I AD I = ..jb2+2a2+c2 • (b)

IADI2 =(b+C)2 = b2+2bc+c2 = b2+2a2+c2, by part (a). By subtraction, 2bc=2a2 so a2=bc.

CHAPTER 3. AREA AND THE PYTHAGORAS THEOREM

27.

35

Given AABC with IAC I = ICB I and CN .1 AB. Prove IAN I = INB I. (See fig. 3.42) IACI 2 = ICNI 2+ IANI2 and ICBI 2 = ICNI 2+ INBI2. Since IAC I = ICB I, by substitution we have ICNI 2+ IANI2= ICNI 2+ INB I2thus IANI2= INBI 2and IANI = INBI·

28.

Given AABC is a right triangle with the right angle at C and point P on AB the foot of a perpendicular from C. (see fig. 3.43) (a)

Prove: IPCI 2 = IAPIIPBI

Let IACI =b, ICBI =a, IAPI =x, IPBI =y and IPCI =h. By Pythagoras, a2+b2=(x+y)2=x2+2xy+y2whilea2=h2 +y2andb2=h2+x2. By substitution, 2h2+X2+y2=X2+2xy+y2. Therefore 2h2=2xy and IPCI 2= IAPIIPBI·

Using the notation above to prove this statement we need to show b2=x(x+y). From the above statements we use a2+b2=x2+2xy+y2 and a2=h2+y2. Subtracting the second equation from the first we obtain b2=x2+2xy-h2. By adding ~=h2+x2 to each side 2b2=2x2+2xy. By dividing by 2 and factoring the right side we note b2=x(x+y) and IACI 2= IAPIIABI.

29.

Given figure 3.44 with IPMI=IMQI and IXM'I=IM'YI andlineL perpendicular to PQ and n. (a)

Prove: IPXI = IQYI

Draw a perpendicular from P to P' on XY and another from Q to Q' on XY. (See fig. 3.45). SincePQQ'P' is a rectangle IPQI = IP'~d with M the midpoint of PQ we also have M' the midpoint of P'Q'. Thus IP'M' I = IM'Q' I and by subtraction we obtain IXP' I = IYQ' I. In the right triangles AXPP' and AYQQ' , IPP' I = IQQ' I and IXP' I = IYQ' I· By the Pythogoras Theorem, IPP'1 2 + IXP'1 2 = IPXI 2 and IQQ'12 + IYQ'1 2 = IQYI 2. Since the left sides of the equations are equal we have IPXI 2= IQYI 2 so IPXI = IQYI.

36

GEOMETRY: SOLUTION MANUAL

(b)

Prove: IPYI = IQXI

'M

G.

y' - -r:rI I

p'

y

Following th~proach of part (a), construct a perpendicular from X meeti1J!..line PQ at point X'. Construct a perpendicular from Y meeting line PQ at Y'. XX'PP' is a rectangle thus IXP' I = IX'P I and IPP'I = lXX' I· YY'QQ' is also a rectangle thus IQY' I =Q'YI and IQQ' I = IYY' I· From part (a) we have IXP' I = IQ'Y I and IPQI = IP'Q' I· Now one can show that IXP' 1= IP'YI. IX'PI = IPQI + IX'PI SEQ = IP'Q' I + IXP' I by substitution = IP'Q' I + IQ'Y I by substitution =IP'YI· Since, IXX' I = IYY' I and IX'P I = IP'Y I the right triangles PY'Y and QX'X are congruent by SAS. Therefore, by corresponding parts of these congruent triangles, IPY I = IQX I· 30.

Let the right triangle ABC have c as the length of the hypotenuse and a and b as the lengths of its legs. Prove: c>a By Pythagoras c2=a2+b2. We can note that c2>a2 since b2>0. Thus, c > a since length is non-negative. Therefore, the length of the hypotenuse of a right triangle is greater than or equal to the length of a leg.

31.

(a)

Given IAF 1= 10 in figure 3.46, the area of square ADEF is 100. toADG has one fourth of this area, thus it is 25 sq. units.

(b)

Given ICEI=18. ICDI=IBHI=IHGI=lhIDEI so IHGI=6. ICDI = IHDI =6 solADI =12. Area of toADG=lh(6)(12)=36 sq. units.

(c)

Given IBDI =3..[2. The area of the square ABDG =(3..[2)2=18. Area of toADG = Ih(18)=9 sq. units.

(d)

Given area of square BCDH 49. IBC1 2=49 so IBCI =7= IHDI = IARI = IHGI. Thus the area of toADG Ih(14)(7)=49 sq. units

CHAPTER 3. AREA AND THE PYTHAGORAS THEOREM

(e)

37

Given area of AGDEF = 27. The area of AGDEF is one-fourth the area of square ADEF. Let A = area of square ADEF. Thus 27= (3/4}A so A=36. AADG is one-fourth ofthe square ADEF so its area is (114}(36}=9 sq. units.

Experiment 3-3 1.

d(P,Q}=9

2.

d(P,R}=9

3.

.tJ>QR is an isosceles right triangle.

5.

Let Z=(-1,-2), then d(X,Z)=9, d(Y,Z)=4 so d(X,Y)= .J92 +42 = .J97.

6.

Let Y=(x 1 ,3}, then d(A,Y)=.J(x l -5? and d(X,Y)=.J(X2-3)2. d(A,X) =.J(x l -5?+ (x2-3)2.

CHAPTER 4

The Distance Formula

§ 1. Exercises (Pages 113,114)

2.

~

3.

../61

4.

.J2

5.

..j'106

6.

--J 106

7.

2..[2

8.

.J58

9.

2.[53

10.

.J5

11.

IAB I =7, IDC I =7, IAD I =J'19, IBC I =.J29 ABCD is a parallelogram.

12.

(a)

B and C determine a horizontal line segment of length 8 units.

(b)

A and B determine a vertical line segment of length 6.

(c)

IACI

38

= 10

CHAPTER 4. THE DISTANCE FORMULA

13.

39

Let x= distance between (3,S) and (-l,Z). x2=(3-(-1))2+(S-Z)2=ZS, x=S. Let y= distance between (3,S) and (3,0). y2=(3-3)2+(S-0)2=ZS, y=S. Thus x=y.

14.

ZS=(Z-(-1))2+(k-1)1=9+k1-Zk+ 1 =kl_Zk+ 10 0=k2-Zk-1S = (k-S)(k+ 3) Thus k=;,S or k=-3. I

IS. I- ~11 I'll

./ i'\. -

16.

/

"-

/

1"-

/

,

CHAPTER 5. SOME APPLICATIONS OF RIGHT TRIANGLES

51

For radii of this circle, let IOPI=IOQI=randlet IOMI=y. SinceP and Q are points of tangency for the two lines, the triangles .t..OPM and .t..OQM are right trian2les. By Pythagoras we have IPMI 2=y2_r = IQMI!· Thus IPMI = IQMI· 10.

In figure 5.69 we are given a quadrilateral whose sides are tangent to the

circle. Prove: IOVI + ILEI = ILOI + IVEI· By the proof in problem 9 we have IOTI=IOAI ILMI=ILAI IEMI=IEHI IVT I = IVH I . If we add these equations we obtain lOTI +ILMI + IEMI +IVTI = IOAI + ILAI + IEHI + IVHI· By a rearrangement of these terms we have (lOTI + IVTi)+(ILMI + IEMi)=(IOAI + ILAI)+(IEHI + IVHI)· By the SEG Postulate we can replace these pairs by IOV I + ILE I = ILO I + IEV I· 11.

In figure 5.70 we are given that all three lines are tangent to the circle

and IRPI =5 cm. IRPI=ITRI=5. IPQI=IQEI and ITSI=IESI· 5 = I RP I = I RQ I + I QP I = I RQ I + I QE I and 5 = ITR I = ITS I +1 SR I = IES I + ISR I. Since IQE I + IES I = IQS I , IRQ I + IQS I + ISR I = 10. Therefore, the perimeter of L QRS is 10 cm. 12.

In figure 5.71 we are given a tangent to the circle from point M and a secant to the circle from M and passing through the center, O. Prove: IPMI 2= IMQIIMRI. Draw OP and let c= IOPI = IORI = IOQI. Also let a= IPMI and b = IMR I. Since radius OP is perpendicular to PM, in the right triangle OPM we get (b+c)2=a2+c2 and b2+2bc+c2 =a2+c2. By cancellation we obtain a2=b2+2bc=b(b+2c). By substitution we have IPMI2= IMRIIMQI since b+2c= IMRI + IORI + IOQI = IMQI.

13.

Let PUNT be a quadrilateral inscribed in a circle. (See fig. 5.72) Prove: The opposite angles of PUNT are supplementary. The angles of the quadrilateral PUNT are inscribed angles of the circle. Consequently,

52

GEOMETRY: SOLUTION MANUAL

m(LP)=lhm(arc UNT) and m( L N)= lhm(arc TPU). Furthermore, m(LP)+m(LN)=lhm(arc UNT)+lhm(arc TPU) =lh[m(arc UNT)+m(arc TPU)] = lh(360) = 180°. Therefore, L P and L N are supplementary. Since a quadrilateral's angles have measures adding up to 36Q2 the remaining two angles, L U and L T, are supplementary. 14.

In figure 5.73 we are given the chords of a circle meet at point 0, not

necessarily the center. Prove: m( L ROP) = lh(sum of the measures of the arcs 1U" and QS): Draw PS. L ROP is an exterior angle of t..POS. By the Exterior Angle Theorem, m( L ROP)=m( L S)+ M( L P). Since L S and L P are inscribed angles of the circle, m( L S)= lhm(arc RP) and m(LP)=lhm(arc~. By substitution we have m(LROP)=lhm(arc PR)+lhm(arc (is) =lh[m(arc PR)+m(arc~

15.

In figure 5.74 we are given RAe and RAE intersecting the circle. Prove: m(LA)='h[m(arc CE)-m(arc

Bi5)]

Draw CD. L CDE is an exterior angle for t..ADC. By the Exterior Angle Theorem, m(LCDE)=m(LA)+m(LC). Solving for m(LA), m( L A) =m( L CDE)-m( L C). Since L CDE and L C are inscribed angles we have mel CDE)=lhm(arc CE) and m( L C)= lhm(arc 1ID). By substitution we obtain m( L A) = lhm( arc CE}-lhm(arc BD) or m(LA)='h[m(arc CE)-m(arc BD)].

CHAPTER 6

Polygons

§ 1. Exercises (Pages 174-176)

1.

2.

3.

4.

(a)

yes, concave

(b)

no

(c)

yes, convex

(d)

no

(e)

yes, concave

(a)

For the octagon (8-2)180= 1080°

(b)

For the pentagon (5-2)180=540°

(c)

For the 12-gon (12-2)180= 1800°

(a)

(6-2)180/6 = 120°

(b)

(11-2)180/11 = 147 3/11°

(c)

(14-2)180/14= 154 2/7°

(d)

(n-2) 180/n

Let E be the measure of each exterior angle of this regular polygon. If each interior angle contains 153°, E= 180-153 =27°. Since the sum of the exterior angles is 360° each one contains 360/n°. Thus 27=360/n and n=360/27=13 113. No polygon can have a fractional number of sides. Thus an interior angle of 153° is impossible. 53

54

GEOMETRY: SOLUTION MANUAL

5.

(n-2)lS0/n=165 or use E= lS0-165= 15. n=360/15=24 sides.

6.

(a)

(n-2)lS0=2700, n-2= 15, n= 17

(b)

(n-2)lS0=10S0, n-2=6, n=S

(c)

(n-2)lS0=d, n-2=dl180, n=2+dl1S0

7.

IfI=140°, E=lS0-140=40. n=360/40=9 sides

S.

(a)

rhombus

(b)

rectangle

9.

The two equal sides could be of lengths 4 or 10. However, 4+4 is less than 10, the base has to be 4 and the equal sides 10 so the perimeter is 24.

10.

(a)

Since two of these equations have to be equal if the triangle is isosceles we let 2n-1=n+S and find n=6 and the side lengths are 10, 11 and 11.

(b)

Try the other two combinations.

2n-1=3n-S, then n=7 and the sides have lengths 12, 13 and 13. 3n-8=n+S, then n=6.S and the side lengths are 11.5, 11.5 and Therefore, three values produce an isosceles triangle, 6, 7, 6.5. (c)

No value of n produces an equilateral triangle.

11.

P=36 so s=12. A=12'1..v3/4=36.[3 sq. units

12.

P=36 so s=9. A=9'1.=81 sq. units

13.

Let this polygon have n sides. (See fig. 6.1S) Prove: Sum of the outside angles equals (n + 2)180°. One of these outside angles and the corresponding interior angle have a sum of measures equal to 3600. The total of the measures of these pairs of angles is n(360)0. The sum of the interior angles is (n-2)lS0 so the sum of the outside angles equals n(360)-(n-2) 180 =2n(lS0)-(n-2)lS0 = (2n-(n-2))lS0 =(n+2)lS0.

CHAPTER 6. POLYGONS

55

14.

For figure 6.19, P=2(5-2x)+2(4-2x)+8x=10-4x+8-4x+8x= 18 where the second term requires that x < 2 (note that one side has length 0 if x=2.)

15.

A=(5)(4)-4x2

16.

Let the regular hexagon have sides of length s. Prove: A = 3s2..j3/2 Draw the radii to the vertices of the hexagon. This produces six equilateral triangles with sides of length s. Thus the area of the hexagon can be computed by A=6(S2...['J/4)=3s2.,f3/2.

17.

P=30cm. sos=30/6=5cm. Thus the area = 3(5'J.j3/2 = 7s.!3/2 sq. cm.

18.

A=3(2).j3/2=6[j sq. units

19.

In figure 6.20 we have a regular hexagon with sides of length s and the perpendicular distance from the center to the side equal to x. Prove: A=3xs. Draw the radii to vertices F and E. The area of this equilateral triangle is Ihxs. The total area of the hexagon equals 6(lhxs)=3xs.

CHAPTER 7

Congruent Triangles

§

1. Exercises

(Pages 188-191)

1.

IABI=IPQ\, IBCI=IQRI, IACI=IPRI, m( L B)=m( L Q) and m( L C)=m( L R).

2.

(a)

Yes, SSS

(b)

No, two possible triangles.

(c)

No, could be any size of the same shape.

(d)

Yes, ASA

(e)

No, The equal sides are not both between the given angles.

(t)

Yes, SAS

3.

m(LA)=m(LP),

We are given that AABC has IAB I = IBC I and segment BN perpendicular to AC. Prove: BN bisects L ABC. By the Isosceles Triangle Theorem, m( L A)=m( L C). Also we have two right angles, L ANB and L CNB. Since IAB I = IBC I we have the ASA conditions met for congruence of AANB and ACNB. m(LABN)=m(CBN) since they are corresponding parts of these congruent triangles. Thus BN bisects L ABC.

4.

56

We are given two right triangles, AABC and AXYZ, wlth right angles at Band Y and IBCI = IYZI and m(LA)=m(LX). Prove: AABC::::: AXYZ

CHAPTER 7. CONGRUENT TRIANGLES

57

Since L B and L Y are right angle we have m( L B)=m( L Y). We are given that m( L A) = m( L X) and IBC I = IYZ I so, by the ASA property, !>ABC =< I>XYZ. 5.

We have APQR with m(LP)=m(LR). Prove: IPQI = I QRI Draw the perpendicular from Q to PR meeting it at point N. Thus m( L QNP)=m( L QNR) since they are right angles. We have m(LP)=m(LR). In addition, I QNI = I QNI. By ASA, we have the two congruent right triangles: APNQ == ARNQ. Since corresponding parts of congruent triangles have the same measure, IPQ I = I QR I·

6.

Let !>ABC have three angles with the same measure. Prove: L ABC is equilateral. Given that the m( L A) = m L B) by problem 5 it follows that IBCI = IACI. Also we have m(LB)=m(LC). Again by problem 5 we have IAC I = IAB I. Thus IAB I = IAC I = IBC I and AABC is equilateral.

7.

Let AAHB and ADHC have I CH I = IHB I and IAH 1 = IHD I· Prove: AB is parallel to CD. Since ICHI=IHBI, IAHI=IHDI and m(LAHB)=m(DHC) by Opposite Angle Theorem we have AAHB =< ADHC by SAS. Regarding the ironing board, we now know that m(LHCD)=m(LABH) because they are corresponding angles in these congruent triangles. That means that AB is parallel to CD since these alternate angles have the same measure (problem 21, page 59). Consequently, the ironing board will remain parallel to the floor.

8.

In Thales triangles we have m(LD'BA)=m(LDBA), m(LD'AB)=m(LDAB). Prove: IAD'I = IADI (See figure 7.21)

We have given that m( L D'BA)=m( L DBA) and m(LD'AB)=m(LDAB). We know that IBAI = IBAI. By the ASA property, !>ABD =< AABD'. Thus, since these are corresponding parts, the land distance, IAD' I equals the sea distance 1AD 1 . 9.

This method of Thales is also based on the ASA congruence case. Here we have m(LXZY')=m(LXZY), m(LZXY')=m(LZXY) and 1 ZX I = 1 ZX I. Thus AZXY =< AZXY'. His land distance, 1XY' 1 equals the sea distance, IXY I.

58

10.

GEOMETRY: SOLUTION MANUAL

For this pliers we are given IAB I = IEC I and IBE I = IED I. Since L ABB and L DEC are opposite angles their measures are equal. By SAS we have t.AEB=ADEC. m(LDEC)=m(LBAE) since they correspond in these congruent triangles. These alternate angles having the same measure implies that AB is parallel to DC.

§ 2. Exercises (Pages 195-200) 1.

In parallelogram ABCD the diagonals intersect at Prove: IAO I = IOC I and IBO I = IOD I·

o.

By definition of a parallelogram, AB is parallel to DC so alternate angles L OAB and L DCO have the same measure. Since BC is parallel to AD we have m(LABO)=m(LCDO). By Theorem 7-1, ICDI = IBAI. By ASA AAOB= ACOD. Finally, IAOI = lOCI and IBOI = 1001 because they are corresponding sides in these congruent triangles. 2.

In quadrilateral MATH we are given that

IMOI = lOTI and IHO I = lOA I where 0 is the intersection point of the diagonals. (See fig. 7.30) Prove: MATH is a parallelogram. In the triangles iliOM and AAOT we are given IMO I = lOT I and

IHO I = lOA I. Since L HOM and L AOT are opposite angles their measures are equal. By SAS it follows that iliOM = AAOT. By corresponding parts we have m(L OHM)=m(L OAT). Since these are alternate angles for HM and AT, HM is parallel to AT. Similarly, m(LHMO)=(LATO) so HT is parallel to MA. By definition, MATH is a parallelogram. 3.

Let ABCD be a quadrilateral with IAD I = IBC I and. IAB I = IDC I· Prove: ABCD is a parallelogram Draw the diagonal BD. In triangles L ABD and L CDB we are given IADI = IBCI and IABI = lOCI. Since lOBI = lOBI we have AABD=ACDB by SSS. By corresponding parts m(LDBA)=m(L COB). These are alternate angles so CD is parallel to AB. Also m(LADB)=m(LDBC) so AD is parallel to BC. Thus ABCD is a parallelogram by definition.

4.

In quadrilateral ABCD assume that AB is parallel to DC and

IAB I = IDC I. Prove: ABCD is a parallelogram.

CHAPTER 7. CONGRUENT TRIANGLES

59

Draw diagonal BD. In triangles 4ABD and ACDB we are given that IAB I = IDC I and AB is parallel to DC. By the alternate angle theorem we have m( L CDB)=m( L ABD). Since IBD I = IBD 1 by SAS we get 4ABD = ACDB. By corresponding parts we have 1AD 1= 1CB I. By Theorem 7-4 we conclude that ABCD is a parallelogram. 5.

Let MATH be a rhombus and let its diagonals be drawn meeting at point o. Prove: The diagonals are perpendicular to each other and they bisect the angles of the rhombus. (See fig.7.30) Since a rhombus is a parallelogram, by problem 1 we have that the diagonals of a parallelogram bisect each other. We consider the triangles tJIOM and AAOM. Since diagonal MT bisects AH we have IHO I = 1AO I. Because MATH is a rhombus 1MA I = IMH I. Finally IMOI = IMOI. By SSS LHOM=AAOM. By corresponding parts, we havem(LHOM)=m(LAOM). BecauseAOHisastraightangle, LHOM and LAMO are right angles. Thus MT is perpendicular to AH. We also havem(LHMO)=m(LAMO). ThusMTbisects LM. Since each of the other triangles formed by a side and half of each diagonal is congruent by the same argument to the first two triangles, we conclude that each angle is bisected by the diagonal to its vertex.

6.

Let ABCD be a parallelogram with a diagonal BD that bisects L B and L D. Prove: ABCD is a rhombus. Since BD bisects L B and L D we have m( L ABD =m( L CBD) and m( L ADB) =m( L CDB). With 1BD I = 1BD I, by ASA we obtain 4ABD=ABCD. The corresponding parts have the same measure so IAB I = IBC I and IAD I = ICD I. Opposite sides of a parallelogram have the same measure so 1AB I = 1CD I. Thus all four sides have the same measure and ABCD is a rhombus by definition.

7.

Let point 0 lie on the angle bisector of LPQR. (see figure 7.31) Prove: 0 is equidistant from the sides of the angle. Draw OX 1. ~p and OY 1. ~R' Thus m( L X)=m( L Y). The segment QO is common to the triangles AQOX and AQOY. Because QO bisects LPQR we have m(LXQO)=m(LYQO). By ASA AQOX= AQOY. lOX 1 = lOY 1 since they are corresponding parts of these triangles. Therefore, 0 is equidistant from the sides of the angle.

8.

Let 0 be a point inside the L PQR and equidistant from the sides of the angle. Prove: 0 lies on the bisector of L PQR.

60

GEOMETRY: SOLUTION MANUAL

From 0 draw perpendiculars to the sides of the angle meeting ~p at X and ~R at Y. Since 0 is equidistant from the sides I OX I = I OY I. Also L X and L Yare right angles so m( L X) = m( L Y). In the right triangles L QOYand L QOX, by the Pythagoras Theorem I QY 12 + I OY 12 = IQO 12 and IQXI 2 +IOXI 2 =IQOI2. By substitution, 2 IQYI +IOYI 2=IQXI 2+IOXI 2. Since IOXI=IOYI we get lOX 12= lOY 12. Subtracting equals we obtain IQY 12= I QX 12. Thus I QY I = I QX I· We also have IQO I = I QO I. By SSS we get AQOY "" AQOX. Since L YQO and L XQO are corresponding parts of these congruent triangles we have m( L XQO) =m( L YQO). Therefore, ® bisects L PQR. 9.

(a)

Let APQA be a triangle with three lines, L, M, and N, that bisect the three angles of the triangle. Let 0 be the point of intersection ofL andM. Prove: Point 0 lies on line N. Let L bisect L P, M bisect L Q and N bisect L A. Since 0 lies on L, Problem 7 assures that it is equidistant from PQ and P A. Since it lies on M, it is equidistant from QP and QA. Thus 0 is equidistant from Q;\ and PA and, by Problem 8, lies on the bisector of L A which is line N.

(b)

Let C be an inscribed circle to the triangle APQA, and let M be its center. Prove: M is the point of intersection of the angle bisectors of APQA. Let X, Y and Z be the points of tangency on the sides of APQA. The radii to X, Y, and Z are perpendicular to the sides. Therefore M is equidistant from the three sides and, by Problem 8, lies on all three angle bisectors and is their point of intersection.

10.

Let AXYZ be equilateral with A, B, and C as the midpoints of its sides. Prove: AABC is equilateral.

4 \

X

,

I

L

I

l.

Since AXYZ is equilateral, IXY I = IYZ I = IXZ I. Because A, B, and C are midpoints of the sides IXAI = IAYI = IYBI = IBZI = I ZCI = ICXI· By the Corollary to Theorem 5-3 (page 141), m(L X)=m( L Y)=m(L Z). Therefore, by SAS AXCA"" AYBA "" AZBC. I AC I = I CB I = IBA I by the

CHAPTER 7. CONGRUENT TRIANGLES

61

corresponding parts theorem for congruence. Thus AABC is equilateral by definition. 11.

Let AABC be an equilateral triangle with line L bisecting LB. Prove: Line L is the perpendicular bisector of AC. (3

~6~ IJ

Name the intersection of Land AC point D. Since AABC is equilateral, IAB I = IBC I. L bisects L B so m( L ABD)=m( L CBD). Using the fact that IBD I = IBD I we get AABD = ACBD by SAS. As corresponding parts, I AD I = IDC I so L bisects AC. Again, as corresponding parts, m( L BDA) = m( L BDC). They form a linear pair so m( L BDA)=m( L BDC)=90°. Thus L is the perpendicular bisector of AC. 12.

In figure 7.33 let POR and QOS be straight line segments and let m( L 1)=m( L 2) and m( L 3)=m( L 4). Prove: 0 is the midpoint of QS. We are given that m(Ll)=m(L2) and m(L3)=m(L4). Since IPR I = IPR I we have APQR = APSR by ASA. By corresponding parts, IPQI=IPSI. Since m(Ll)=m(L2), and IPOI=IPOI we get APQO=ASPO. Therefore, IQOI = ISOl by corresponding parts. We can conclude that 0 is the midpoint of QS.

13.

In figure 7.34, ECBF is a parallelogram and m(LA)=m(LD). Also, EF A and DCB are straight line segments. Prove: AEDC=AFAB. Since ECBF is a parallelogram, m( L ECB)=m( L EFB) because opposite angles have equal measure. Because they form linear pairs of angles, m( L DCE) + m( L ECB) = 180° and m( L BFA) + m( L EFB) = 180°. Therefore, m( L DCE) +m( L ECB) = m( L BFA) +m( L EFB). Subtracting eguals from each side gives us m( L DCE) = m( L BFA). Opposite sides of a parallelogram are equal so IEC I = IBF I. Given that m( L A)=m( L D), by ASA we have AEDC = ABF A.

14.

In the construction of an angle equal to a given angle (p. 199), we have 10SI=IPQI, 10TI=IPRI and ISTI=IQRI. BySSS, AOST=APQR. Thus m(LO)=m(LP).

62

§

GEOMETRY: SOLUTION MANUAL

3. Exercises

(pages 206-209) 1.

In figure 7.47, weare given IACI =4, ICBI =4, IABI =4./2, IYZI =4,

m(LZ)=90" andm(LY}=45°. Prove:

~C==AXYZ.

In AXYZ we have a 45° right triangle with IYZ I = 4 so IZX I =4 and

IXYI=4./2. Thus IBCI=IYZI, IACI=IXZI and IABI=IXYI. Therefore, ~C == AYXZ by SSS. 2.

In ATOR draw ON perpendicular to RT. ATON is a 30-60-90 triangle.

lOTI =8 so IONI =4 and INTI =4/3 and ITRI =sJ3. The area of AROT is (SV3)(4)/2= 16J3.

3.

In the diagram above note that L A=45° and the perpendiculars from B

and C to AD are the height of the trapezoid, h. h= 61.J'i= (6J2)("[2/.J2) = 3/2. The long side of the trapezoid has length 3.j2+6 + 3.j2=6 +6J'1. A = (112)(~(6 + (6 + 6J'l> = (112)(~(12 + 6./T) = (112)(36Jl"+ 36) =1sJ!+18. 4.

In this trapezoid the altitude is h=3J2/.J2=3. The long side's length is

3+6+3=12. Therefore the area is (112)(3)(6+12)=(112)(3)(18)=27. 5.

In figure 7.51,note the 30-60-90 triangle in which the longer leg is 3 m. The ratio of legs is ...(3:1 so the shorter leg is 3/...fj=..[3 m. Build the

walls...(3 m from the sides.

CHAPTER 7. CONGRUENT TRIANGLES

6.

63

~

~

J

I~

Y

In the 45-45-90 triangle AHOY IBOI = IBYI =61...['I=(6./!)(..[2/..[2)=3,fi. IJB1 2=122 - (3/2)2=144 - 18=126 so IJBI =..j126='J.[f4. Area of ABOY =(1I2)(3./T)(3o/'l)=9 Area of ABJY = (1I2)(3.[f4)(3../T) =(CJ./28)/2= (1s..{f)/2=9J7. Area of AJOY=Area of ABJY - Area of ABOY = CJ../7-9

7.

In figure 7.53 we are given that AABC is equilateral and L BAD is a right angle. Prove: IBC I = ICD I·

Since AABC is equilateral, by the Isosceles Triangle Theorem we get m(LB)=60°. Thus AABD is a 30-60-90 right triangle where IABI=(1I2)IBDI orIBDI=2IABI. By the SEG Property, IBDI = IBCI + ICDI. By substitution we get 21ABI = IBCI + ICDI. Since IAB I = IBC I in the equilateral triangle, we have 21 BC I = IBC I + ICD I. By subtraction we conclude IBC I = ICD. 8.

In figure 7.54 we are given ill parallel to QE and QU parallel to EI. By definition, QUIE is a parallelogram. Thus IQU I = IEI I and IQE I = IUI I· The area of QUIE is bh=6(10)=60 sq. cm. IQEI 2=1Q2+52=100+25=125. lUll = IQEI =..[125=s..[5 cm.

9.

In figure 7.55 we have XM parallel to SA, MA parallel to XS and MO perpendicular to OS.

By definition, XMAS is a parallelogram. Hence, IXMI = ISAI =7. the area of XMAS =bh=(7)(5)=35 sq. units

CHAPTER 8

Dilations and Similarities

§

1. Exercises

(Pages 214-216) 1.

2.

Dr(P)=P when r=1.

3.

Points for this image set have 3 as a sUbscript.

4.

Points for this image set have 4 as a subscript and the triangle is a dashed line.

64

65

CHAPTER 8. DILATIONS AND SIMILARITIES

5.

,

- f-.

1",1-- f-'I

'-b~'-

...-1- - --"" - '-\---- -ltQ .';:f 1 '-_ -- f-- -- "'f-"~

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~)~ -"- - f - -

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f-f-

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1--

} . r>' r- t3 oo! -~ ... -- . .. c-' ~

~

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,--

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- '\~

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,

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6.

-I--;-!---I-.l-'-I- - -

---

66

7.

GEOMETRY: SOLUTION MANUAL

Original area = 9, New area = 36

(a)

J 1\

-

l IJ

--

t~

i ,

-- f - -- I -1-- f---

..... -

,-I-.

--:- ..

-

f-- -.

-

'-

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(b)

-- f-- l -

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Original area = 8, New area = 32 i

'" ,~

-,

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- !-/--

~

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67

CHAPTER 8. DILATIONS AND SIMILARITIES

8.

(a)

Original area = 4, New area = 9 I ~ {~

• .- - 4- -r-;

. L

t

.'" )1-

(b)

Original area = 10, New area = 22.5 I

I I

~ r-

,I I

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-- -'- r-- _.

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r

.

-

GEOMETRY: SOLUTION MANUAL

68

§ 2. Exercises (Pages 227-231) 1.

Area of a 30 by 60 rectangle = 1800 mm2 • If the sides are tripled in length, r=3 and the area = (32)(1800)=16,200 mm2 •

2.

Area of original square = 10 cm2• Doubling the lengths of the sides means r=2 and the new area = (22)(10)=40 cm2 •

3.

When the sides of any square are doubled in length, the area is multiplied by -±...

4.

If the radius of a disk is doubled, its area is multiplied by

5.

Area of original table = 1.52 =2.25 m2• New area = 4.5=r(2.25) so r=2 and r=..[2. Each side should be (~(1.5) m.

6.

If you wish to double the area of a square, you must multiply the sides by...:£I....

7.

The smaller lot is one-fourth the size of the larger lot. A fair price for the smaller lot would be (114)(2000)=$500.

8.

Your count should be near to 2827.

9.

About 26 km2 •

10.

(a)

A = (60/360)(71")(42) = (813)71"

(b)

A = (90/360)(71")(62) = 971"

(c)

A = area of sector - area of triangle = (116)(71")(36) - (6 2/4)(..[3)=671" - 9V3.

(d)

A=(1/4)(7I")(25) - (112)(5)(5)=(25/4)(71") - 25/2

~.

CHAPTER 8. DILATIONS AND SIMILARITIES

11.

12.

(a)

(114)( 71")(1 ~ = 71"/4

(b)

71"/3

(c)

71"/12

(d)

(x/360)7I"

(a)

[(02-01)/360]rr

(b)

7ff} -

ll"Il

(c)

[(0 2

(1)/360][7I"(r22

-

69

= 7I"(rl - rl~ -

rl~

13.

In figure 8.32 draw PQ. Because of the symmetry in the figure, we can compute the area of S (one-half of the total area) as the area of the sector in the left circle minus the area of the triangle OPQ. Area = (114)(71")(36)-(112)(6)(6)=971"-18. Thus the total area = 1871"-36.

14.

The area of the shaded region is the area of the square minus the areas of SI,S2 and S3. Area Area Area Area

16.

of square = 64 of Sl = area of S2 = (114)71"(16)=471" of S3 = 42 = 16 of shaded region = 64 - (471"+471"+ 16)=48-871"

In figure 8.35, m(LBOA)=360/8=45°. IBOI = IOAI =r and IBXI = rl-/2 = (r.J2)/2. Area of .c.BOA = (112) I OA II BX I = (1I2)(r)(r.J2/2) = (rV2)/2 The 8 triangles of the octagon are congruent by SAS thus the area of the octagon = 8[(rV2)/4)=2rV2.

70

§

GEOMETRY: SOLUTION MANUAL

3. Exercises

(Pages 234-235) 1.

Area of the dilated square = (32)(25)=225. Perimeter of dilated square = (3)(20)=60

2.

Let T have sides 2,4,5. Perimeter ofT = 2+4+5=11. The dilation of the short side, 2, equals 7 so 2d=7 and d=7/2. Thus, the perimeter of the dilated triangle = (7/2)(11)=77/2=38 112.

3.

If r is the dilation factor, rP I=Pz. r= 10/16 (or 16/10). rAI=Az so AzlAI = (10/16)z= 100/256.

§

4. Exercises

(pages 240-243) 1.

(a)

For the inscribed square, the side of the triangle is approximately 7 cm and the height is about 3.5 cm. The area of the triangle is (112)(7)(3.5)= 12.25. The area of the square is (4)(12.25)=49 cmz. For the circle the area is 2511'"=78.5 cmz. The perimeter of the square is 28 cm and the circumference of the circle is 31.4 cm.

(b-e) Follow similar procedures. 2.

Perform the measurements.

3.

(a)

A=2411'"=rr so r=24 and r=.J24=2"f6. Thus C=(2)(1I'")(2.J6)=411'".J6 cm.

(b)

A=3611'"=rr so r=36 and r=6. Thus C= (2)(11'")(6) = 1211'" cm.

(a)

C=1511'"=2n so r=15/2. Thus A=1I'"(1512)z=(225/4)1I'" cmz.

(b)

C=911'"=2n so r=9/2. Thus A=1I'"(9/2)z=(8114)1I'".

4.

CHAPTER 8. DILATIONS AND SIMILARITIES

(c)

C=20'll"

SO

71

r=10. A=l00'll" cm2.

5.

The surface area of a cylinder is the same as a rectangle with length C=2'11"(3)=6'11" m and height = 5 m. A=5(6'11")=30'll" m2.

6.

Total area = half the circumference times the height. C=2'11"(4)=8'11". (1I2)C=4'11". Thus A=4'11"(1l)=44'11" m2.

7.

Area of roof = 10 times half the circumference = 10«112)(9'11")=45'11". Area of the sides = 4 rectangular areas + 2 semicircles. A=(2)(8)(9) +(2)(8)(10) +2(112)('11")(9/2)2= 144+ 160+(8114)'11" =304+(8114)'11"=304+(20 114)1\ m2. Thus the total area =304+(65 114)'11" m2.

8.

Area of the wasted space = area of square - area of four circles. A=(12)2 - (4)('II")(3?=24 - 36'11" cm2.

9.

If the radius is doubled, the circumference is doubled.

10.

Let S = length of the strip. S=(2)(12) +(2)(6) +(4)(114)(6'11")=24+ 12+6'11"=36+6'11" cm.

11.

Let S = speed of the bicycle = 15 kmIhr. Let N = number of revolutionslhr. Circumference of the wheel = 1.2'11" m=(.0001)(1.2'11") km= .0012'11" km. N=S/C=(15)/(0.0012'11")=(12500)/'II" revolutions/hr. This is approximately 3979 revolutionslhr.

13.

(a)

V = cross section area times the average length. V=('II"b~(2'11"a)=2a~2

(b)

SA = circumference of the torus times its length. SA=(2'11"b)(2'11"a)=4ab~.

14.

Let C'=C+1. Then 2n'=2n+1 so r'=r+1I(2'11"). The increase in the radius is 11(2'11") m. In other words, the increase in the radius is independent of the original r.

72

GEOMETRY: SOLUTION MANUAL

Additional Exercises on Circles (Pages 243-244) 1.

In the figure below, draw AC perpendicular to OB . .t.AOC is a 45-45-90

right trianglewithhypotenusex. lOCI = IACI = (x.,f2)12. Thus ICB I =x(x.,f2)12 by Pythagoras. lAB 12 = [(x..j2)/2]2+ [x-(x..j2)12]2 = [x2(2/4)]+[x2(1...J2+(1I2)] =x2[(1I2)+ 1 +(112)...[2) =x2(2 -..j2) Therefore, IAB I =x.,f=2--

:n--.

Ft

2.

To construct a segment of length ./5, form a right triangle with legs of lengths 1 and 2. The hypotenuse is of length -J5. To construct the length (1I2)(.,f5-1), mark a segment of length -J5 on a line and subtract the length 1. Then bisect the segment of length (.J5-1). One half is the required length.

3.

In the figure below draw OB, CO', 00' and AO' parallel to BC.

IAOI=25-9=16. 100'1=25+9=34. Since ABCO' is a rectangle IBCI = IAO'I· IBC 12= 342-16 2 == 1156-256=900 so IBCI =30.

CHAPTER 8. DILATIONS AND SIMILARITIES

4.

73

(a)

We are given that IOMI =10. t.AOM is a 30-60-90 right triangle. The ratio of the legs is 1I-JJ so V.[3=x/10 so x=10/..j3. Thus IABI =20/..j3. Therefore the perimeter, P = 120/..j!. The area of the hexagon = 6(1I2)(201,J3)(10)=600j3. (b)

.6.ABC == t.ABO by ASA since m(LCAB)=m(LCBA)=m(LBAO)=m(LABO)=60° and AB is a common side. IAOI =21AMI =20/.j'f so IAB I = IAO I =20/.[3. The perimeter of the star is equal to P = (12)(20j..J3) = 240/..j3. The area twice the area of the hexagon = 1200J3.

5.

C= 167r=2u so r= 8. One side of the hexagon = r. The perimeter = 6r=48.

6.

C=807r=2u so r=40. In the 45-45-90 triangle, s=4- f;'_ I

(c)

A-B=(O,-l)

~I

_ r\

r

~\

\

I,

I

1'1 -c

87

CHAPTER 10. VECTORS AND·DOT PRODUCT

(d)

,

A-B = (0,4)

.

1\ I

-

12

1/

,

I ~

B V

-

(e)

A-B=(-3,-1)

(t)

A-B = (-2,-1)

(c)

c(A-B)=c(al-bl,~-b0 =(c(al-bl),C(~-b0)

=(cal-cbl,~-cb0

=cA-cB by part (a).

9.

Let A={1,2) and B={3,1) A+B=(4,3) A+2B=(7,4) A+3B=(10,5) A-2B={-5,0) A-3B=(-8,-1) A+4B=(13,6) A-4B = (-11,-2)

i I

- fl +1

i

t

p

-t _ ~- ~G

~.2Jf I

A ~~ .~

~.

88

10.

11.

GEOMETRY: SOLUTION MANUAL

Let A=(2,-1) and B=(-l,l) A+B=(l,O) A+2B=(0,1) A+3B=(-1,2) A+4B=(-2,3) A-B=(3,-2) A-2B=(4,-3) A-3B = (5,-4) A-4B=(6,-5) A + (1I2)B =(1.5,-0.5) A-(1I2)B=(2.5,-1.5) (a)

A=(3,1,-2) and B=(-1,4,5) A + B=(3 +(-1),1 +4,-2+5)=(2,5,3)

(b)

A=(al'~'8.)

-r-

7

--i",

"

-~,

I

~Id -.«.'

,

_L

I

I

I

2. Exercise

2.

A' B =(1)(-1) + (3)(5)= 14

(b)

A . B=(-4)(-3) +(-2)(6)=0

(c)

A . B=(3)(2)+(7)(-5)=-29

(d)

A' B=(-5)(-4)+(2)(3)=26

(e)

A' B =(-6)(-5) +(3)(-4)= 18

...

+-

)"'..

:(;.~ I1\ ,.

~ ')1

and B=(bl>b2,b3)

(a)

- - :-- I-

t

D.

(Page 304) 1.

i

I

A+B=(al+bl,~+b2,8.)+b3)

§

1-"" -t- l-

Let A=(al'~'8.)

and B=(bl,b2,b3) A . B=a1b 1+~b2+~b3 SP 1: It is commutative. A . B=a1b1+~b2+~b3 = b1a1+b2~+b3~ =B . A SP 2: If A, B and C are points, then distributivity holds. Let C=(Cl>C2'~). A' (B+C)=al(bl+cl)+~(b2+CJ+~(b3+C3) =a1b1+a1c1+~b2+~C2+~b3+a3c3 = (a1b1+~b2+~b3)+(alcl +~~+~C3) =A'B + A' C

I

I\''f~ I

I

"

I

I

CHAPTER 10. VECTORS AND DOT PRODUCT

SP 3: If x is a number, then (xA) , B=x(A ' B)=A ' (xB). and xB=(xb 1,xbz,xb3) (xA) 'B=(xa1)b1+(X8z)bz+(x~)b3 =al(xbl)+8z(xb0+~(xb3)=A ' (xB) =x(albl)+x(8zb0+x(~b3)=x(A . B).

xA=(xal'x8z,x~)

3.

(a)

A' B=(3)(-2) +(1)(3)+(-1)(4)=-7

(b)

A' B=(-4)(2)+(1)(1)+(1)(1)=-6

(c)

A' B=(3)(-2)+(-1)(4)+(5)(2)=O

(d)

A' B=(4)(3) +(1)(2) +(-2)(7)=0

(A+B)2 =(A+B) , (A+B) =A' (A+B)+B ' (A+B) by distributivity =A'A+A'B+B'A+B'B by distributivity =A2+A' B+A' B+B2 by commutativity =N+2A' B+B2 (A+B)"(A-B)=A' (A-B)+B ' (A-B) by distributivity =A' A-A' B+B ' A-B 'B by distributivity by commutativity =A2_A 'B+A' B-B2 =A2 _ B2.

4.

(a)

IA I = [(5)(5) + (3)(3)]1f2=y'34

(b)

IAI=.J34

(c)

IAI=v'34

(d)

IAI=v'34

(e)

IAI

(f)

IA I =..fi6 = 2.,[5

(g)

IAI =V20=2..f5

(h)

IAI =.JW=2$

=.JW=2.j5

89

90

5.

GEOMETRY: SOLUTION MANUAL

(a)

(b) -.,-~-

~ '}J

\

I-- --

"\

-

_ '1 ,)

---

_\_~ - II.

IV>

--- ---

-~ J

(c)

(d) -

---

f--+- .-- t: -- J-

I

n

\0. Therefore x=O. If one dots both sides by B, it follows that y=O. If one dots both sides by C, it follows that z=O. Therefore, x=y=z=O.

7.

Let A=(al'~'~)' El=(l,O,O), E2=(0,1,0) and E3=(0,0,1). AEI =al(l)+~(O)+~(O)=al. AE 2 =a1(0) +~(1) +~(O)=~. AE3=al(O) +~(O) +a3(1)=a3.

8.

Let A=(al>az,~,a4) and B=(bl,b2,b3,b4) be points in R4. A + B=(a1+bl,~+b2,a3+b3,a4+b4) Let x be a number, xA=(xal>x~,X~,xa4) AB=a1b 1+~b2+a3b3+a4b4 For SP1 it is clear that each pair commutes as real numbers so SP 1 holds for the dot product in R4 (and in Rn). SP2 follows since the real number distributive property holds in the products in the dot product definition for all Rn. SP3 follows since the real number products are associative and commutative. This holds in all Rn. SP4 follows from the observation that the squares of non-zero real numbers are greater than O. Thus, if a non-zero A is in Rn, AA>O. If A=O, all of the squares are zero so AA=O. Therefore, SP1-SP4 hold in all Rn.

CHAPTER 10. VECTORS AND DOT PRODUCT

§

97

7. Exercises

(Page 320) 1.

2.

(a)

XN=PN where P=(-4,S,I) and N=(2,3,1) (x,y,z)"(2,3, 1) = (-4,S, 1)"(2,3, 1) =-8 + IS + 1=8 2x+3y+z=8

(b)

(x,y ,z)"(-2,1,4) =(1,-3,-7)"(-2,1,4) -2x+y+4z=-33

(c)

(x,y,z)"(I,-I,I)=(-2,1,3)"(1,-1,1) x-y+z=O

(d)

(x,y ,z)"(2,1,3)=(3,1,-1)"(2, 1,3) 2x+y+3z=4

(a)

Let PI: 3x-2y+Sz=0 and P 2: 4x-y+7z=1 Let NI=(3,-2,S) and N 2=(4,-1,7) N IN 2= (3)(4) +(-2)(-1) + (S)(7) =49. Since NI"N2>0, NI is not perpendicular to N2 and PI is not perpendicular to P 2'

(b)

N IN 2=(3,-2,S)"(7,-3,2)=37. Thus PI is not perpendicular to P 2.

(d) (e)

N IN 2= (3,4, 11)"(2,-7,2) =0. Thus P I .lP2. N I"N2= (7,-1,8)"(8,2,-1) =46. Thus PI is not perpendicular to P 2.

CHAPTER 11

Transformations

§

1. Exercises

(page 323)

1.

2.

98

(a)

IXZI 2 =102 +242 =676 so IXZI =26

(b)

The sides of aXYZ are twice the lengths of the corresponding sides of aABC.

(c)

P(aABC) =5 + 12+ 13 =30 cmandP(aXYZ) = 10+24+26=60cm. The perimeter of aXYZ is twice the perimeter of aABC.

(d)

A(aABC) = (1/2)(5)(12) = 30 c m 2 and A(aXYZ)=(1I2)(10)(24)=120 cm2 • The area of aXYZ is four times the area of aABC.

(e)

The corresponding angles of the two triangles have the same measure.

(a)

TOT, HAH and TOOT. (There are other correct choices.)

(b)

These words work because each letter has a vertical axis of symmetry and the word reads the same backwards and forwards.

99

CHAPTER 11., TRANSFORMATIONS

Experiment 11-2 (pages 326,327) 1. I

A: ---8--- C I

--

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D _

.... - -

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F C -l-H~ --- I'

I

I

---E~ -I

i

I

,

-

-

---

1

1

None

None

JKL

I

I

N

PO RST lWYW},«- Y None

Infinite No.

I

z :

None

None

None

None

None

1: I

None

None

I

I

I

None

I

100

GEOMETRY: SOLUTION MANUAL

2. You cannot read TOO in a mirror. You cannot read EYE in the mirror. The word MOM has letters with a vertical line of symmetry and the word has a vertical line of symmetry. The word TOO has letters with a vertical line of symmetry; but, the word does not have a vertical line of symmetry. The word EYE has one letter with a vertical line of symmetry; but, the letter E only has a horizontal line of symmetry.

""· · N

3.

\

v'"

.......

......

...... \

\

4.

,-1

, I

I

)I

I I

,...

-

)

I

I

",

\

I

I

E

~

I

5. A figure has line symmetry if each point can be paired with another point such that the segment joining these points is perpendicular to the line of symmetry and the midpoint of the segment lies on the line of symmetry.

101

CHAPTER 11. TRANSFORMATIONS

§ 3. Exercises (Pages 330, 331) 1.

TOOT TOT TOOT, WOW MOM WOW, WOW TOT WOW, ...

2.

Write up your fmdings.

3.

Read this chapter in One. Two. Three ... Infinity.

4.

(a)

S I i n~.5 .3 ~

/

01:

rY\. M.e -tr--~

(b)

/ ,

0 I lnes .......

I

(c)

\

0+

J

_3 ~ rn n1 e tr

102

GEOMETRY: SOLUTION MANUAL

§ 4. Exercises (Pages 334-337) 1.

(a)

" QI f' •

(b)

pi

, R\ R.

(Q.

IfRL(P)=P', then RdP')=P.

2.

RL(P)=P holds for all points on line L.

3.

The reflection through a line L, RL(P), maps point P to the point P' such that L is the perpendicular bisector of PP'. If P is on line L, RL(P) = P.

4.

Given RL(P)=P' and Q is on line L, prove d(p ,Q)=d(P' ,Q). Since RL(P)=P', by the definition in exercise 3 we know L is the perpendicular bisector of PP'. Therefore, by the Perpendicular Bisector Theorem (Theorem 5-1) we conclude that d(p,Q)= d(p' ,Q).

5.

6.

Using the x-axis as the line of reflection: (a)

P=(3,7) and P'=(3,-7)

(b)

P=(-2,5) and P'=(-2,-5)

(c)

P=(5,-7) and P'=(5,7)

(d)

P=(-5,-7) and P'=(-5,7)

CHAPTER 11. TRANSFORMATIONS

7.

103

Using the y-axis as the line of reflection: (a)

P=(3,7 ) and P'=(-3 ,7)

(b)

P=(-2, 5) and P'=(2, 5)

(c)

P=(5,- 7) and P'=(-5, -7)

(d)

P=(-5, -7) and P'=(5,- 7)

(e)

P=(Pl,Pz) and P'=(-Pl 'Pz)

8.

When the line x=3 is reflected in the x-axis it is its own image. In other words, if p is on the line x=3, Rx(P)=P' where P' is also on the line x=3.

9.

The image of the line x = 3 when reflected in the y = axis is the line x = -3.

10.

In figure 11.25, let L be the line corresponding to the river bank. Let RL(A)=A' and let X be the intersection of A'B with L. From the SEG postulate we obtain: IA'XI + IXBI = IA'BI· From exercise 4 we have IA'XI = IAXI. By substitution we have IAX I + IXB I = IA'B I· Let Y be any other point on L. Since A', B and Yare non-collinear, the Triangle Inequality Postulate implies 1A'Y I + 1YB 1> 1AB I. By exercise 4 we obtain IAYI = IA'YI. By substitution, IAYI + IYBI > IA'BI. Therefore, any location different from X will cause the driver to drive further than the path A to X to B.

11.

In figure 11.26 RL is the reflection in line L. Also, RdP)= P' and RdQ)= Q'. Prove: d(P,Q) =d(P' ,Q'). Draw the line through P perpendicular to QQ' at R and another line through P' perpendicular to QQ'at R'. Thus, PR is parallel to P'R'. Since Pi" and 00' are perpendicular to line L, they are parallel. A parallelogram with one right angle is a rectangle. Thus PRR'P ' is a rectangle and 1PRI = IP'R' I. Since RdP)= P', L is the perpendiCUlar bisector of PF' and is also the perpendicular bisector of RR'. In addition , L is the perpendicular bisector of QQ'. Let Y be the intersection of L with the line QQ'. From each of the above we obtain, IRY 1 == 1R'Y I and 1QY I = IQ'Y I· By the SEG Postulate IQYI = IQRI + IRYI and 1Q'Y I = 1Q'R' I + 1R'Y I· By substitution,

104

GEOMETRY: SOLUTION MANUAL

IQRI+IRYI=IQ'R'I+IR'YI. Since IRYI=IR'YI we can use the subtraction property of equality and obtain 1QRI = 1Q'R' I. In the right triangles APQR and AP'Q'R' we have 1PR 1= 1P'R' 1 and 1QR 1= 1Q'R' I. By the Right Triangle Postulate 1PQ 1=P'Q' I· 12.

Let P=(PhP2,P3) be a point in three-space. (a)

R,.y(P)= (PhPl,-P3)

(b)

Rxz(P) = (Pl,-P2,P3)

(c)

~(P)=(-PhP2,P3)

§ 5. Exercises (pages 339-340)

1.

2.

(a)

B

p:z}, then P'=(-2+Pl,5+p:z}. 2.

Let A=(-3,2). Let S be a triangle with vertices (2,5), (-3,7), and (3,6). TA(2,5)=(2,5) +(-3,2)=(-1,7) T A(-3,7)=(-6,9) and TA(3,6)=(O,8). Thus, TA(S) is a triangle with vertices (-1,7), (-6,9), and (0,8).

3.

Let A=(-3,2) and L be the line x=4. TA(L) is the line x=1.

4.

Let A=(-3,2) and K be the line y=4. TA(K) is the line y=6.

5.

Let A=(-3,2) and C be a circle centered at the origin with radius 3. TA(C) is a circle centered at (-3,2) with radius 3.

6.

Let A=(O,O), B=(2,3), C=(4,6), D=(-2,-3), E=(3,2), and F=(5,5). (a)

TAB=TBA is false since B-A=(2,3) and A-B=(-2,-3).

(b)

TAB=TBC is true since B-A=(2,3) and C-B=(2,3).

(c)

TBC=TAD is false since C-B=(2,3) and D-A=(-2,-3).

(d)

TBC=TDA is true since C-B=(2,3) and A-D=(2,3).

(e)

TAB=TEC is false since B-A=(2,3) and C-E=(1,4).

(f)

TBC=TEF is true since C-B=(2,3) and F-E=(2,3).

112

GEOMETRY: SOLUTION MANUAL

(g)

TEC=TEF is false since C-E=(1,4) and F-E=(2,3).

(h)

TDB=TAC is true since B-D=(4,6) and C-A=(4,6).

Additional Exercises for Chapter 11 (Pages 353-355) 1.

(a)

For the identity mapping, all points of the plane are fixed points.

(b)

Ro has the point 0

(c)

RL has the line L as the set of fixed points.

(d)

Gx.o(P) has one fixed point, point O.

(e)

A translation has no fixed points unless it is of 0 length in which case the entire plane of points is fixed.

(t)

The constant mapping whose value is a given point X has X as its only fixed point.

as its only fixed point.

2.

I

I

3 KJV\.

---------+~

8 Km

Note right triangle AA'BC where IA'B1 2 =82 +8 2 =128 so IA'BI ='W2. Observing that IAX I = IA'X I and IA'B I = IA'X I + IXB I we obtain by substitution that IAX I + IXB I = 'W2 km. 3.

Use your creativity to make up three mappings.

4.

(a)

5.

T AB (6,3)=(9,3)=B'

Rx.axis(3,3)=(-3,3)=A'

CHAPTER 11. TRANSFORMATIONS

6.

G90,B(9,3)=(6,6)=B"

113

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1-1: I .1 I

I

+-

.:..

r-1-: I vI'

~ /1',

'~It'I ~{/.m

I

.

-I"'i 'lP

1---

. "-

n( It) v'

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+-- 1---

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l It~

.

.J,

.-

_.f- / - - --I-

1

-/-f-

,

7.

dCA' ,B")2=d[(3,-3),(6,6)]2= (6-3)2+ (6-(-3»2=9 + 81 =90 d(A',B")=3V10.

8.

Area of .tAB"B' = (112)(6)(3)=9 sq. units.

9.

m(LAB"B')=90°.

10.

In figure 11.35, reflect A through line Ll and reflect B through line~. Let X be the point where A'B intersects Ll and Y be the point where AB' intersects~. The path would be A to X to Y to B.

CHAPTER 12

Isometries

§

1. Exercises

(Pages 360-363)

1.

I

,

,

,

\ 2.

114

CHAPTER 12. ISOMETRIES

115

3.

4.

(a)

Note that Rp(C)=C but each point maps to the point on the other side. (b)

116

GEOMETRY: SOLUTION MANUAL

(c)

(d)

o

~t

r--\2J-I

fj;\, ~L ~ Go~7D,O~c)~~

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