Answers to Exercises
CHAPTER 11 • CHAPTER
11 CHAPTER 11 • CHAPTER
Answers to Exercises
USING YOUR ALGEBRA SKILLS 11
3 3
1. 8; 5
AC 3 CD 5 BD
8
, , 2. CD 5 BD 8 BC 13
3
3a. 1
9
3b. 1
4. a 6
5. b 16
6. c 39
7. x 5.6
8. y 8
9. x 12
10. z 6
11. d 1
12. y 5
13. 318 mi
14. 2.01
15. 12 ft by 15 ft
124
ANSWERS TO EXERCISES
16. almost 80 years old
17a. true
3 1
17b. false; 6 2, but 3 1 6 2
17c. true
17d. true
3 1
3 1
17e. false; 6 2, but 2 6
17f. true
18a. arithmetic: 10, 25, 40, 65; geometric: 10, 20,
40, 80 or 10, 20, 40, 80
18b. arithmetic: 2, 26, 50, 74; geometric: 2, 10, 50,
250 or 2, 10, 50, 250
18c. arithmetic: 4, 20, 36, 52; geometric: 4, 12, 36,
108 or 4, 12, 36, 108
19a. Add the numbers and divide by 2.
19b. Possible answer: Multiply the numbers and
take the positive square root of the result.
19c. c ab
; This formula holds for all positive
values of a and b.
19d. c 2 • 50 100
10;
c 4 • 36 144
12
3 9
15. 1; 1
LESSON 11.1
y
1. A
2. B
3. possible answer:
Y' (6, 15)
Y (2, 5)
R' (6, 6)
R (2, 2)
O' (18, 6)
O (6, 2)
x
16. Yes, they are similar.
3. possible answer:
y
6
(0, 6)
(5, 6)
4
(7, 3)
2
(1, 1)
5
6. Figure A is similar to Figure C. Possible answer:
If
A
B
and
B
C
, then
A
.
C
7. AL 6; RA 10; RG 4; KN 6
8. No; the corresponding angles are congruent,
but the corresponding sides are not proportional.
9. NY 21; YC 42; CM 27; MB 30
10. Yes; the corresponding angles are congruent,
and the corresponding sides are proportional.
11. x 6 cm, y 3.5 cm
2
12. z 103 cm
13. Yes, the corresponding angles are congruent.
Yes, the corresponding sides are proportional.Yes,
AED ABC.
9
9
14. m 2 cm 4.5 cm, n 4 cm 2.25 cm
17. Possible answer: Assuming an arm is about
three times as long as a face, each arm would be
about 260 ft.
18. Possible answer: Not all isosceles triangles are
similar because two isosceles triangles can have
different angle measures. A counterexample is
shown at right. Not all right triangles are similar
because they can have different side ratios, as in a
triangle with side lengths 3, 4, and 5 and a triangle
with side lengths 5, 12, and 13. All isosceles right
triangles are similar because they have angle
measures 45°, 45°, and 90°, and the side lengths have
the ratio 1:1: 2.
d
21. c
1825
22. Possible answers: Jade might get 4475 of the
2650
profits, or $2,773.18, and Omar might get 4475 of
the profits, or $4,026.82. Or they take out their
investments and they divide the remaining $2,325:
Jade, $1825 $1,162.50 $2,987.50; and Omar,
$3,812.50.
23a.
23b.
19. 36
20. bc
60°
60°
24. approximately 92 gallons
ANSWERS TO EXERCISES
125
Answers to Exercises
5. possible answer:
x
LESSON 11.2
Answers to Exercises
1.
2.
3.
4.
6 cm
40 cm; 40 cm
28 cm
54 cm; 42 cm
37 35
5. No, 3
0 28 .
6. Yes, MOY NOT by SAS.
7. Yes, PHY YHT because YH 12 and
20
16
12
(SSS).Yes, PTY is a right triangle
15
12
9
because 202 152 25 2.
8. TMR THM MHR by AA.
x 15.1 cm, y 52.9 cm, h 28.2 cm
9. Yes,QTA TUR and QAT ARU.
QTA QUR by AA; 632 cm
10. 24 cm; 40 cm
11. Yes, THU GDU and HTU DGU;
52 cm; 42 cm
12. SUN TAN by AA; 13 cm; 20 cm
13. Yes, RGO FRG and GOF RFO.
GOS RFS by AA; 28 cm; 120 cm
14. 20 cm; 21 cm
15. x 50, y 9
126
ANSWERS TO EXERCISES
2 1
16. r R R
2 12
2 1
17. She should order approximately 919 lb every
three months. Explanations will vary.
18. 448
19. The corresponding angles are congruent; the
ratio of the lengths of corresponding sides is 13; the
dilated image is similar to the original.
20. Yes, ABCD ABCD. The ratio of the
perimeters is 12. The ratio of the areas is 14.
y
D' C'
A'
B'
x
21. 118 square units
22. The statue was about 40 ft, or 12 m, tall. To
estimate, you need to approximate the height of a
person (or some part of a person) in the picture,
measure a part of the statue in the picture, calculate
the approximate height of that statue piece, and
assume that the statue has the same proportions
as the average person.
LESSON 11.3
1. 16 m
2. 4 ft 3 in.
3. 30 ft
4. 10.92 m
5. 5.46 m
6. Thales used similar right triangles. The height of
the pyramid and 240 m are the lengths of the legs of
one triangle; 6.2 m and 10 m are the lengths of the
corresponding legs of the other triangle; 148.8 m.
7. 90 m; R and O are both right angles and
P is the same angle in both triangles, so
PRE POC by AA.
8. 300 cm
6
13. GHF FHK GFK by AA; h 181
3,
9
4
x 71
3 , y 44 13 .
14. sample answer:
1
A 2(8.2)(1.7) 6.97 cm2
1
A 2(3)(4.6) 6.9 cm2
8.2 cm
3 cm
1.7 cm
4.6 cm
45 cm
3 cm
30 cm
2 cm
d
A
1
Answers to Exercises
2
15. 53
16.
B
2
20 cm
4
9.
3
D
t
h
y
x
Possible answer: Walk to the point where the guy
wire touches your head. Measure your height, h;
the distance from you to the end of the guy wire, x;
and the distance from the point on the ground
directly below the top of the tower to the end of the
guy wire, y. Solve a proportion to find the height of
the tower, t: ht xy. Finally, use the Pythagorean
Theorem to find the length of the guy
wire: t 2 y 2 .
10. The triangles are similar by AA (because the
ruler is parallel to the wall), so Kristin can use the
length of string to the ruler, the length of string to
the wall, and the length of the ruler to calculate the
height of the wall; 144 in., or 12 ft.
2
11. MUN MSA by AA; x 313.
12. BDC AEC by AA; y 63.
C
Given: Parallelogram ABCD
CD
and AD
BC
Show: AB
ABCD is a parallelogram
Given
BC
AD
CD
AB
Definition of
parallelogram
Definition of
parallelogram
BD
BD
2 4
1 3
Same segment
AIA Conjecture
AIA Conjecture
ABD CDB
ASA
BC
AD
CD
AB
CPCTC
CPCTC
17a. 4.6.12
17b. 3.12.12 or 3.12 2
ANSWERS TO EXERCISES
127
1 + 5
18. The golden ratio is , or approximately
2
1.618. Here is one possible construction:
C
D
A
X
B
mB 90°
1
Construct BC 2AB
5
AC 2 AB
1
5
AB
AX AD 2 AB 2
5
1
AB
AX 2
Answers to Exercises
AB
1 + 5
2
AX 5 1 2
Therefore, X is the golden cut.
128
ANSWERS TO EXERCISES
19a. Answers will vary.
19b. Possible answer: The shape is an irregular
curve.
19c. Answers will vary. Possible answers: As the
circular track becomes smaller, the curve becomes
more circular; as the track becomes larger, the
curve becomes more pointed near the fixed point.
As the rod becomes shorter, the curve becomes
more pointed near the fixed point; as the rod
becomes longer, the curve becomes more like an
oval. As the fixed point moves closer to the traced
endpoint, the curve becomes more pointed near
the fixed point; as the fixed point moves closer to
the circular track, the curve begins to look like a
crescent moon.
a
c
17. b 1 d 1
a b c
d
b b d d
LESSON 11.4
1.
2.
3.
4.
5.
6.
7.
8.
18 cm
12 cm
21 cm
15 cm
2.0 cm
126 cm2; 504 cm2
16 cm
60 cm
4
9. 49 cm
p b
10. q; q
ab cd
b
d
18. The ratio will be the same as the ratio for the
original rectangle. The ratio is 21 if it can be divided
like this:
It might be any ratio if divided like this:
L
O
V
M
A
19. Yes, by the SSS or the SAS Similarity
Conjecture.
20a. 2a b
20b. 2a b
20c. all values of a and b
20d. no values of a and b
21. AB 3 cm, BC 7.5 cm
Answers to Exercises
11. 6
3 cm
12. 6 cm
5
13
1
2
13. x 33 cm, y 3 cm, z 83 cm
k 7
3
14. B (3, 5), R 14, 7 ; h 4
2
15. 1
H
16.
E
T
Consider similar triangles LVE and MTH with
and HA
. To
corresponding angle bisectors EO
show that the corresponding angle bisectors are
proportional to corresponding sides, for example
EO
EL
, show by AA that LOE MAH,
HA
HM
EL
and then you can show that HEOA HM . You
know that L M. Use algebra to show that
LEO MHA.
ANSWERS TO EXERCISES
129
LESSON 11.5
1. 18 cm2
4. 27 cm2
2. 18
1
5. 4
9
3. 5; 10
6. 1: 3
9
12
,
16d. x 9, y 16, 1
(9)(16) 2 16 h 144
12
17.
m2
m 2
7. or n
2
n
8. Possible answer: Assuming the ad is sold by
area, Annie should charge $6000.
9
9. 5000 tiles
10. 1
11.
7.1 m 14.2 m
14.2 m 6.5 m
10 m
42 m
42 m
18.
1
2
2
4
3
6
4
8
5
10
6
12
y
5
5
x
12b. A(x) 2x 2
y
x
Area
in cm2
70
1
2
60
2
8
50
3
18
4
32
5
50
6
72
40
30
20
10
5
x
12c. The equation for a(x) is linear, so the graph is
a line. The equation for A(x) is quadratic, so the
graph is a parabola.
13. Possible proof: The area of the first rectangle
is bh. The area of the dilated rectangle is rh rb, or
r 2bh. The ratio of the area of the dilated rectangle
r2bh
2
to the area of the original rectangle is bh , or r .
16 20
15. 8
14. 3, 3
16a. (i) 40°; 50°; 40°
(ii) 60°; 30°; 60°
(iii) 22°; 68°; 22°
Conjecture: similar; right triangle
s
h
h
b
16b. (i) h
(ii) x
(iii) m or a
p h
Conjecture: q
h
16c. h pq
130
ANSWERS TO EXERCISES
H
E
L
10
Area in cm2
x
Area
in cm2
Area in cm2
Answers to Exercises
12a. a(x) 2x
O
V
M
A
T
Consider similar triangles LVE and MTH with
and HA
. Show
corresponding altitudes EO
EO
EL
that LOE MAH, then HA HM . You know
and that L M. Because EO
HA are altitudes,
LOE and MAH are both right angles, and
LOE MAH.
EL
So, LOE MAH by AA. Thus HEOA HM ,
which shows that the corresponding altitudes
are proportional to corresponding sides.
19. x 92°; Explanations will vary but should
reference properties of linear pairs, isosceles
triangles, and alternate interior angles formed
by parallel lines.
20. 105.5 cm2
21. 60 ft2
22. true
23. Two pairs of angles are congruent, so the
triangles are similar by the AA Similarity
Conjecture. However, the two sets of corresponding
105
sides are not proportional 8600 135 , so the
triangles are not similar.
24.
Top
Front
Right side
Top: 6 square units; front: 4 square units; right side:
4 square units. The sum of the areas is half the
surface area. The volume of the original solid is
8 cubic units. The volume of the enlarged solid is
512 cubic units. The ratio of volumes is 614 .
LESSON 11.6
1. 1715 cm3
x
1
Surface area in cm2
2
3
4
5
22
88
198
352
550
Volume in cm 3
6
48
162
384
750
18. yes, because 182 242 302 (Converse of the
Pythagorean Theorem)
19a. Possible answer: Fold a pair of corresponding
vertices (any vertex in the original figure and the
corresponding vertex in the image) together and
crease; repeat for another pair of corresponding
vertices; the intersection of the two creases is the
center of rotation.
19b. Possible answer: Draw a segment (a chord)
between a pair of corresponding vertices and
construct the perpendicular bisector; repeat for
another pair of corresponding vertices; the
intersection of the two perpendicular bisectors
is the center of rotation.
20e. Label the third vertex C. Construct segment
D
2x
2
CD, which bisects C. AD
B 3x 3 , or 2:3
y
800 Volume
600
Surface area
400
200
5
x
ANSWERS TO EXERCISES
131
Answers to Exercises
2. 16 cm, 4 cm; 768 cm3 2412.7 cm3;
64
12 cm3 37.7 cm3; 1
3 125
3
3. 5; 2;
7 1500 cm
H 3 64
;
4. 1944 ft3 6107.3 ft3; 2
4 27 32 ft
125
5. 2
7
2
6. 2:5
7. 3
8. $1,953.13
9. 2432 lb
10. Possible answer: No, a 4-foot chicken, similar
to a 14-inch 7-pound chicken, would weigh
3
approximately 282 pounds 41843 7x. It is unlikely
that the legs of the giant chicken would be able to
support its weight.
11. Possible answer: Assuming the body types of
the African goliath frog and the Brazilian gold frog
are similar, the gold frog would weigh about
0.0001 kg, or 0.1 g.
12. surface area ratio 116 , volume ratio 614 . The
dolphin has the greater surface area to volume
ratio.
13. The ratio of the volumes is 217 . The ratio of the
surface areas is 91.
14.
S(x) 22x 2 and V(x) 6x 3.
Possible answer: The surface area equation is
quadratic, so the graph is a parabola, and the
volume equation is cubic, so the graph is a cubic
graph.
15. Possible proof: The volume of the first
rectangular prism is lwh. The volume of the second
rectangular prism is rl rw rh, or r 3lwh. The ratio
r3lwh
3
of the volumes is lwh , or r .
3
16. 9,120 m or approximately 28,651 m3
s
s s2 s
17. 4 or 4
2
2
LESSON 11.7
1
2. 333 cm
3. 45 cm
4. 21 cm
5. 28 cm
6. no
7. José’s method is correct. Possible explanation:
Alex’s first ratio compares only part of a side of the
larger triangle to the entire corresponding side of the
smaller triangle, while the second ratio compares
entire corresponding sides of the triangles.
8. yes
9. 6 cm; 4.5 cm
10. 13.3 cm; 21.6 cm
11. yes; no; no
12. yes; yes; yes
13. 3
2 cm; 6
2 cm
14.
Answers to Exercises
1. 5 cm
E
15.
F
I
J
So two pairs of corresponding sides of XYZ and
XAB are proportional. X X, so
XYZ XAB by the SAS Similarity Conjecture.
Because XYZ XAB, XAB XYZ.
by the Converse of the Parallel
YZ
Hence, AB
Lines Conjecture.
20. Set the screw so that the shorter lengths of the
styluses are three-fourths as long as the longer
lengths.
21. x 4.6 cm, y 3.4 cm
22. x 45 ft, y 40 ft, z 35 ft
343
23. 729
24. She is incorrect. She can make only nine 8 cm
diameter spheres.
25. 6x 2; 24x 2; 54x 2
1
26. 3r
27a. Possible construction method: Use the
triangle-and-circle construction from Lesson 11.3,
.
Exercise 18, to locate the golden cut, X, of AB
Then use perpendicular lines and circles to create a
rectangle with length AB and width AX.
27b. Possible construction method: Construct
golden rectangle ABCD following the method from
by constructing
27a. For square AEFD, locate EF
circle A and circle D each with radius AD. Repeat
the process of cutting off squares as often as
desired. For the golden spiral from point D to point
E, construct circle F with radius EF; select point D,
point E, and circle F and choose Arc On Circle from
the Construct menu.
28. possible answer:
B
16. Extended Parallel/Proportionality Conjecture
17. You should connect the two 75-marks. By the
Extended Parallel/Proportionality Conjecture,
drawing a segment between the 75-marks will
5
form a similar segment that has length 170
0 , or 75%,
of the original.
18. 2064 cm3 6484 cm3
19. possible proof:
a b
c
d
ad cb
ad ab cb ab
a(d b) b(c a)
a(d b) b(c a)
ab
ab
db ca
b
a
132
ANSWERS TO EXERCISES
P
Q
A
C
S
R
D
Given: Circumscribed quadrilateral ABCD, with
points of tangency P, Q, R, and S
Show: AB DC AD BC
Use segment addition to show that each sum of
lengths of opposite sides is composed of four lengths:
AB DC (AP BP) (DR CR) and AD BC (AS DS) (BQ CQ). Using the Tangent
Segments Conjecture, AP AS, BP BQ,
CR CQ, and DR DS, and the four lengths in
each sum are equivalent. Here are the algebraic
steps to show that the whole sums are equivalent.
29.
AB DC (AP BP) (DR CR) Segment
addition.
(AS BQ) (DS CQ) Substitute AS
for AP, BQ
for BP, DS for
DR, and CQ
for CR.
(AS DS) (BQ CQ) Regroup the
measurements
by common
points of
tangency.
Answers to Exercises
AB DC AD BC
Use segment
addition to
rewrite the
right side as
the other sum
of opposite
sides.
ANSWERS TO EXERCISES
133
CHAPTER 11 REVIEW
1. x 24
2. x 66
3. x 6
4. x 17
5. 6 cm; 4.5 cm; 7.5 cm; 3 cm
1
1
6. 46 cm; 72 cm
7. 13 ft 2 in.
8. It would still be a 20° angle.
9.
Answers to Exercises
K
P
L
10. Yes. If two triangles are congruent, then
corresponding angles are congruent and
corresponding sides are proportional with ratio
1
, so the triangles are similar.
1
11. 15 m
12. 4 gal; 8 times
13. Possible answer: You would measure the
height and weight of the real clothespin and the
height of the sculpture.
Wsculpture
Hsculpture 3
Wclothespin
Hclothespin
If you don’t know the height of the sculpture, you
could estimate it from this photo by setting up a
ratio, for example,
Hperson
Hsculpture
Hperson’s photo Hsculpture’s photo
134
ANSWERS TO EXERCISES
14. 9:49
5 125
15. 4; 6
4
16. $266.67
17. 640 cm3
18. 32; 24; 40; 126
19. 841 coconuts
1
20a. 1 to 4 to 2, or 4 to to 2
20b. 3 to 2 to 1
20c. Answers will vary.
21. Possible answer: If food is proportional to
1
body volume, then 8000 of the usual amount of food
is required. If clothing is proportional to surface
1
of the usual amount of clothing is
area, then 400
required. It would take 20 times longer to walk a
given distance.
22. The ice cubes would melt faster because they
have greater surface area.