Answers to Exercises

Transcription

CHAPTER 0 • CHAPTER
0
LESSON 0.1
1. possible answers: snowflakes and crystals;
flowers and starfish
2. Answers might include shapes and patterns,
perspective, proportions, or optical illusions.
3. Answers will vary. Bilateral symmetry.
4. A, B, C, and F
5. A, B, D, and E
6. 4 of diamonds; none
7. The line of reflection is a line along the
surface of the lake.
8. One line of symmetry is a vertical line through
the middle of the Taj Mahal, and the other is a
horizontal line between the Taj Mahal and the
reflecting pool. The pool reflects the building,
giving the scene more reflectional symmetry than
the building itself has.
9. Designs will vary, but all should have 2-fold
rotational symmetry.
2 lines of reflectional
symmetry
It is impossible to have
symmetry without
rotational symmetry.
10. Answers will vary.
ANSWERS TO EXERCISES
1
LESSON 0.2
1. compass and straightedge
2. Answer will be the Astrid or 8-Pointed Star,
possibly with variations.
3. Answers will be a design from among the three
choices.
4. The first design has 4-fold symmetry and four
lines of reflectional symmetry. The triangle has
3-fold rotational symmetry and three lines of
reflectional symmetry. The third design has 6-fold
rotational symmetry if you ignore color and 2-fold
rotational symmetry if you don’t.
5. possible answer:
2 lines of
reflectional
symmetry
2
6. possible answer:
7.
symmetry
3-fold rotational
symmetry: rotated
1208, 2408, 3608
LESSON 0.3
4. possible answer:
1. Design with reflectional symmetry is drawn on
this background.
2. Design should have rotational symmetry.
Possible answer:
5. Drawing should resemble the hexagons in the
given figure, except that each radius, and the side
of each hexagon, should measure 1 inch.
3. possible answer:
3
LESSON 0.4
1. Possible answer: The designs appear to go in
and out of the page (they appear 3-D). The squares
appear to spiral (although there are no curves in
the drawing). The spiral appears to go down as if
into a hole. Lines where curves meet look wavy.
Bigger squares appear to bulge out.
2. Possible answer: When zebras group together,
their stripes make it hard for predators to see
individual zebras.
4
3. Designs will vary.
5. Possible answers: 2-D: rectangle, triangle,
trapezoid; 3-D: cylinder, cone, prism. The palace
facade has one line of reflectional symmetry
(bilateral symmetry).
LESSON 0.5
5.
1. possible answers: Scotland, Nigeria
2. possible answer:
3. possible answer:
6. Answers will be a lusona from among the three
choices.
7. It means to solve a problem boldly and
decisively or in a creative way not considered by
others.
8. The square knot has reflectional symmetry
across a horizontal line. The less secure granny
knot has 2-fold rotational symmetry.
9. The result is a regular pentagon.
4. possible answer:
Cut the middle ring.
5
LESSON 0.6
1. possible answers: Morocco, Iran, Spain,
Malaysia
2. Alhambra
3. 2.1 cm
4. in terms of the original square tile: 4-fold in the
center (center of orange) or corner (center of
white); 2-fold in the midpoint of the edge of the
tile (where two orange shapes meet)
6
7. Designs will vary, but should contain a regular
hexagon.
8. Tessellations will vary.
CHAPTER 0 REVIEW
1. possible answer: Islamic, Hindu, Celtic
2. Sometimes the large hexagons appear as blocks
with a corner removed, and sometimes they appear
as corners with small cubes nestled in them.
3. Compass: A geometry tool used to construct
circles.
Straightedge: A geometry tool used to construct
straight lines.
4. possible answer:
8. Wheel A has four lines of reflectional symmetry;
Wheel C has five lines of reflectional symmetry.
Wheels B and D do not have reflectional symmetry.
9. Wheels B and D have only rotational symmetry.
Wheels A and B have 4-fold, Wheel C has 5-fold,
and Wheel D has 3-fold rotational symmetry.
10, 11. Drawing should contain concentric circles
and symmetry in some of the rings.
12. The mandala should contain all the required
elements.
13a. The flag of Puerto Rico is not symmetric
because of the star and the colors.
13b. The flag of Kenya does not have rotational
symmetry because of the spearheads.
13c. possible answers:
Japan
5.
6. possible answers: hexagon: honeycomb,
snowflake; pentagon: starfish, flower
7. Answers will vary. Should be some form of an
interweaving design.
Nigeria
7
1
LESSON 1.1
1. sample answers: point: balls, where lines cross;
segment: lines of the parachute, court markings;
collinear: points along a line of the court structure;
coplanar: each boy’s hand, navel, and kneecap,
points along two strings
@#$, TP
@#$
2. PT
@#$, RA
@#$, AT
@#$, TA
@#$, RT
@#$, TR
@#$
3. any two of the following: AR
@#$
@#$
@#$
@#$
@#$
@#$
4. any two of the following: MA,MS,AS,AM,SA,SM
5.
6.
A
#$, PN
#$
22. PM
24. A
25.
26.
Y
7.
M
N
X
27.
C
D
B
A
28. 10
29–31.
y
y
5
3
M
D
–3
L
3
C
x
S
11
O
–5
–3 E
# or CA
#
# or QP
#
8. AC
9. PQ
# or RT
#, RI
# or IR
#, and TI
# or IT
#
10. TR
y
11. A
12.
B
A (4, 0)
x
D
–5
B
32. Yes, P9(6, 2), Q9(5, 22), R9(4, 26); the slope
between any two of the points is }14}.
y
R
Q
R
P´
x
x
Q´
# 5 14.3 cm 14. mCD
# 5 6.7 cm
13. mAB
15–17. Check each length to see if it is correct.
Refer to text for measurements.
#. X is the midpoint of
18. R is the midpoint of PQ
##
#. No midpoints are
WY . Y is the midpoint of XZ
shown in nABC.
19. possible answers:
or
A
B
C
D
P
R´
33. possible answer:
y
Q
N
x
P
E
#### > CE
####
AC
####
####
BC > CD
20.
34.
P
R
A
T
G
Y
35. Answers will vary. Possible answers:
S
M
T
8 cm
8 cm
11 cm
11 cm
Q
8
B
B
K
#$, AC
#$
21. AB
#$, XZ
#$
23. XY
USING YOUR ALGEBRA SKILLS 1
1. (3, 4)
2. (29, 1.5)
3. (5.5, 5.5)
4. (26, 44)
5. Yes. The coordinates of the midpoint of a
segment with endpoints (a, b) and (c, d ) are found
a1c
}
by taking the average of the x-coordinates, }
2 ,
b1d
and the average of the y-coordinates, }2}. Thus the
a1c b1d
} }}
midpoint is 1}
2 , 2 2.
6. (3, 2) and (6, 4). To get the first point of
trisection, sum the coordinates of points A and B
to get (9, 6), then multiply those coordinates by }31} to
get (3, 2). To get the second point of trisection, sum
the coordinates of points A and B to get (9, 6), then
multiply those coordinates by }32} to get (6, 4). This
works because the coordinates of the first point
are (0, 0).
7. Find the midpoint, then find the midpoint of
each half.
8a. Midpoints for Figure 1 are (5.5, 6.5); for
Figure 2, (16, 6.75); and for Figure 3, (29.75, 5.5).
8b. For these figures the midpoints of the two
diagonals are the same point.
9
LESSON 1.2
24.
C
1358
1. /TEN, /NET, /E; /FOU, /UOF, /1;
/ROU, /UOR, /2
N
2.
A
D
25.
T
67.58
D
26. no
B
G
67.58
228
228
A
3.
E
C
Y
D
I
4.
N
S
A
M
27.
L
5. /S, /P, /R, /Q; none in the second figure
6. possible answer:
B
28.
C
A
D
7. 90°
8. 120°
9. 45°
10. 135°
11. 45°
12. 135°
13. 30°
14. 90°
15. Yes; m/XQA 1 m/XQY 5 45° 1 90° 5 135°,
which equals m/AQY.
16. 69°
17. 110°
18. 40°
19. 125°
20. 55°
21. /SML has the greater measure because
m/SML < 30° and m/BIG < 20°.
22.
A
29.
30. One possibility is 4:00.
H
31.
908
8
6
O
T
32.
T
H
R
33.
S
A
T
G
448
A
23.
I
908
10
B
N
34.
W
O
B
T
45. 180 km. The towns can be represented by
three collinear points, P, S, and G. Because S is
between P and G, PS 1 SG 5 PG.
12 cm
46.
6 cm
47.
I
35.
37.
39.
41.
43.
MY; CK; m/I
x 5 54°
z 5 32°
242°
no
36.
38.
40.
42.
/SEU; /EUO; MO
y 5 102°
288°
They add to 360°.
6 cm
4.36 cm
2.18 cm
2.18 cm
48. MS 5 DG means that the distance between
M and S equals the distance between D and G.
MS > DG means that segment MS is congruent
to segment DG. The first statement equates two
numbers. The second statement concerns the
congruence between two geometric figures.
However, they convey the same information and
are marked the same way on a diagram.
49. Answers will vary. Possible answer.
C
708
4 cm
408
A
B
F
44. You will miss the target because the incoming angle is too big.
408
D
708
6 cm
E
Target
Mirror
Laser light
source
11
LESSON 1.3
1–3. possible answers:
D
1.
458
O
2.
G
T
E
R
3.
11. The measures of complementary angles sum
to 90°, whereas the measures of supplementary
angles sum to 180°.
12. No, supplementary angles can be unconnected, while a linear pair must share a vertex and
a common side.
13a. an angle; measures less than 90°
13b. angles; have measures that add to 90°
13c. a point; divides a segment into two congruent
segments
13d. a geometry tool; is used to measure the sizes
of angles in degrees
14.
E
B
P
C
B
I
G
4.
A
G
@#$ and CD
@#$ intersect at point P so that P is
If AB
between A and B, and P is between C and D, then
/APC and /BPD are a pair of vertical angles.
15. true
D
S
M
5.
A
16. true
P
6.
E
R
D
A
17. true
B
E
C
7.
408
A
18. false
508
B
8.
19. false
1408
408
C D
9. B is a Zoid. A Zoid is a creature that has in its
interior a small triangle with a large black dot
taking up most of its center.
10. A good definition places an object in a class
and also differentiates it from other objects in that
class. A good definition has no counterexamples.
12
D
20. false
28. One possible answer is A(28, 8), B(24.5, 6.5),
C(211, 1).
508
y
A
B
508
808
6
S
C
21. true
–12
x
–6
T
22. false; Though the converse is true, a
counterexample is
C
T
R
29. 12 cm
30. 36°
A
23. false
32.
1408
0 pt.
C
1 pt.
D
408
24. false
2 pt.
A
3 pt.
33.
C
2 pt.
1 pt.
T
25. possible answers: (23, 0), (0, 2), or (6, 6)
26. possible answers: (2, 23) or (5, 21)
27. The reflected ray and the ray that passes
through (called the refracted ray) are mirror
images of each other. Or they form congruent
angles with the mirror.
A
Reflected
segment
B
Mirror
4 pt.
3 pt.
5 pt.
6 pt.
Infinitely
many points
120° 1 2
} }}, }}
34a. }
360° 5 3 3 left
1 1
60°
} }}, }}
34b. }
360° 5 6 6 missing
360°
34c. }9} 5 40°
13
LESSON 1.4
1.
# > FI
# and /IVE > /ANC
18. PA
1308
1308
1308
5 cm
3.
4. octagon
5. hexagon
6. heptagon
7. pentagon
8–10. One possible answer for each is shown.
8. pentagon FIVER
9. quadrilateral FOUR
10. equilateral quadrilateral BLOC
11a. a polygon; has eight sides
11b. a polygon; has at least one diagonal outside
of the polygon
11c. a polygon; has 20 sides
11d. a polygon; has all sides of equal length
12. One possibility is /C and /Y are consecutive
# and YN
# are consecutive sides.
angles; CY
13. 9; possible answer:
#, AD
##, BD
#, BE
#, CE
#
14. AC
15. nTIN
16. nWEN
17a. a 5 44, b 5 58, c 5 34
17b. m/T 5 87° and m/I 5 165°
14
21. 84 in.
22. 5.25 cm
23. AB 5 14 m, CD 5 25 m
24. complementary angles: /AOS and /SOC;
vertical angles: /OCT and /ECR or /TCE and
/RCO
25.
E
A
B
C
F
D
5
20
27. All are possible except two points.
0 pts.
1 pt.
3 pts.
4 pts.
5 pts.
6 pts.
LESSON 1.5
1. D
3. C
5.
2. A
4. B
6. C
C
C
14 cm
10 cm
408
A
7.
T
S
M
L
A
B
A
F
10 cm
R
408
C
A
8. possible answer:
2a
2a
b – 2a
b
6 cm
B
408
6 cm
A
B
408
y
C
C
(–3, 5)
(2, 4)
C
(–4, 1)
(1, 1)
x
4 cm
22. (24,1) → (23,22) (1,1) → (2, 22) (2, 4) →
(3, 1) (23, 5) → (22, 2) Yes, the quadrilaterals are
congruent.
23. Find the midpoint of each rod. All the
midpoints lie on the same line; place the edge
of a ruler under this line.
24–26. Sample answers for 24–26
P
P
24.
25.
808
A
4 cm
B
A
4 cm
1208
P
4 cm
Z
12. possible answers: (21, 21), (21, 0), or (24, 3)
13. possible answers: (3, 23), (3, 4), (211, 4),
(211, 23), (20.5, 0.5), or (27.5, 0.5)
14. possible answers: (23, 1), (21, 29), (2, 2),
(4, 28), (0, 21), or (1, 26)
C
E
A
N
26.
E
T
A
N
Q
U
D
A
T
C
7 cm
12 cm
458
A
9 cm
B
458
A
9 cm
B
15
9. possible answer:
A
E
17. true
18. true
19. False, a diagonal connects nonconsecutive
vertices.
20. False, an angle bisector divides an angle into
two congruent angles.
21. true
3a
3a
10 cm
LESSON 1.6
1. first figure: quadrilateral ABCD with two
congruent sides and one right angle; second
figure kite EFGH; third figure: trapezoid IJKL
with two right angles; fourth figure parallelogram
MNPQ
2. B
3. D
4. F
5. C
6. A, D, F
I
7. D
Z
19b. Possible answer: Rotate one triangle so that
a congruent pair of sides forms a diagonal of a
parallelogram.
20. There are three possibilities: a rhombus, a
concave kite, or a parallelogram.
O
8.
N
E
B
F
9.
L
A
E
21a. right triangles
21b. isosceles right triangles
22.
U
Q
10.
T
H
23.
I
R
5 cm
5 cm
G
11.
24.
12. square
13. possible answers: (2, 6), (22, 4); (6, 22),
(2, 24); (3, 21), (1, 3)
14. 90 cm
15. (5, 6)
16. S(9, 0), I(4, 22)
17. S(23, 0), I(21, 25)
18. A(7, 6), N(5, 9) or A(25, 22), N(27, 1)
19a. Possible answer: Flip one triangle and align it
so that a congruent pair of sides forms a diagonal
of a kite.
16
728 728
728 728
728
25. no
y
C (0, 5)
B (4, 4)
x
A (5, 0)
LESSON 1.7
1. Answers will vary. Sample answers:The green
area in the irrigation photo is a circle,the water is a
radius,and a path on the far side of the circle appears
to be tangent to the circle.The wood bridge is an arc of
a circle,and the railings are arcs of concentric circles.
The horizontal support beam under the bridge is
a chord.
#, BD
#, EC
#, EF
#
2. three of the following: AB
#
3. EC
#, EP
#, FP
#, BP
#, CP
#
4. AP
X, AE
X, AB
X, BC
X, CD
X, DF
X,
5. five of the following: EF
X
X
X
X
X
X
X
X
EB , ED , FC , AC , DB , AF , AD , BF
X or EFC
X, EBC
X or EAC
X
6. EDC
X
X, FEC
X, DEC
X, . . .
7. two of the following: ECD , EDF
B
P
Q
A
17. equilateral; 3 to 1
s
18.
19. yes; yes
@#$, HB
@#$
8. FG
9. either F or B
10. possible answers: cars, trains, motorcycles;
washing machines, dishwashers, vacuum cleaners;
compact disc players, record players, car racing,
Ferris wheel
X 5 110°; mPRQ
X 5 250°
11. mPQ
12.
16. Equilateral quadrilateral. (The figure is
actually a rhombus, but students have not yet
learned the properties needed to conclude that the
sides are parallel.)
y
5
–5
5
x
658
–5
2158
20. yes; no
13. possible answers: concentric rings on cross
sections of trees (annual rings),bull’s-eye or target,
ripples from a rock falling into a pond.
14.
y
5
–5
5
x
–5
21. no; no
15.
y
8
P
Q
8
16
x
17
22. 80°. The slices of pizza do not overlap, so
60° 1 x 5 140°, where x is the measure in
degrees of the angle of the second slice.
23.
24. not possible
25.
C
608
608
32. not possible
33. R
G
T
34.
A
B
27. about 0.986°
28. 15°
29.
4a + 2b
35.
T
1008
508
36.
1208
B
C
2p
2p
P
E
A
37.
508 708
708
I
F
30.
1208
608
1208
A
26.
A
31.
U
2p
558
2p
Q
E
1208
T
R
K
T
I
18
1208
608
22. False. The two lines are not necessarily in the
same plane, so they might be skew.
LESSON 1.8
1.
2.
3.
4.
23. true
5.
6.
7.
8.
24. true
25. true
2
9. 60 boxes
3
4
10.
26. False. They divide space into seven or eight
parts.
5
3
4
11.
12.
27. true
13. B, D
15. C
17. A
18.
19.
20. true
21. true
28. (23, 1)
14. B
16. D
29. perimeter 5 20.5 cm; m(largest angle) 5 100°
30.
8 cm
1208
13 cm
19
18. pyramid with hexagonal base
LESSON 1.9
1. Sample answer: Furniture movers might
visualize how to rotate a couch to get it up a
narrow staircase.
2. yes
###$
3. W
O M E N ; Nadine is ahead.
4. 28 posts
5. 28 days
6. 0 ft (The poles must be touching!)
B
7.
20. pyramid with square base
A
B
A
21. x 5 15, y 5 27
22. x 5 12, y 5 4
23.
8.
19. prism with hexagonal base
A
9.
A
3 cm
B
3 cm
24.
3 cm
A
B
10.
Polygons
Quadrilaterals
Trapezoids
11.
12.
13.
14.
15.
16.
Triangles
Obtuse
Isosceles
no
C9(22, 3), A9(0, 0), B9(0, 5), D9(22, 1)
Y9(24, 1), C9(3, 21), N9(0, 3)
(x, y) → (x 1 3, y 1 2)
@#$ ' CP
@#$, EF
@#$ i GH
@#$; i ' k, j ' k
AB
perimeter 5 34 cm
25.
26.
28.
30.
32.
34.
point, line, plane
@#$
AB
vertex
@#$ i CD
##
AB
/ABC
congruent to
20
X
AB
#$
AB
protractor
# ' CD
@#$
AB
35. The distance is two times the radius.
r
r
P
Q
PQ = 2r
36. They bisect each other and are perpendicular.
A
17. Left photo: Three points determine a plane.
Middle photo: Two intersecting lines determine a
plane. Right photo: A line and a point not on the
line determine a plane.
27.
29.
31.
33.
P
Q
B
CHAPTER 1 REVIEW
1.
2.
3.
4.
5.
6.
7.
8.
true
#$.
False; it is written as QP
true
False; the vertex is point D.
true
true
False; its measure is less than 90°.
false; two possible counterexamples:
T
Y
E
A
P
30.
31.
A
D
A
P
R
C
D
A
29.
P
P
B
C
T
/APD and /APC
are a linear pair.
/APD and /APC
are the same angle.
E
Y
A
C
B
9. true
10. true
11. False; they are supplementary.
12. true
13. true
14. true
15. False; it has five diagonals.
16. true
17. F
18. G
19. L
20. J
21. C
22. I
23. no match
24. A
25. no match
T
26.
32.
D
33.
34.
2 in.
5 in.
3 in.
35.
1258
36.
K
408
27.
S
5
3
37.
T
7
28.
O
P
A
N
R
I
E
G
21
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
114°
x 5 2, y 5 1
x 5 12, y 5 4
x 5 4, y 5 2.5
x 5 10, y 5 8
AB 5 16 cm
96°
105°
30°
51. He will get home at 5:46 (assuming he goes
inside before he gets blown back again).
52. (2, 3)
53.
Quadrilateral
54.
Rectangle Square Rhombus
55.
Trapezoid
48. 66 inches
49. 3 feet
50. The path taken by the midpoint of the ladder is
an arc of a circle or a quarter-circle if the ladder
slides all the way from the vertical to the horizontal.
57.
Shed
Path of flashlight
22
56.
2
LESSON 2.1
13.
14.
28.
30. sample answer:
31.
33.
35.
37.
39.
41.
collinear
dodecagon
protractor
diagonal
90°
sample answer:
32.
34.
36.
38.
40.
42.
1. “All rocks sink.” Stony needs to find one rock
that will not sink.
2. If two angles are formed by drawing a ray from a
line, then their measures add up to 180°.
3. 10,000, 100,000, . . . . Each term is 10 times the
previous term.
4. }65}, 1, . . . . (Written with the common
denominator 6, the pattern becomes
1 2 3 4 5 6
}}, }}, }}, }}, }}, }}, . . . .)
6 6 6 6 6 6
5. 217, 221
6. 28, 36
7. 21, 34
8. 49, 64
9. 210, 224
10. 64, 128
11.
12.
27.
isosceles
parallel
radius
regular
perpendicular
sample answer:
I
15.
16.
A
17. 1, 4, 7, 10, 13
18. 15, 21, 28, 36, 45
20. sample answers: 3, 6, 12, 24, 48, . . . and 4, 8, 12,
16, 20, . . .
22. 7th term: 56; 10th term: 110; 25th term: 650
23. Conjecture is false; 142 5 196 but 412 5 1681.
24. 11,111 ? 11,111 5 123,454,321. For the sixth
line, if 111,111 is multiplied by itself, then the
product will be 12,345,654,321. But 1,111,111,111 ?
1,111,111,111 5 1,234,567,900,987,654,321.
25.
26.
G
N
T
43. sample answer:
A
C
####
Clearly AC
does not
bisect /A
or /C.
45. Methods will vary. It is not possible to draw a
second triangle with the same angle measures and
side length that is not congruent to the first.
C
408
A
608
9 cm
B
23
4.
5.
6.
7.
LESSON 2.2
1. See table below.
2. See table below.
3. See table below.
See table below.
See table below.
See table below.
See table below.
1. (Lesson 2.2)
n
1
2
3
4
5
6
…
n
…
20
f (n)
3
9
15
21
27
33
… 6n 2 3 …
117
…
20
…
256
…
20
…
148
2. (Lesson 2.2)
n
1
2
3
4
5
6
…
f (n)
1
22
25
28
211
214
…
n
23n 1 4
3. (Lesson 2.2)
n
f (n)
1
2
3
4
5
6
…
24
4
12
20
28
36
…
n
8n 2 12
4. (Lesson 2.2)
Number of sides
3
4
5
6
…
n
…
35
Number of triangles formed
1
2
3
4
…
n22
…
33
5. (Lesson 2.2)
Figure number
1
2
3
4
5
6
…
n
…
200
Number of tiles
8
16
24
32
40
48
…
8n
…
1600
Figure number
1
2
3
4
5
6
…
n
…
200
Number of tiles
1
5
9
13
17
21
… 4n 2 3 …
797
6. (Lesson 2.2)
7. (Lesson 2.2)
24
Figure number
1
2
3
4
5
6
…
Number of matchsticks
5
9
13
17
21
25
…
Number of matchsticks
in perimeter of figure
5
8
11
14
17
20
n
…
200
4n 1 1 …
801
… 3n 1 2 …
602
8. 8n is steeper; the coefficient of n.
13.
R
E
f (n)
T
8n
C
50
14.
40
30
4n 1 1
4n 2 3
20
3n 1 2
2
1
2
1
3
3
10
1
2
4
6
1
n
8
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
Octane (C8H18)
3
10. y 5 }2}x 1 3
11.
1
2
1
2
9. CnH2n12
H
1
1
L
H
15. Márisol should respond by noticing that all
the triangles José drew were isosceles, but it is
possible to draw a triangle with no two sides
congruent. In other words, she should show him a
counterexample.
16. Methods will vary. It is not possible to draw
another triangle with the given side lengths and
angle measure that is not congruent to the first.
C
9 cm
T
458
A
E
12.
Q
O
M
L
8 cm
B
17. She could try eating one food at a time;
inductive.
18. 2600
Y
25
n(n 2 1)
8. Use }2} to get 780 direct lines. Use a central
hub with a line to each house to get 40 lines. The
art shows the direct-line solution and the practical
solution for six houses.
LESSON 2.3
1.
See table below.
2. n 1 1, 36
9. Use points for the 10 teams and segments
connecting them to represent four games played
n(n 2 1)
between them. So, use 4 ? }2} to get 180 games
played.
n(n2 1)
10. If }2} 5 66, then n(n 2 1) 5 132. What
two consecutive numbers multiply to equal 132?
12 times 11. Thus, there were 12 people at the
party.
11. true
12. true
13. False; an isosceles right triangle has two sides
congruent.
14. false
3. You can draw a diagonal from one vertex to all
the other vertices except three: the two adjacent
vertices and the vertex itself.
n(n 2 3)
4. }2}, 560
n(n 2 1)
5. }2}, 595
E
B
C
A
/AED and /BED are
not a linear pair.
n(n 2 1)
6. }2}, 595
15. False; they are parallel.
16. true
17. False; a rectangle is a parallelogram with all of
its angles congruent.
18. False; a diagonal is a segment in a polygon
connecting any two nonconsecutive vertices.
19. true
20. 5049
7. Answers will include relationships between
points in Exercises 5 and 6 and vertices of a polygon.
The total number of segments connecting n random
points is the number of diagonals of the n-sided
n(n 2 3)
polygon, }2}, plus the number of sides, n.
Thus, the total number of segments connecting n
n(n 2 3)
2n
n(n 2 3) 1 2n
random points is }2} 1 }2} 5 }2} 5
n2 2 3n 1 2n
n2 2 n
}
}5}
}
2
2
D
n(n 2 1)
5 }2} .
1. (Lesson 2.3)
26
Lines
1
2
3
4
5
…
n
…
35
Regions
2
4
6
8
10
…
2n
…
70
LESSON 2.4
1. inductive; deductive
2. m/B 5 65°; deductive
3. inductive
25
18.
19.
36
4. DG 5 258 cm; deductive
5. 180°; 180°; The sum of the five angles is 180°;
inductive.
20.
208
328
648
278
378
I
E
L
33.
D
B
1408
O
G
34.
C
T
O
D
35.
K
D
N
27
6. LNDA is a parallelogram; deductive.
7. Possible answer: AB 5 CD, so AB 1 BC 5
CD 1 BC. Because AB 1 BC 5 AC and CD 1
# > BD
#.
BC 5 BD, AC 5 BD, so AC
8. 3; 10; 7
9. Just over 45°; if m . 90°, then }21}m . 45°.
10. 48°; 17°; 62°; m/CPB
11. Possible answers: Because m/APC 5
m/BPD, add the same measure to both sides to get
m/APC 1 m/CPD 5 m/BPD 1 m/CPD.By
angle addition, m/APC 1 m/CPD 5 m/APD
and m/BPD 1 m/CPD 5 m/CPB. Therefore,
m/APD 5 m/CPB.
13. The pattern cannot be generalized because
once the river is straight, it cannot get any shorter.
14. 900, 1080
15. 75, 91
4
16. }5}, 12
17.
21. Sample answer: There were three sunny days
in a row, so I assumed it would be sunny the fourth
day, but it rained.
22. L
23. M
24. A
25. B
26. E
27. C
28. G
29. D
30. H
31. I
W
32.
15.
LESSON 2.5
1. a 5 60°, b 5 120°, c 5 120°
2. a 5 90°, b 5 90°, c 5 50°
3. a 5 77°, b 5 52°, c 5 77°, d 5 51°
4. a 5 60°, b 5 c 5 120°, d 5 f 5 115°, e 5 65°,
g 5 i 5 125°, h 5 55°
5. a 5 90°, b 5 163°, c 5 17°, d 5 110°, e 5 70°
6. The measures of the linear pair of angles add up
to 170°, not 180°.
7. The angles at which he should cut measure 45°.
8. Greatest: 120°. Smallest: 60°. One possible
explanation: The tree is perpendicular to the
horizontal. The angle of the hill measures 30°.
The smaller angle and the angle between the hill
and the horizontal form a pair of complementary
angles, so the smaller angle equals 90° 2 30° 5 60°.
The smaller angle and larger angle form a linear
pair, so the larger angle equals 180° 2 60° 5 120°.
9. The converse is not true. ; sample counterexample:
A
A
O
T
16.
17.
6
3
4
18. Possible answer: All the cards look exactly as
they did, so it must be the 4 of diamonds, because
it has rotational symmetry while the others do not.
19. 22.5°
20.
558
1258
B
10. each must be a right angle
21. CnH2n
H
H
11. Let the measures of the congruent angles be x.
They are supplementary, so x 1 x 5 180°, 2x 5 180°,
x 5 90°. Thus each angle is a right angle.
12. The ratio is 1. The ratio does not change as
long as the lines don’t coincide. Because the
demonstration does not explain why, it is not
a proof.
13. P
14.
5
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
22. See table below.
n(n 2 1) 80(79)
23. handshake problem: }2} ; }2} 5 3160
pieces of string
24. 3160 intersections
n(n 22)
25. }2} yields 760 handshakes.
26. 21; 252
n(n 23)
27. }2} 5 560; there are 35 vertices.
#; deductive.
28. M is the midpoint of AY
3
A
H
T
22. (Lesson 2.5)
Rectangle
1
2
3
4
5
6
…
…
200
Perimeter of rectangle
10
14
18
22
26
30
… 4n 1 6 …
806
Number of squares
6
12
20
30
42
56
…
n
… 40,602
(n 1 1)(n 1 2)
28
13. No, tomorrow could be a holiday. Converse:“If
tomorrow is a school day, then yesterday was part
of the weekend;” false.
14. 42°
15. 20°
16. No, the lines are not parallel.
17. isosceles triangle
18. a parallelogram that is not also a rectangle or a
rhombus
19. 18 cm
20. 39°
21. 3486
22. 30 squares (one 4-by-4, four 3-by-3, nine
2-by-2, and sixteen 1-by-1)
23. The triangle moved to the left 1 unit.Yes,
congruent to original.
LESSON 2.6
1. 63°
2. 90°
3. no
4. 57°
5. yes
6. 113°
7. a 5 d 5 64°, b 5 c 5 116°, e 5 g 5 i 5 j 5
k 5 108°, f 5 h 5 s 5 72°, m 5 105°, n 5 79°,
p 5 90°, q 5 116°, t 5 119°; Possible explanation:
Using the Vertical Angles Conjecture, n 5 79°.
Using the Linear Pair Conjecture, p 5 90°. Using the
Corresponding Angles Conjecture and b 5 116°,
q 5 116°.
8. Possible answer: In the
k
<
diagram, lines , and m are parallel
1
and intersected by transversal k.
m
Using the Corresponding Angles
2
3
Conjecture, /1 > /2. Using the
Vertical Angles Conjecture,
/2 > /3. Because /1 and /3 are both congruent
to /2, they must be congruent to each other. So
/1 > /3. Therefore,if two parallel lines are cut by a
transversal, then alternate exterior angles are
congruent.
9. 56° 1 114° 5 170° Þ 180°. Thus, the lines
marked as parallel cannot really be parallel.
10. Alternate interior angles measure 55°, but
55° 1 45° Þ 180°.
11. The incoming and
458
908
outgoing angles measure
458
45°. Possible explanation:
Yes, the alternate interior
angles are congruent, and
thus, by the Converse of the
458
Parallel Lines Conjecture,
908
the mirrors are parallel.
458
24. The quadrilateral was reflected across both
axes or rotated 180° about the origin.Yes, congruent
to original.
y
4
x
4
25. The pentagon was reflected across the line
x 5 y.Yes, congruent to original.
y
N
5
O'
L
M'
Q
P
x
5
O
N'
x
E
B
A
6
E'
M
L'
26. (Lesson 2.6)
Figure number
1
2
3
4
5
6
…
n
…
35
Number of yellow squares
2
3
4
5
6
7
…
n11
…
36
Number of blue squares
3
5
7
9
11
13
… 2n 1 1 …
71
Total number of squares
5
8
11
14
17
20
… 3n 1 2 …
107
29
12. Explanations will vary.
Sample explanation:“I used the
protractor to make corresponding angles congruent when I
drew line PQ.”
y
5
1. 22
12
2. 2}1}
3
97
3. }4}
6 < 2.1
4. y 5 23
5. x 5 4
6. Any point of the form (3p, 24p).Possible answer:
(6,28),(9,212),and (12,216).To find another
point go right 3 and down 4.
30
7. < 66.7 mi/h
8. At 6 m/s, Skater 1 is 4 m/s faster than Skater 2 at
2 m/s.
9. A 100% grade has a slope of 1. It has an
inclination of 45°, so you probably could not
drive up it.You might be able to walk up it. Grades
higher than 100% are possible, but the angle of
inclination would be greater than 45°.
10. The slope of the adobe house flat roof is
approximately 0. The Connecticut roof is steeper,
with a slope of about 2, and will shed the snow.
19d. possible answers: /AFG and /FGD or
/CFG and /BGF
20. possible answers: /GFC by the Vertical
Angles Conjecture, /BGF by the Corresponding
Angles Conjecture, and /DGH, using the
Alternate Exterior Angles Conjecture
21. True. Converse:“If two polygons have the
same number of sides, then the two polygons are
congruent”; false. Counterexamples may vary;
students might draw a concave quadrilateral and a
convex quadrilateral.
22.
CHAPTER 2 REVIEW
1. poor inductive reasoning, but Diana was
probably just being funny
4. 19, 230
5. S, 36
6. 2, 5, 10, 17, 26, 37
7. 1, 2, 4, 8, 16, 32
8.
9.
568
1248
7 cm
568
4.5 cm
4.5 cm
568
1248 568
7 cm
900
930
See table below.
See table below.
n2, 302 5 900
n(n 1 1) 100(101)
15. }2}, }2} 5 5050
n(n 21)
16. }2} 5 741; therefore, n 5 39
n(n 21)
17. }2} 5 2926; therefore, n 5 77
18. n 2 2, 54 2 2 5 52
19a. possible answers: /EFC and /AFG, /AFE
and /CFG, /FGD and /BGH, or /BGF and
/HGD
19b. possible answers: /AFE and /EFC, /AFG
and /CFG, /BGF and /FGD, or /BGH and
/HGD (there are four other pairs)
19c. possible answers: /EFC and /FGD, /AFE
and /BGF, /CFG and /DGH, or /AFG and
/BGH
@#$ i RX
@#$ and SU
@#$ i VX
@#$; explanations will vary.
23. PV
24. The bisected angle measures 50° because of
AIA. So each half measures 25°. The bisector is a
transversal, so the measure of the other acute angle
in the triangle is also 25° by AIA. However, this
angle forms a linearpairwiththeanglemeasuring
165°,and 25° 1 165° Þ 180°, which contradicts the
Linear Pair Conjecture.
25. a 5 38°, b 5 38°, c 5 142°, d 5 38°, e 5 50°,
f 5 65°, g 5 106°, h 5 74°; The angle with measure e
forms a linear pair with an angle with measure 130°
because of the CorrespondingAngles Conjecture.So
e measures 50° because of the Linear Pair Conjecture.
The angle with measure f is half of the angle with
measure 130°, so f 5 65°.The angle with measure g is
congruent to the angle with measure 106° by the
CorrespondingAngles Conjecture,so g 5 106°.
12. (Chapter 2 Review)
n
1
2
3
4
5
6
…
f(n)
2
21
24
27
210
213
…
n
…
20
…
255
…
20
…
210
23n 1 5
n
1
2
3
4
5
6
…
f(n)
1
3
6
10
15
21
…
n
n(n 1 1)
}}
2
31
10.
11.
12.
13.
14.
3
LESSON 3.1
1.
B
A
D
C
F
E
2.
9. For Exercise 7, trace the triangle. For Exercise 8,
trace the segment onto three separate pieces of
patty paper. Lay them on top of each other, and
slide them around until the segments join at the
endpoints and form a triangle.
##. Copy /Q and
10. One method: Draw DU
construct nCOY > nQUD. Duplicate /DUA
at point O. Construct nOYP > nUDA.
CD
AB
U
O
A
AB 1 CD
3.
AB
EF
EF
Q
AB 1 2EF 2 CD
CD
D
C
Y
11. One construction method is to create congruent
circles that pass through each other’s center. One
side of the triangle is between the centers of the
circles; the other sides meet where the circles
intersect.
12. a 5 50°, b 5 130°, c 5 50°, d 5 130°, e 5 50°,
f 5 50°, g 5 130°, h 5 130°, k 5 155°, l 5 90°,
m 5 115°, n 5 65°
13. west
14. An isosceles triangle is a triangle that has at
least one line of reflectional symmetry.Yes, all
equilateral triangles are isosceles.
15.
4.
5. possible answer:
L
G
P
E
copy
6. m/3 5 m/1 1 m/2; possible answer:
/2
16. new coordinates: A9(0, 0), Y9(4, 0), D9(0, 2)
y
1
3
2
6
/1
Y
7. possible answer:
D'
Y'
C
–6
AC
D A' A
6
x
BC
–6
A
B
AB
8.
17. Methods will vary. It isn’t possible to draw a
second triangle with the same side lengths that is
not congruent to the first.
11 cm
8 cm
10 cm
32
LESSON 3.2
1.
A
B
2.
Q
3.
D
Edge of the paper
Original segment
4.
D
1 CD
2
1 CD
2
AB
AB
I
1 CD
2
2AB – 1 CD
2
5.
A
AB
M
L
CD
N
MN = 1 (AB + CD)
2
8. The medians all intersect in one point.
C
6. Exercises 1–5 with patty paper:
Exercise 1 This is the same as Investigation 1.
Exercise 2
##.
Step 1 Draw a segment on patty paper. Label it QD
Step 2 Fold your patty paper so that endpoints Q
and D coincide. Crease along the fold.
Step 3 Unfold and draw a line in the crease.
Step 4 Label the point of intersection A.
Step 5 Fold your patty paper so that endpoints Q
and A coincide. Crease along the fold.
Step 7 Label the point of intersection B.
Step 8 Fold your patty paper so that endpoints A
and D coincide. Crease along the fold.
Step 10 Label the point of intersection C.
N
A
M
L
B
## appears to be parallel to EF
#, and its length
9. GH
#
is half the length of EF .
F
G
D
H
E
33
C
Exercise 3 This is the same as Investigation 1.
Exercise 4
Step 1 Do Investigation 1 to get }12}CD.
#
Step 2 On a second piece of patty paper, trace AB
two times so that the two segments form a segment
of length 2AB.
Step 3 Lay the first piece of patty paper on top of
the second so that the endpoints coincide and the
shorter segment is on top of the longer segment.
Step 4 Trace the rest of the longer segment with a
different colored writing utensil. That will be the
answer.
Exercise 5
Step 1 Trace segments AB and CD so that the two
segments form a segment of length AB 1 CD.
Step 2 Fold your patty paper so that points A and
D coincide. Crease along the fold.
7. The perpendicular bisectors all intersect in one
point.
10. The quadrilateral appears to be a rhombus.
V
E
O
C
R
D
14. One way to balance it is along the median. The
two halves weigh the same.
sample figure:
C
S
I
11.
D
Ness Station
Umsar
Station
A
Any point on the perpendicular bisector of the
segment connecting the two offices would be
equidistant from the two post offices. Therefore,
any point on one side of the perpendicular bisector
would be closer to the post office on that side.
12. It is a parallelogram.
F
L
Area CDB = 158 in2
Area DAB = 158 in2
B
Ruler
15. F
16. E
17. B
18. A
19. D
20. C
21. B, C, D, E, H, I, O, X (K in some fonts, though
not this one)
22. Methods will vary.
C
T
10 cm
A
408
13. The triangles are not necessarily congruent,
but their areas are equal. A cardboard triangle
would balance on its median.
34
A
708
B
It is not possible to draw a second triangle with the
same angle and side measures that is not congruent
to the first.
LESSON 3.3
1.
6. The two folds are parallel.
P
B
P
Q
I
G
The answer depends on the angle drawn and where
P is placed.
C
2.
D
A
B
7. Fold the patty paper through the point so that
two perpendiculars coincide to see the side closest
to the point. Fold again using the perpendicular
of the side closest to the point and the third
perpendicular; compare those sides.
8. Draw a line. Mark two points on it, and label
them A and C. Construct a perpendicular at C.
# congruent to CA
#. The altitude CD
## is
Mark off CB
also the angle bisector, median, and perpendicular
bisector.
A
D
C
B
9.
U
O
B
T
O
T
3.
B
U
10.
E
Two altitudes fall outside the triangle, and one falls
inside.
4. From the point, swing arcs on the line to
construct a segment whose midpoint is that point.
Then construct the perpendicular bisector of the
segment.
L
A
B
11.
12.
Complement
of /A
/A
5. Construct perpendiculars from point Q and
# and RE
# congruent to QR
##.
point R. Mark off QS
Connect points S and E.
Q
R
S
E
35
14.
18. not congruent
6 cm
6 cm
408
408
8 cm
15.
F
8 cm
Y
I
16.
A
T
E
C
D
5 cm
F
9 cm
B
5 cm
17.
9 cm
13. (Lesson 3.3)
Rectangular pattern with triangles
Rectangle
1
2
3
4
5
6
…
Number of shaded
triangles
2
9
20
35
54
77
…
n
…
35
…
2484
(2n 2 1)(n 1 1)
36
LESSON 3.4
1.
2.
3.
4.
5.
6.
D
F
A
C
E
11.
12. The angle bisectors are perpendicular. The
sum of the measures of the linear pair is 180°. The
sum of half of each must be 90°.
z
z
z
7.
A
B
M
R
P
8.
P
A
P
E
M
S
M
M
E
U
M
O
S
9a, b.
18.
G
458
908 458
N
9c.
I
A
S
L
1358
O
T
458
10.
Altitude
Angle
bisector
Median
37
A
R
R
13. If a point is equally distant from the sides of
an angle, then it is on the bisector of an angle. This
is true for points in the same plane as the angle.
Mosaic answers: Square pattern constructions:
perpendiculars, equal segments, and midpoints;
The triangles are not identical, as the downward
ones have longer bases.
14. y 5 110°
15. m/R 5 46°
16. Angle A is the largest; m/A 5 66°,
m/B 5 64°, m/C 5 50°.
17. STOP
19.
22a. A web of lines fills most of the plane, except
a U-shaped region and a V-shaped region. (The
U-shaped region is actually bounded by a section
# were
of a parabola and straight lines. If AB
@#$
extended to AB , the U would be a complete
parabola.)
B
5.6 cm
1308
A 3.5 cm C
20.
A
A
B
6.5 cm
C
21. No, the triangles don’t look the same.
608
408
608
408
8 cm
8 cm
38
D
B C
Parabola
# and half the
22b. a line segment parallel to AB
length (The segment is actually the midsegment
of nABD.)
LESSON 3.5
1.
2.
3. Construct a segment with length z. Bisect the
segment to get length }2z}. Bisect again to get a
segment with length }4z}. Construct a square with
each side of length }4z}.
8. /1 > /S and /2 > /U by of the Alternate
Interior Angles Conjecture
9. The ratios appear to be the same.
10. /1 > /3 and /2 > /4 by the
Corresponding Angles Conjecture
11. A parallelogram
12. Using the Converse of the Parallel Lines
Conjecture, the angle bisectors are parallel:
@#$ i BC
@#$.
/DAB > /ABC, so AD
D
B
A
1z
4
13. Construct the perpendicular bisector of each
of the three segments connecting the fire stations.
1z
2
z
x
A
x
x
A
Eliminate the rays beyond where the bisectors
intersect. A point within any region will be closest
to the fire station in that region.
O
B
M
14. Z
15.
x
x
5. sample construction:
R
P
T
D
A
16.
I
R
R
608
O
E
W
T
K
R
## perpendicular to
6. Draw a line and construct ML
it. Swing an arc from point M to point G so that
MG 5 RA. From point G, swing an arc to construct
#. Finish the parallelogram by swinging an arc of
RG
length RA from R and swinging an arc of length GR
from M. There is only one possible parallelogram.
M
G
L
A
R
RC = KE = 8 cm
C
17. a 5 72°, b 5 108°, c 5 108°, d 5 108°, e 5 72°,
f 5108°, g 5 108°, h 5 72°, j 5 90°, k 5 18°, l 5 90°,
m 5 54°, n 5 62°, p 5 62°, q 5 59°, r 5 118°;
Explanations will vary. Sample explanation:
c is 108° because of the Corresponding Angles
Conjecture. Using the Vertical Angles Conjecture,
2m 5 108°, so m 5 54°. p 5 n because of the
Corresponding Angles Conjecture. Using the
Linear Pair Conjecture, n 5 62°, so p 5 62°.
Using the Linear Pair Conjecture, r 1 p 5 180°.
Because p 5 62°, r 5 118°.
7. /1 > /S, /2 > /U
39
4.
C
1. perpendicular
2. neither
3. perpendicular
4. parallel
5. possible answer: (2, 5) and (7, 11)
6. possible answer: (1, 25) and (22, 212)
7. Ordinary; no two slopes are the same, so no
# ' EM
## because their
sides are parallel, although TE
slopes are opposite reciprocals.
8. Ordinary, for the same reason as in Exercise 7—
none of the sides are quite parallel.
# i RO
#
9. trapezoid: KC
##
## 5 }61};
10a. Slope HA 5 slope ND
## 5 slope NA
# 5 26. Quadrilateral HAND
slope HD
is a rectangle because opposite sides are parallel
and adjacent sides are perpendicular.
40
## 5 midpoint AD
## 5 1}21}, 32. The
10b. Midpoint HN
diagonals of a rectangle bisect each other.
11a. Yes, the diagonals are perpendicular.
# 5 1; slope VR
# 5 21.
Slope OE
# 5 midpoint OE
# 5 (22, 4).
11b. Midpoint VR
The diagonals of OVER bisect each other.
11c. OVER appears to be a rhombus. Slope
# 5 slope RE
# 5 2}51} and slope OR
# 5 slope
OV
#
VE 5 25, so opposite sides are parallel. Also, all of
the sides appear to have the same length.
12a. Both slopes equal }21}.
12b. The segments are not parallel because they
are coincident.
12c. distinct
13. (3, 26)
LESSON 3.6
1. Sample description: Construct one of the
segments, and mark arcs of the correct length from
the endpoints. Draw sides to where those arcs meet.
M
angle at each end of that segment congruent to one
of the angles in the book. Where they meet is the
third vertex of the triangle.
5.
C
A
A
B
C
A
M
A
A
S
S
M
S
2.
O
D
B
Sample description: Construct /A and mark off
the distance AB. From B swing an arc of length BC
to intersect the other side of /A at two points.
Each gives a different triangle.
C
6.
A
y 2x
____
2
O
O
T
x
T
y
T
Sample description: Construct /O. Mark off
distances OD and OT on the sides of the angle.
Connect D and T.
I
3.
Sample description: Mark the distance y, mark
back the distance x, and bisect the remaining
length of y 2 x. Using an arc of that length, mark
arcs on the ends of segment x. The point where
they intersect is the vertex angle of the triangle.
7.
I
Y
G
Y
I
Y
#. Construct /I at
Sample description: Construct IY
I and /Y at Y. Label the intersection of the rays
point G.
4. Yes, students’ constructions must be either
larger or smaller than the triangle in the book.
Sample description: Draw an angle. Mark off equal
segments on the sides of the angle. Use a different
compass setting to draw intersecting arcs from the
ends of those segments.
8. Sample description: Draw an angle and mark
off unequal distances on the sides. At the endpoint
of the longer segment (not the angle vertex), swing
an arc with the length of the shorter segment. From
the endpoint of the shorter segment, swing an arc
the length of the longer segment. Connect the
endpoints of the segments to the intersection
points of the arcs to form a quadrilateral.
Sample description: Draw one side with a different
length than the lengths in the book. Duplicate an
41
O
D
y 2x
____
2
x
9.
12. new coordinates: E (4, 26), A (7, 0), T (1, 2)
y
–5
Sample description: Draw a segment and draw an
angle at one end of the segment. Mark off a
distance equal to that segment on the other side of
the angle. Draw an angle at that point and mark off
the same distance. Connect that point to the other
end of the original segment.
10.
E
T'
A'
–5
A
–5 T
E'
13.
Reflectional
symmetries
Rotational
symmetries
Trapezoid
0
0
Kite
1
0
Parallelogram
0
2
Rhombus
2
2
Rectangle
2
2
Figure
Sample description: Draw an angle and mark off
equal lengths on the two sides. Use that length to
determine another point that distance from the
points on the sides. Connect that point with the
two points on the side of the angle.
11. Answers will vary. The angle bisector lies
between the median and the altitude. The order of
the points is either M, R, S or S, R, M. One possible
conjecture: In a scalene obtuse triangle the angle
bisector is always between the median and the
altitude.
m/ABC = 1118
14. half a cylinder
15. 503
16.
1108
E
R
B
3.2 cm
R
A
S
Altitude
42
Median
M
Angle
bisector
1108
C
x
A
5.5 cm
1108
C
LESSON 3.7
1. incenter
Because the station needs to be equidistant from
the paths, it will need to be on each of the angle
bisectors.
2. circumcenter
3. incenter
8. Yes, any circle with a larger radius would not
fit within the triangle. To get a circle with a larger
radius tangent to two of the sides would force the
circle to pass through the third side twice.
9. No, on an obtuse triangle the circle with the
largest side of the triangle as the diameter of the
circle creates the smallest circular region that
contains the triangle. The circumscribed circle of
an acute triangle does create the smallest circular
region that contains the triangle.
Stove
Fridge
Sink
To find the point equidistant from three points,
find the circumcenter of the triangle with those
points as vertices.
5. Circumcenter. Find the perpendicular bisectors
of two of the sides of the triangle formed by the
classes. Locate the pie table where these two lines
intersect.
6.
10. For an acute triangle, the circumcenter is
inside the triangle; for an obtuse triangle, the
circumcenter is outside the triangle. The
circumcenter of a right triangle lies on the
midpoint of the hypotenuse.
11. For an acute triangle, the orthocenter is inside
the triangle; for an obtuse triangle, the orthocenter
is outside the triangle. The orthocenter of a right
triangle lies on the vertex of the right angle.
##
12. The midsegment appears parallel to side MA
and half the length.
A
S
M
7.
H
T
13. The base angles of the isosceles trapezoid
appear congruent.
A
T
O
M
43
The center of the circular sink must be equidistant
from the three counter edges, that is, the incenter of
the triangle.
4. circumcenter
14. The measure of /A is 90°. The angle inscribed
in a semicircle appears to be a right angle.
21.
Y
408 408
A
4.8 cm
6.4 cm
M
T
K
15. The two diagonals appear to be perpendicular
bisectors of each other.
A
T
E
22. construction of an angle bisector
T
16.
y
9
23. construction of a perpendicular line through a
point on a line
x+y=9
9
x
17.
Construct the incenter by bisecting the two angles
shown. Any other point on the angle bisector of the
third angle must be equidistant from the two
unfinished sides. From the incenter, make congruent
arcs that intersect the unfinished sides. The
intersection points are equidistant from the incenter.
Use two congruent arcs to find another point that
is equidistant from the two points you just
constructed. The line that connects this point and
the incenter is the angle bisector of the third angle.
18. Answers should describe the process of
discovering that the midpoints of the altitudes are
collinear for an isosceles right triangle.
19. a triangle
M
20.
6.0 cm
R
6.0 cm
608 6.0 cm 608
O
608
608
6.0 cm
6.0 cm
H
44
24. construction of a line parallel to a given line
through a point not on the line
25. construction of an equilateral triangle
26. construction of a perpendicular bisector
LESSON 3.8
1. The center of gravity is the centroid. She needs
to locate the incenter to create the largest circle
within the triangle.
2. AM 5 20; SM 5 7; TM 5 14; UM 5 8
3. BG 5 24; IG 5 12 4. RH 5 42; TE 5 45
5. The points of concurrency are the same point
for equilateral triangles because the segments are
the same.
9. circumcenter
10. The shortest chord through P is a segment
perpendicular to the diameter through P, which is
the longest chord through P.
O
P
11.
A
B'
C
6.
B
12. rule: 2n 2 2, possible answer:
H
Centroid
Incenter
7. ortho-/in-/centroid/circum-; the order changes
when the angle becomes larger than 60°. The
points become one when the triangle is equilateral.
Orthocenter
Centroid
Circumcenter
8. Start by constructing a quadrilateral, then make
a copy of it. Draw a diagonal in one, and draw a
different diagonal in the second. Find the centroid
of each of the four triangles. Construct a segment
connecting the two centroids in each quadrilateral.
Place the two quadrilaterals on top of each other
matching the congruent segments and angles.Where
the two segments connecting centroids intersect is
the centroid of the quadrilateral.
H
C
C
H
H
H
C
C
H
H
H
C
C
C
H
H
H
H
14. a 5 128°, b 5 52°, c 5 128°, d 5 128°,
e 5 52°, f 5 128°, g 5 52°, h 5 38°,
k 5 52°, m 5 38°, n 5 71°, p 5 38°
15. Construct altitudes from the two accessible
vertices to construct the orthocenter. Through the
orthocenter, construct a line perpendicular to the
southern boundary of the property. This method
will divide the property equally only if the southern
boundary is the base of an isosceles triangle.
Altitude to
missing vertex
16. 1580 greetings
C
C
D
D
M4
M2
M3
M1
A
H
C
13.
Orthocenter
Incenter
H
B A
B
45
Circumcenter
CHAPTER 3 REVIEW
1. False; a geometric construction uses a straightedge and a compass.
2. False; a diagonal connects two non-consecutive
vertices.
3. true
4. true
5. false
B
A
28.
_1 z
2
y
y
Segment
29.
5x
P
C
D
R
3x
6. False; the lines can’t be a given distance from a
segment because the segment has finite length and
the lines are infinite.
7. false
C
x
_1 z
2
4x
Q
30. m/A 5 m/D.You must first find /B.
m/B 5 180° 2 2(m/A).
2y
B
A
D
B
8. true
9. true
10. False; the orthocenter does not always lie
inside the triangle.
11. A
12. B or K 13. I
14. H
15. G
16. D
17. J
18. C
19.
20.
2y
/D
/B
/A
A
31.
21.
4x
A
Copy
B
y
22.
y
D
F
32.
23. Construct a 90° angle and bisect it twice.
24.
I
y
25. incenter
26. Dakota Davis should locate the circumcenter
of the triangular region formed by the three stones,
which is the location equidistant from the stones.
B
27.
A
46
z
C
R
x
T
33. rotational symmetry
34. neither
35. both
36. reflectional symmetry
37. D
38. A
39. C
40. B
41. False; an isosceles triangle has two congruent
sides.
42. true
43. False; any non-acute triangle is a
counterexample.
44. False; possible explanation: The orthocenter is
the point of intersection of the three altitudes.
45. true
46. False; any linear pair of angles is a
counterexample.
47. False; each side is adjacent to one congruent
side and one noncongruent side, so two consecutive
sides may not be congruent.
48. false;
58c. no
59.
Q
Q'
49. False; the measure of an arc is equal to the
measure of its central angle.
50. false; TD 5 2DR
51. False; a radius is not a chord.
52. true
53. False; inductive reasoning is the process of
observing data, recognizing patterns, and making
generalizations about those patterns.
54. paradox
55a. /2 and /6 or /3 and /5
55b. /1 and /5
55c. 138°
56. 55
58a. yes
58b. If the month has 31 days, then the month is
October.
n
f(n)
1
2
3
4
5
6
…
21
2
5
8
11
14
…
n
…
20
…
56
f(n) 5 3n 2 4
n
1
2
3
4
5
6
…
f(n)
0
3
8
15
24
35
…
n
f(n) 5
n2
…
20
…
399
21
47
62. a 5 38°, b 5 38°, c 5 142°, d 5 38°, e 5 50°,
f 5 65°, g 5 106°, h 5 74°.
Possible explanation: The angle with measure c is
congruent to an angle with measure 142° because
of the Corresponding Angles Conjecture, so
c 5 142°. The angle with measure 130° is
congruent to the bisected angle by the
Corresponding Angles Conjecture. The angle with
measure f has half the measure of the bisected
angle, so f 5 65°.
63. Triangles will vary. Check that the triangle is
scalene and that at least two angle bisectors have
been constructed.
64. m/FAD 5 30° so m/ADC 5 30°, but
its vertical angle has measure 26°. This is a
contradiction.
65. minimum: 101 regions by 100 parallel lines;
maximum: 5051 regions by 100 intersecting,
noncurrent lines
4
angles P and D can be drawn at each endpoint
using the protractor.
LESSON 4.1
Q
1. The angle measures change, but the sum
remains 180°.
2. 73°
3. 60°
4. 110°
5. 24°
6. 3 ? 360° 2 180° 5 900°
7. 3 ? 180° 2 180° 5 360°
8. 69°; 47°; 116°; 93°; 86°
9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50°
558
P
10.
/R
/M
/A
11.
/G
/L
12. First construct /E, using the method used in
Exercise 10.
408
858
7 cm
17. The third angles of the triangles also have the
same measures; are equal in measure
18. You know from the Triangle Sum Conjecture
that m/A 1 m/B 1 m/C 5 180°, and m/D 1
m/E 1 m/F 5 180°. By the transitive property,
m/A 1 m/B 1 m/C 5 m/D 1 m/E 1 m/F.
You also know that m/A 5 m/D, and m/B 5
m/E. You can substitute for m/D and m/E in the
longer equation to get m/A 1 m/B 1 m/C 5
m/A 1 m/B 1 m/F. Subtracting equal terms
from both sides, you are left with m/C 5 m/F.
19. For any triangle, the sum of the angle measures
is 180°, by the Triangle Sum Conjecture. Since the
triangle is equiangular, each angle has the same
measure, say x. So x 1 x 1 x 5 180°, and x 5 60°.
20. false
E
R
A
13.
Fold
/M /A
21. false
E
/R
/G
/L
A
R
22. false
14. From the Triangle Sum Conjecture
m/A 1 m/S 1 m/M 5 180°. Because /M is a
right angle, m/M 5 90°. By substitution,
m/A 1 m/S 1 90° 5 180°. By subtraction,
m/A 1 m/S 5 90°. So two wrongs make a right!
15. Answers will vary. See the proof on page 202.
To prove the Triangle Sum Conjecture, the Linear
Pair Conjecture and the Alternate Interior Angles
Conjecture must be accepted as true.
16. It is easier to draw nPDQ if the Triangle Sum
Conjecture is used to find that the measure of
# can be drawn to be 7 cm, and
/D is 85°. Then PD
48
D
23. false
24. true
25. eight; 100
LESSON 4.2
F
D
E
C
A
B
16.
P
M
N
K
G
H
Fold 1
Fold 3
Fold 2
Fold 4
1058 608
458
18.
19.
20.
21.
22.
23.
0
perpendicular
parallel
parallel
neither
parallelogram
40
8
16
24
32
40
24. New: (6, 23), (2, 25), (3, 0). Triangles are
congruent.
25. New: (3, 23), (23, 21), (21, 25). Triangles
are congruent.
49
1. 79°
2. 54°
3. 107.5°
4. 44°; 35 cm
5. 76°; 3.5 cm
6. 72°; 36°; 8.6 cm
7. 78°; 93 cm
8. 75 m; 81°
9. 160 in.; 126°
10. a 5 124°, b 5 56°, c 5 56°, d 5 38°, e 5 38°,
f 5 76°, g 5 66°, h 5 104°, k 5 76°, n 5 86°,
p 5 38°; Possible explanation: The angles with
measures 66° and d form a triangle with the angle
with measure e and its adjacent angle. Because d,
e, and the adjacent angle are all congruent,
3d 1 66° 5 180°. Solve to get d 5 38°. This is
also the measure of one of the base angles of the
isosceles triangle with vertex angle measure h.
Using the Isosceles Triangle Conjecture, the other
base angle measures d, so 2d 1 h 5 180°, or
76° 1 h 5 180°. Therefore, h 5 104°.
11. a 5 36°, b 5 36°, c 5 72°, d 5 108°, e 5 36°;
none
12a. Yes. Two sides are radii of a circle. Radii must
be congruent; therefore, each triangle must be
isosceles.
12b. 60°
13. NCA
14. IEC
15.
1.
3.
5.
7.
false
not a solution
not a solution
y 5 24
true
solution
x57
x 5 28
1
10. n 5 2}2}
9. x 5 4.2
2.
4.
6.
8.
11. x 5 2
12. t 5 18
2
9
13. n 5 }5}
14a. x 5 }4}
14b. x 5 }94}; The two methods produce identical
results. Multiplying by the lowest common
denominator (which is comprised of the factors of
both denominators) and then reducing common
factors (which clears the denominators on either
side) is the same as simply multiplying each numerator
by the opposite denominator (or cross multiplying).
Algebraically you could show that the two methods
are equivalent as follows:
a
c
}} 5 }}
b d
a
c
bd }b} 5 bd }d}
abd bcd
}} 5 }}
b
d
ad 5 bc
The method of “clearing fractions” results in the
method of “cross multiplying.”
12
50
12
15. You get an equation that is always false, so
there is no solution to the equation.
16. Camella is not correct. Because the equation
0 5 0 is always true, the truth of the equation does
not depend on the value of x. Therefore, x can be
any real number. Camella’s answer, x 5 0, is only
one of infinitely many solutions.
17.
2x
x
2x
If x equals the measure of the vertex angle, then
the base angles each measure 2x. Applying the
Triangle Sum Conjecture results in the following
equation:
x 1 2x 1 2x 5 180°
5x 5 180°
x 5 36°
The measure of the vertex angle is 36° and the
measure of each base angle is 72°.
LESSON 4.3
1. yes
2. no
4
5
9
3. no
5
6
12
4. yes
5. a, b, c
6. c, b, a
7. b, a, c
8. a, c, b
9. a, b, c
10. v, z, y, w, x
11. 6 , length , 102
12. By the Triangle Inequality Conjecture, the
sum of 11 cm and 25 cm should be greater than
48 cm.
13. b 5 55°, but 55° 1 130° . 180°, which is
impossible by the Triangle Sum Conjecture.
14. 135°
15. 72°
16. 72°
17. a 1 b 1 c 5 180° and x 1 c 5 180°. Subtract c
from both sides of both equations to get x 5 180 2 c
and a 1 b 5 180 2 c. Substitute a 1 b for 180 2 c
in the first equation to get x 5 a 1 b.
18. 45°
19. a 5 52°, b 5 38°, c 5 110°, d 5 35°
20. a 5 90°, b 5 68°, c 5 112°, d 5 112°, e 5 68°,
f 5 56°, g 5 124°, h 5 124°
21. By the Triangle Sum Conjecture, the third
angle must measure 36° in the small triangle, but it
measures 32° in the large triangle. These are the
same angle, so they can’t have different measures.
22. ABE
23. FNK
24. cannot be determined
51
13. Cannot be determined. SSA is not a congruence
conjecture.
14. AIN by SSS or SAS
15. Cannot be determined. Parts do not correspond.
16. SAO by SAS
17. Cannot be determined. Parts do not correspond.
18. RAY by SAS
# and PR
# is (0, 0).
19. The midpoint of SD
Therefore, nDRO > nSPO by SAS.
20. Because the LEV is marking out two triangles
that are congruent by SAS, measuring the length
of the segment leading to the finish will also
approximate the distance across the crater.
21.
22.
LESSON 4.4
1. Answers will vary. Possible answer: If three
sides of one triangle are congruent to three sides of
another triangle, then the triangles are congruent
(all corresponding angles are also congruent).
2. Answers will vary. Possible answer: The picture
statement means that if two sides of one triangle
are congruent to two sides of another triangle, and
the angles between those sides are also congruent,
then the two triangles are congruent.
If you know this:
then you also know this:
3. Answers will vary. Possible answer:
23. a 5 37°, b 5 143°, c 5 37°, d 5 58°,
e 5 37°, f 5 53°, g 5 48°, h 5 84°, k 5 96°,
m 5 26°, p 5 69°, r 5 111°, s 5 69°; Possible
explanation: The angle with measure h is the vertex
angle of an isosceles triangle with base angles
measuring 48°, so h 1 2(48) 5 180°, and h 5 84°.
The angle with measure s and the angle with
measure p are corresponding angles formed by
parallel lines, so s 5 p 5 69°.
24. 3 cm , third side , 19 cm
13
3
26a. y 5 6
b. y 5 }}
c. y 5 2}4}x 1 2
3
27. (25, 23)
4. SAS
5. SSS
7. SSS
8. SAS
9. SSS (and the Converse of the Isosceles Triangle
Conjecture)
10. yes, nABC > nADE by SAS
11. Possible answer: Boards nailed diagonally in
the corners of the gate form triangles in those
corners. Triangles are rigid, so the triangles in the
gate’s corners will increase the stability of those
corners and keep them from changing shape.
12. FLE by SSS
25. (Lesson 4.4)
Side length
1
2
3
4
5
Elbows
4
4
4
4
4
T’s
0
4
8
12
16
Crosses
0
1
4
9
16
…
4n 2 4
52
n
…
4
20
4
76
361
(n 2 1)2
LESSON 4.5
1. If two angles and the included side of one triangle
are congruent to the corresponding side and angles
of another triangle, then the triangles are congruent.
2. If two angles and a non-included side of one
triangle are congruent to the corresponding side
and angles of another triangle, then the triangles
are congruent.
If you know this:
then you also know this:
3. Answers will vary. Possible answer:
22. Construction will show a similar but larger
(or smaller) triangle constructed from a drawn
triangle by duplicating two angles on either end of a
new side that is not congruent to the corresponding
side.
23. Draw a line segment. Construct a perpendicular.
Bisect the right angle. Construct a triangle with
two congruent sides and with a vertex that
measures 135°.
24. 125
25. False. One possible counterexample is a kite.
26. None. One triangle is determined by SAS.
L
K
M
27.
28a. about 100 km southeast of San Francisco
28b. Yes. No, two towns would narrow it down
to two locations. The third circle narrows it down
to one.
ORE
GO
N
Boise
IDAH
Eureka
Sacramento
San
Francisco
Reno
NEV
O
Elko
ADA
CA
U TA H
LI
FO
RN
Las
Vegas
IA
Los Angeles
Miles
0 50 100
0
100
Kilometers
200
ARIZ
ONA
400
53
4. ASA
6. SAA
8. ASA
10. FED by SSS
11. WTA by ASA or SAA
12. SAT by SAS
13. PRN by ASA or SAS; SRE by ASA
14. Cannot be determined. Parts do not
correspond.
15. MRA by SAS
16. Cannot be determined.AAA does not guarantee
congruence.
17. WKL by ASA
18. Yes, nABC > nADE by SAA or ASA.
# 5 slope CD
## 5 3 and slope #
19. Slope AB
BC 5
1
## 5 2}3}, so AB
# ' BC
#, CD
## ' DA
##, and
slope DA
# i DA
##. nABC > nCDA by SAA.
BC
20.
21. The construction is the same as the
construction using ASA once you find the third
angle, which is used here. (Finding the third angle
is not shown.)
14. KEI by ASA
15. UTE by SAS
16.
LESSON 4.6
# > BD
# (same segment), /A > /C
1. Yes. BD
(given), and /ABD > /CBD (given), so nDBA >
# > CB
# by CPCTC.
nDBC by SAA. [ AB
## > WN
## and /C > /W (given), and
2. Yes. CN
/RNC > /ONW (vertical angles), nCNR >
## > ON
## by CPCTC.
nWON by ASA. [ RN
3. Cannot be determined. The congruent parts
lead to the ambiguous case SSA.
# > IT
#
4. Yes. /S > /I, /G > /A (given), and TS
(definition of midpoint), so nTIA > nTSG by
# > IA
# by CPCTC.
SAA. [ SG
# > FR
# and UO
## > UR
## (given), and UF
#
5. Yes. FO
# (same segment), so nFOU > nFRU by SSS.
> UF
[ /O > /R by CPCTC.
## > MA
## and ME
## > MR
## (given), and
6. Yes. MN
/M > /M (same angle), so nEMA > nRMN by
SAS. [ /E > /R by CPCTC.
# > EU
# and BU
# > ET
# (given), and
7. Yes. BT
#
#
UT > UT (same segment), so nTUB > nUTE by
SSS. [ /B > /E by CPCTC.
8. Cannot be determined. nHLF > nLHA by
## and HF
# are not corresponding sides.
ASA, but HA
9. Cannot be determined. AAA does not guarantee congruence.
10. Yes. The triangles are congruent by SAS.
11. Yes. The triangles are congruent by SAS, and
the angles are congruent by CPCTC.
# and DF
# to form nABC and nDEF.
12. Draw AC
#
#
#
# because all were drawn with
AB > CB > DE > FE
# > DF
# for the same reason.
the same radius. AC
nABC > nDEF by SSS. Therefore, /B > /E by
CPCTC.
17.
18. a 5 112°, b 5 68°, c 5 44°, d 5 44°, e 5 136°,
f 5 68°, g 5 68°, h 5 56°, k 5 68°, l 5 56°, m 5 124°;
Possible explanation: f and g are measures of base
angles of an isosceles triangle, so f 5 g. The vertex
angle measure is 44°, so subtract 44° from 180° and
divide by 2 to get f 5 68°. The angle with measure
m is the exterior angle of a triangle. Add the remote
interior angle measures 56° and 68° to get
m 5 124°.
19. ASA. The “long segment in the sand” is a
shared side of both triangles
20. (24, 1)
22. Value C is always decreasing.
23. x 5 3, y 5 10
21. (Lesson 4.6)
54
Number of sides
3
4
5
6
7
…
12
…
n
Number of struts needed
to make polygon rigid
0
1
2
3
4
…
9
…
n23
3. See flowchart below.
LESSON 4.7
1. (Lesson 4.7)
1
SE > SU
?
Given
nESM > nUSO
2
/E > /U
?
3
4
Given
5
n ? > n ?
MS > SO
? CPCTC
ASA Congruence
Conjecture
/1 > /2
? Vertical Angles Conjecture
2. (Lesson 4.7)
1
3
CI > IM
Definition
of midpoint
Given
2
I is midpoint of CM
4
I is midpoint of BL
6
IL > IB
? Definition of
midpoint
? Given
7
n ? >n ?
? CL
# > MB
##
? nCIL > nMIB
by SAS
CPCTC
5 /1 > /2
? Vertical Angles Conjecture
3. (Lesson 4.7)
1
3
NS is a median
Given
2
NS > NS
Same segment
4
S is a midpoint
Definition
of median
6
WS > SE
? SSS
Definition
of midpoint
5
nESN
nWSN > n ?
7
/E
/W > / ?
?
CPCTC
WN > NE
? Given
4. (Lesson 4.7)
1 NS is an
2
angle bisector
/1 > / ?
/2
Definition of ?
angle bisector
Given
3
/W > /E
? Given
4
5
nWNS > nENS
n ? >n ?
? SAA
6
WN > NE
? CPCTC
7 nNEW is
isosceles
? Definition of
isosceles triangle
NS > NS
? Same segment
55
12. ACK by SSS
13. a 5 72°, b 5 36°, c 5 144°, d 5 36°, e 5 144°,
f 5 18°, g 5 162°, h 5 144°, j 5 36°, k 5 54°,
m 5 126°
14. The circumcenter is equidistant from all three
vertices because it is on the perpendicular bisector
of each side. Every point on the perpendicular
bisector of a segment is equidistant from the
endpoints. Similarly, the incenter is equidistant
from all three sides because it is on the angle
bisector of each angle, and every point on an angle
bisector is equidistant from the sides of the angle.
15. ASA. The fishing pole forms the side.
“Perpendicular to the ground” forms one angle.
“Same angle on her line of sight” forms the other
angle.
2
16. }7}
17.
# > BC
#, CD
## > AD
##
6. Given: /ABC with BA
# is the angle bisector of /ABC.
Show: BD
A
D
1
2
B
C
BA > BC
CD > AD
BD > BD
Given
Given
Same segment
nABD > nCBD
SSS
/1 > /2
CPCTC
→
BD bisects /ABC
Definition of angle
bisector
7. The angle bisector does not go to the midpoint
of the opposite side in every triangle, only in an
isosceles triangle.
#, because it is across from the smallest angle
8. NE
#, which is across
in nNAE. It is shorter than AE
from the smallest angle in nLAE.
9. The triangles are congruent by SSS, so the two
central angles cannot have different measures.
10. PRN by ASA; SRE by ASA
11. Cannot be determined. Parts do not
correspond.
18.
y
4 Y
X
B
O
B' Y'4
O'
x
X'
5. (Lesson 4.7)
1
SA i NE
3
? Given
}
2
SE i NA
/3 > /4
AIA Conjecture
nESN > nANS
4 ? 1 > /2
}
? AIA Conjecture
}
? Given
}
5
SN > SN
Same segment
56
6
? >n ?
n }
}
? ASA
}
7 ? SA > NE
#
}
?
} CPCTC
This proof shows that in a parallelogram,
opposite sides are congruent.
7.
LESSON 4.8
1.
2.
3.
4.
5.
6.
6
90°; 18°
45°
See flowchart below.
1
Isosceles nABC
with AC > BC
and CD bisects
/C
2
AD > BD
AC > BC
2
Given
Same segment
D is
midpoint
of AB
AD > BD
Def. of midpoint
/ACD > /DCB
CPCTC
7
CD is angle
bisector of /ACB
Def. of angle bisector
nADC > nBDC
8
CD is altitude
of nABC
Conjecture B
(Exercise 5)
CPCTC
4
nADC > nBDC
SSS
6
Conjecture A
(Exercise 4)
4
3
CD > CD
Given
5
Given
3
1
9
CD ' AB
Def. of altitude
8. Yes. First show that the three exterior triangles
are congruent by SAS.
CD is a median
Def. of median
4. (Lesson 4.8)
1
2 ?
CD is the bisector of /C
Given
/1 > /2
Definition of
angle bisector
3
nABC is isosceles
with AC > BC
5
nADC > nBDC
? SAS
Given
4
CD > CD
Same segment
5. (Lesson 4.8)
1
nABC is isosceles with
AC > BC, and CD is
the bisector of /C
2
nADC > nBDC
4
/1 > /2
? CPCTC
Conjecture A
Given
3
/1 and /2 form
a linear pair
Definition of
linear pair
5
6
Congruent
supplementary
angles are 908
/1 and /2 are
supplementary
Linear Pair
Conjecture
/1 and /2 are
right angles
7
####' AB
####
CD
Definition of ?
perpendicular
# is an altitude 8 ?
CD
Definition of
altitude
57
9.
C
A
B
D
Drawing the vertex angle bisector as an auxiliary
segment, we have two triangles. We can show them
to be congruent by SAS, as we did in Exercise 4.
Then, /A > /B, by CPCTC. Therefore, base angles
of an isosceles triangle are congruent.
10. The proof is similar to the one on page 245,
but in reverse, and using the Converse of the
Isosceles Triangle Conjecture.
11.
5
14. y 5 2}3}x 1 16
15. 120
16. (4, 6) or (4, 0) or any point at which the
x-coordinate is either 1 or 7 and the y-coordinate
does not equal 3
17. Hugo and Duane can locate the site of the
fireworks by creating a diagram using SSS.
Fireworks
340 m/s • 3 s
= 1.02 km
340 m/s • 5 s
= 1.7 km
Hugo
1.5 km
Duane
18. Cn H2n
H H
H
308
H
H
12. a 5 128°, b 5 128°, c 5 52°, d 5 76°, e 5 104°,
f 5 104°, g 5 76°, h 5 52°, j 5 70°, k 5 70°, l 5 40°,
m 5 110°, n 5 58°
13. between 16 and 17 minutes
58
H
H
C
C
C
C
C
H
H
C
C
H
H
H
H
H
CHAPTER 4 REVIEW
P
/L
A
y
/A
/P
L
34. Construct /P. Mark off the length PB on one
ray. From point B, mark off the two segments that
intersect the other ray of /P at distance x.
S2
S1
x
x
P
z
B
59
1. Their rigidity gives strength.
2. The Triangle Sum Conjecture states that the
sum of the measures of the angles in every triangle
is 180°. Possible answers: It applies to all triangles;
many other conjectures rely on it.
3. The angle bisector of the vertex angle is also the
median and the altitude.
4. The distance between A and B is along the
segment connecting them. The distance from A
to C to B can’t be shorter than the distance from A
to B. Therefore, AC 1 CB . AB. Points A, B, and C
form a triangle. Therefore, the sum of the lengths
of any two sides is greater than the length of the
third side.
5. SSS, SAS, ASA, or SAA
6. In some cases, two different triangles can
be constructed using the same two sides and
non-included angle.
8. ZAP by SAA
9. OSU by SSS
11. APR by SAS
12. NGI by SAS
14. DCE by SAA or ASA
15. RBO or OBR by SAS
## > UT
# by
16. nAMD > nUMT by SAS, AD
CPCTC
# > AL
# by
19. nTRI > nALS by SAA, TR
CPCTC
# > KV
## by
20. nSVE > nNIK by SSS, EL
overlapping segments property
23. nLAZ > nIAR by ASA, nLRI > nIZL by
ASA, and nLRD > nIZD by ASA
24. yes. nPTS > nTPO by ASA or SAA
25. nANG is isosceles, so /A > /G. However,
the sum of m/A 1 m/N 1 m/G 5 188°. The
measures of the three angles of a triangle must sum
to 180°.
26. nROW > nNOG by ASA, implying that
## > OG
##. However, the two segments shown are
OW
not equal in measure.
27. a 5 g , e 5 d 5 b 5 f , c. Thus, c is the
longest segment, and a and g are the shortest.
28. x 5 20°
29. Yes. nTRE > nSAE by SAA, so sides are
congruent by CPCTC.
30. Yes. nFRM > nRFA by SAA. /RFM >
/FRA by CPCTC. Because base angles are
congruent, nFRD is isosceles.
31. x 5 48°
32. The legs form two triangles that are congruent
by SAS. Because alternate interior angles are
congruent by CPCTC, the seat must be parallel to
the floor.
33. Construct /P and /A to be adjacent. The
angle that forms a linear pair with the conjunction
of /P and /A is /L. Construct /A. Mark off the
length AL on one ray. Construct /L. Extend the
unconnected sides of the angles until they meet.
Label the point of intersection P.
37. Possible method: Construct an equilateral
triangle and bisect one angle to obtain 30°.
Adjacent to that angle, construct a right angle
and bisect it to obtain 45°.
38. d, a 5 b, c, e, f
36. Given three sides, only one triangle is possible;
therefore, the shelves on the right hold their shape.
The shelves on the left have no triangles and move
freely as a parallelogram.
/TMI > /RME
2
/ ? >/ ?
Vertical angles ?
## > ME
##
nTMI > nEMR
TM
3 ?
5
M is midpoint
> ?
n ? >n ?
# and IR
#
of TE
? SAS
? Definition
1 ?
of midpoint
? Given
4 ?
> ?
# > MR
##
IM
? Definition of midpoint
60
/T > /E
or
/R > /I
6
/ ? >/ ?
? CPCTC
# i RE
#
TI
7
? i ?
? Converse of
AIA Conjecture
5
13. 17
14. 15
15. the twelfth century
16. The angles of the trapezoid measure 67.5° and
112.5°; 67.5° is half the value of each angle of a regular
octagon, and 112.5° is half the value of 360° 2 135°.
LESSON 5.1
1. See table below.
2. See table below.
3. 122°
4. 136°
5. 108°; 36°
6. 108°; 106°
7. 105°; 82°
8. 120°; 38°
9. The sum of the interior angle measures of the
quadrilateral is 358°. It should be 360°.
10. The measures of the interior angles shown sum
to 554°. However, the figure is a pentagon, so the
measures of its interior angles should sum to 540°.
11. 18
12. a 5 116°, b 5 64°, c 5 90°, d 5 82°, e 5 99°,
f 5 88°, g 5 150°, h 5 56°, j 5 106°, k 5 74°,
m 5 136°, n 5 118°, p 5 99°; Possible explanation:
The sum of the angles of a quadrilateral is 360°, so
a 1 b 1 98° 1 d 5 360°. Substituting 116° for a
and 64° for b gives d 5 82°. Using the larger
quadrilateral, e 1 p 1 64° 1 98° 5 360°.
Substituting e for p, the equation simplifies to
2e 5 198°, so e 5 99°. The sum of the angles of a
pentagon is 540°, so e 1 p 1 f 1 138° 1 116° 5
540°. Substituting 99° for e and p gives f 5 88°.
1358
67.58
1. (Lesson 5.1)
Number of sides of polygon
Sum of measures of angles
7
8
9
10
11
20
55
100
900° 1080° 1260° 1440° 1620° 3240° 9540° 17640°
2. (Lesson 5.1)
Number of sides of
equiangular polygon
Measure of each angle
of equiangular polygon
5
6
7
8
9
10
12
16
100
108° 120° 128}74}° 135° 140° 144° 150° 157}1}° 176}2}°
2
5
61
17. Answers will vary; see the answer for
Developing Proof on page 259. Using the Triangle
Sum Conjecture, a 1 b 1 j 1 c 1 d 1 k 1 e 1 f 1
l 1 g 1 h 1 i 5 4(180°), or 720°. The four angles in
the center sum to 360°, so j 1 k 1 l 1 i 5 360°.
Subtract to get a 1 b 1 c 1 d 1 e 1 f 1 g 1
h 5 360°.
18. x 5 120°
19. The segments joining the opposite midpoints
of a quadrilateral always bisect each other.
20. D
21. Counterexample: The base angles of an
isosceles right triangle measure 45°; thus they are
complementary.
LESSON 5.2
360°
72°; 60°
15
43
a 5 108°
1
6. b 5 45}3}°
3
5
7. c 5 51}7}°, d 5 115}7}°
8. e 5 72°, f 5 45°, g 5 117°, h 5 126°
9. a 5 30°, b 5 30°, c 5 106°, d 5 136°
10. a 5 162°, b 5 83°, c 5 102°, d 5 39°, e 5 129°,
f 5 51°, g 5 55°, h 5 97°, k 5 83°
12. Yes. The maximum is three. The minimum is
zero. A polygon might have no acute interior
angles.
1.
2.
3.
4.
5.
13. Answers will vary. Possible proof using
the diagram on the left: a 1 b 1 i 5 180°, c 1 d 1
h 5 180°, and e 1 f 1 g 5 180° by the Triangle
Sum Conjecture. a 1 b 1 c 1 d 1 e 1 f 1 g 1
h 1 i 5 540° by the addition property of equality.
Therefore, the sum of the measures of the angles of
a pentagon is 540°. To use the other diagram,
students must remember to subtract 360° to
account for angle measures k through o.
14. regular polygons: equilateral triangle and
regular dodecagon; angle measures: 60°, 150°,
and 150°
15. regular polygons: square, regular hexagon, and
regular dodecagon; angle measures: 90°, 120°,
and 150°
## > CR
# by
16. Yes. nRAC > nDCA by SAS. AD
CPCTC.
17. Yes. nDAT > nRAT by SSS. /D > /R by
CPCTC.
11. (Lesson 5.2)
1
a 1 b 5 1808
?
Linear Pair Conjecture }
2
c 1 d 5 1808
?
3
e 1 f 5 1808
?
62
4
540°
?8
a+b+c+d+e+f= }
Addition property
of equality
180°
5
?8
a+c+e=}
?
} Triangle Sum Conjecture
6
360°
?8
b+d+f=}
Subtraction property
of equality
LESSON 5.3
W
1. 64 cm
2. 21°; 146°
3. 52°; 128°
4. 15 cm
5. 72°; 61°
6. 99°; 38 cm
7. w 5 120°, x 5 45°, y 5 30°
8. w 5 1.6 cm, x 5 48°, y 5 42°
10. Answers may vary. This proof uses the Kite
Angle Bisector Conjecture.
S
2
1
N
X
H
## is the other base. /S and /H are a pair of
OW
base angles. /O and /W are a pair of base angles.
## > HO
##.
SW
14. Only one kite is possible
F
because three sides determine
N
B
a triangle.
E
15.
S
H
Y
B
O
I
E
16. infinitely many, possible construction:
B
E
Z
Q
N
17. 80°, 80°, 100°, 100°
18. Because ABCD is an isosceles trapezoid, /A
# > BH
##
> /B. nAGF > nBHE by SAA. Thus, AG
by CPCTC.
19. a 5 80°, b 5 20°, c 5 160°, d 5 20°, e 5 80°,
f 5 80°, g 5 110°, h 5 70°, m 5 110°, n 5 100°;
Possible explanation: Because d forms a linear pair
with e and its congruent adjacent angle,d 1 2e 5
180°.Substituting d 5 20° gives 2e 5 160°,so e 5 80°.
Using theVerticalAngles Conjecture and d 5 20°, the
unlabeled angle in the small right triangle measures
20°, which means h 5 70°. Because g and h are a
linear pair, they are supplementary, so g 5 110°.
K
E
O
I
U
#. /Q and /U are a pair of base
The other base is ZI
angles. /Z and /I are a pair of base angles.
9. (Lesson 5.3)
1 BE > BY
Given
2 EN >YN
Given
# > BN
#
BN
3
?
? >}
}
Same segment
nBYN
4
?
nBEN > }
? Congruence
}
shortcut
SSS
/2
5 /1 > ? and
}
? /4
/3 > }
?
} CPCTC
# bisects /B, BN
# bisects /N
BN
6
?
? and }
}
Definition of
angle bisector
63
Given: Kite BENY with vertex angles /B and /N
# is the perpendicular bisector
Show: Diagonal BN
#.
of diagonal YE
# > BY
#. From the
From the definition of kite, BE
#>
Kite Angle Bisector Conjecture, /1 > /2. BX
# because they are the same segment. By SAS,
BX
# > XE
#.
nBXY > nBXE. So by CPCTC, XY
Because /YXB and /EXB form a linear pair, they
are supplementary, so m/YXB 1 m/EXB 5
180°. By CPCTC, /YXB > /EXB, or m/YXB 5
m/EXB, so by substitution, 2m/YXB 5 180°, or
m/YXB 5 90°. So m/YXB 5 m/EXB 5 90°.
# > XE
# and /YXB and /EXB are right
Because XY
# is the perpendicular bisector of YE
#.
angles, BN
I
11. possible answer: /E > /I
T
W
LESSON 5.4
1. three; one
2. 28
3. 60°; 140°
4. 65°
5. 23
6. 129°; 73°; 42 cm
7. 35
9. Parallelogram. Draw a diagonal of the original
quadrilateral. The diagonal forms two tri-angles.
Each of the two midsegments is parallel to the
diagonal, and thus the midsegments are parallel to
each other. Now draw the other diagonal of the
original quadrilateral. By the same reasoning, the
second pair of midsegments is parallel. Therefore,
the quadrilateral formed by joining the midpoints is
a parallelogram.
10. The length of the edge of the top base
measures 30 m. We know this by the Trapezoid
Midsegment Conjecture.
11. Ladie drives a stake into the ground to create a
triangle for which the trees are the other two
vertices. She finds the midpoint from the stake to
each tree. The distance between these midpoints is
half the distance between the trees.
13. If a quadrilateral is a kite, then exactly one
diagonal bisects a pair of opposite angles. Both the
original and converse statements are true.
14. a 5 54°, b 5 72°, c 5 108°, d 5 72°, e 5 162°,
f 5 18°, g 5 81°, h 5 49.5°, i 5 130.5°, k 5 49.5°,
m 5 162°, n 5 99°; Possible explanation: The third
angle of the triangle containing f and g measures
81°, so using the Vertical Angles Conjecture, the
vertex angle of the triangle containing h also
measures 81°. Subtract 81° from 180° and divide by
2 to get h 5 49.5°. The other base angle must also
measure 49.5°. By the Corresponding Angles
Conjecture, k 5 49.5°.
15. (3, 8)
16. (0, 28)
17. coordinates: E(2, 3.5), Z(6, 5); the slope of
# 5 }83}, and the slope of YT
# 5 }83}
EZ
18.
R
F
N
Cabin
K
There is only one kite, but more than one way to
construct it.
12. Explanations will vary.
80
60 cm
40
60
8. (Lesson 5.4)
1 nFOA with
3
midsegment LN
Given
?
Triangle Midsegment Conjecture
2 nIOA with
midsegment RD
Given
64
LN i OA
# i RD
##
OA
4
?
Triangle Midsegment
Conjecture
# i RD
##
LN
5
?
Two lines parallel to the
same line are parallel
10.
LESSON 5.5
1.
2.
3.
4.
5.
6.
7.
34 cm; 27 cm
132°; 48°
16 in.; 14 in.
63 m
80
63°; 78°
Y
V
b
T
Y
V
c
S
L
A
11. (b 2 a, c)
8.
R
a
O
D
b
c
d
P
P
9.
R
14. The parallelogram linkage is used for the
sewing box so that the drawers remain parallel to
each other (and to the ground) so that the contents
cannot fall out.
Y
Vh
15. a 5 135°, b 5 90°
16. a 5 120°, b 5 108°, c 5 90°, d 5 42°, e 5 69°
Yw
V
13. (Lesson 5.5)
1 LEAN is a
parallelogram
?
Given }
2
EA i LN
?
} Definition of
parallelogram
3 /AEN > /LNE
AIA Conjecture
/EAL > /NLA
4
?
}
AIA Conjecture
# > LN
#
AE
5 ?
}
7
nAET > nLNT
6
ET > NT
# and LA
# bisect
EN
each other
CPCTC
9
?
}
ASA
# > LT
#
AT
8
?
}
CPCTC
?
}
Definition of
segment bisector
Opposite sides
congruent
65
b d
c a
17. x 5 104°, y 5 98°. The quadrilaterals on the
left and right sides are kites. Nonvertex angles are
congruent. The quadrilateral at the bottom is an
isosceles trapezoid. Base angles are congruent, and
consecutive angles between the bases are
supplementary.
18. a 5 84°, b 5 96°
19. No. The congruent angles and side do not
correspond.
20.
21. Parallelogram. Because the triangles are
congruent by SAS, /1 > /2. So, the segments are
66
parallel. Use a similar argument to show that the
other pair of opposite sides is parallel.
1
2
22. Kite or dart. Radii of the same circle are
congruent. If the circles have equal radii, a rhombus
is formed.
1.
y
(0, 1)
(1, –1)
2.
x
y
(3, 8)
(0, 4)
x
3.
y
4. y 5 2x 1 2
6
74
}}
5. y 5 2}1}
3 x 1 13
6. y 5 x 1 1
7. y 5 23x 1 5
2
8
8. y 5 }5}x 2 }5}
9. y 5 80 1 4x
10. y 5 23x 1 26
1
11. y 5 2}4}x 2 3
6
12. y 5 }5}x
13. y 5 x 1 1
2
43
14. y 5 2}9}x 1 }9}
(2, 9)
(0, 6)
x
67
LESSON 5.6
17.
E
V
L
O
1. Sometimes true; it is true only if the
parallelogram is a rectangle.
false
true
18. Constructions will vary.
2. Always true; by the definition of rectangle, all
the angles are congruent. By the Quadrilateral Sum
Conjecture and division, they all measure 90°, so
any two angles in a rectangle, including consecutive
angles, are supplementary.
3. Always true by the Rectangle Diagonals
Conjecture.
4. Sometimes true; it is true only if the rectangle is
a square.
A
B
B
E
false
5. Always true by the Square Diagonals Conjecture.
6. Sometimes true; it is true only if the rhombus is
equiangular.
false
7. Always true; all squares fit the definition of
rectangle.
8. Always true; all sides of a square are congruent
and form right angles, so the sides become the legs
of the isosceles right triangle and the diagonal is
the hypotenuse.
9. Always true by the Parallelogram Opposite
Angles Conjecture.
10. Sometimes true; it is true only if the
parallelogram is a rectangle. Consecutive angles of
a parallelogram are always supplementary, but are
congruent only if they are right angles.
11. 20
12. 37°
13. 45°, 90°
14. DIAM is not a rhombus because it is not
equilateral and opposite sides are not parallel.
15. BOXY is a rectangle because its adjacent sides
are perpendicular.
16. Yes. TILE is a rhombus, and a rhombus is a
parallelogram.
68
E
E
I
S
20. Converse: If the diagonals of a quadrilateral
are congruent and bisect each other, then the
quadrilateral is a rectangle.
Given: Quadrilateral ABCD with diagonals
# > BD
#. AC
# and BD
# bisect each other
AC
Show: ABCD is a rectangle
A
8
7
true
K
19. one possible construction:
P
true
A
K
D
1
2
E
6
5
B
3
4
C
Because the diagonals are congruent and bisect
# > BE
# > DE
# > EC
#. Using the
each other, AE
Vertical Angles Conjecture, /AEB > /CED and
/BEC > /DEA. So nAEB > nCED and nAED
> nCEB by SAS. Using the Isosceles Triangle
Conjecture and CPCTC, /1 > /2 > /5 > /6,
and /3 > /4 > /7 > /8. Each angle of the
quadrilateral is the sum of two angles, one from
each set, so for example, m/DAB 5 m/1 1 m/8.
By the addition property of equality, m/1 1
m/8 5 m/2 1 m/3 5 m/5 1 m/4 5 m/6 1
m/7. So m/DAB 5 m/ABC 5 m/BCD 5
m/CDA. So the quadrilateral is equiangular. Using
# i CD
#.
/1 > /5 and the Converse of AIA, AB
Using /3 > /7 and the Converse of AIA,
# i AD
##. Therefore ABCD is an equiangular
BC
parallelogram, so it is a rectangle.
21. If the diagonals are congruent and bisect each
other, then the room is rectangular (converse of the
Rectangle Diagonals Conjecture).
90°. Because /4 and /5 form a linear pair,
m/4 1 m/5 5 180°. Substitute 90° for m/4 and
solve to get m/5 5 90°. By definition of congruent
angles, /5 > /3, and /5 and /3 are alternate
## i BC
# by the Converse of the
interior angles, so AD
Parallel Lines Conjecture. Similarly, /1 and /5 are
# i CD
# by the
congruent corresponding angles, so AB
Converse of the Parallel Lines Conjecture. Thus,
ABCD is a parallelogram by the definition of
parallelogram. Because it is an equiangular
parallelogram, ABCD is a rectangle.
28. a 5 54°, b 5 36°, c 5 72°, d 5 108°, e 5 36°,
f 5 144°, g 5 18°, h 5 48°, j 5 48°, k 5 84°
29. possible answers: (1, 0); (0, 1); (21, 2); (22, 3)
8
86
30. y 5 }9}x 1 }9} or 8x 2 9y 5 286
12
7
}}
31. y 5 2}1}x
0 2 5 or 7x 1 10y 5 224
32. velocity 5 1.8 mi/h; angle of path 5 106.1°
clockwise from the north
22. The platform stays parallel to the floor
because opposite sides of a rectangle are parallel
(a rectangle is a parallelogram).
23. The crosswalks form a parallelogram: The
streets are of different widths, so the crosswalks are
of different lengths. The streets would have to meet
at right angles for the crosswalks to form a rectangle.
The corners would have to be right angles and the
streets would also have to be of the same width for
the crosswalk to form a square.
24. Place one side of the ruler along one side of
the angle. Draw a line with the other side of the
ruler. Repeat with the other side of the angle. Draw
a line from the vertex of the angle to the point
where the two lines meet.
25. Rotate your ruler so that each endpoint of the
segment barely shows on each side of the ruler.
Draw the parallel lines on each side of your ruler.
Now rotate your ruler the other way and repeat the
process to get a rhombus. The original segment is
one diagonal of the rhombus. The other diagonal
will be the perpendicular bisector of the original
segment.
27. Yes, it is true for rectangles.
Given: /1 > /2 > /3 > /4
Show: ABCD is a rectangle
By the Quadrilateral Sum Conjecture, m/1 1
m/2 1 m/3 1 m/4 5 360°. It is given that all
four angles are congruent, so each angle measures
2 mi/h
608
1.5 mi/h
26. (Lesson 5.6)
1
QU > AD
Given
2
QD > AU
? Given
}
3
4
nQUD > nADU
? SSS
}
5
/1 > /2
/3 > /4
? CPCTC
}
DU > DU
Same segment
8
QU > UA > AD > DQ
Given
6
7 QUAD is a
QU i AD
QD i AU
parallelogram
Converse of the
Parallel Lines
Conjecture
Definition of
parallelogram
9 QUAD is a
rhombus
?
Definition of }
rhombus
69
5. parallelogram
6. sample flowchart proof:
LESSON 5.7
1. See flowchart below. 2. See flowchart below.
2
5
4. 1 SP i OA
nSOP ! nAPO
SP > OA
Given
Given
3
2
IY > GO
Opposite sides of
rectangle are congruent
SAS
/1 > /2
6
AIA Conjecture
4
1
3
/YOG > /OYI
Definition of rectangle
4
YO > OY
Same segment
/YOG > nOYI
SAS
/3 > /4
CPCTC
5
PO > PO
7
Same segment
PA i SO
CPCTC
Converse of
AIA Conjecture
8
YG > IO
SOAP is a parallelogram
Definition of parallelogram
1. (Lesson 5.7)
2
4 /3 > /4
SO i KA
AIA
Definition of
parallelogram
#
SK
1
3
SOAK is a
parallelogram
Given
5
?
OA i }
/1 > /2
? Definition of
? AIA
}
}
parallelogram
6
7
?
nSOA > }
nAKS
? ASA
}
#
? SA
SA > }
? Same segment
}
2. (Lesson 5.7)
1
Parallelogram BATH
with diagonal BT
Given
4
Parallelogram BATH
with diagonal HA
? Given
}
2
nBAT > nTHB
Conjecture proved in
Exercise 1
nTHA
5
?
nBAH = n }
?
}
Conjecture proved
in Exercise 1
3
/BAT > /THB
CPCTC
/ATH
?
/HBA > / }
?
CPCTC }
6
3. (Lesson 5.7)
## > RT
#
WA
1
2
3
70
?
? >}
}
?
} Given
## > AT
#
WR
?
? >}
}
?
} Given
## > WT
##
WT
nWRT > nTAW
4
?
? >}
}
?
} Same segment
?
? >n}
n}
?
} SSS
/2
5 /1 > / ?
}
# i WA
##
RT
6
? CPCTC
}
/3
7 /4 > / ?
}
? CPCTC
}
? i ?
}
}
?
}
## i TA
#
RW
8
by Converse of the Parallel
Lines Conjecture
9 WATR is a
parallelogram
? Definition of
?
?
}
}i }
parallelogram
?
Converse
of
the
}
Parallel Lines Conjecture
7.
1 BEAR is a parallelogram
Given
2
3 BR
### > EA
###
### > AB
###
RE
Given
4 AR
### > EB
###
Parallelogram Opposite
Sides Conjecture
5
Parallelogram Opposite
Sides Conjecture
nEBR > nARB > nRAE > nBEA
SSS
6
/EBR > /ARB > /RAE > /BEA
CPCTC
7
BEAR is a rectangle
Definition of rectangle
9
6
12
3
6 inches
# is parallel to ZT
#, corresponding
8. Because AR
/3 and /2 are congruent. Opposite sides of
parallelogram ZART are congruent so AR 5 TZ.
Because the trapezoid is isosceles, AR 5 PT, and
substituting gives ZT 5 PT making nPTZ
isosceles and /1 and /2 congruent. By
substitution, /1 and /3 are congruent.
9. See sample flowchart below.
10. If the fabric is pulled along the warp or the weft,
nothing happens. However, if the fabric is pulled
along the bias, it can be stretched because the
rectangles are pulled into parallelograms.
11. 30° angles in 4-pointed star, 30° angles in
6-pointed star; yes
12. He should measure the alternate interior
angles to see whether they’re congruent.If they are,
the edges are parallel.
@#$: y 5 22x 2 3; QI
@#$: y 5 }21}x 1 2
13. ES
14. (12, 7)
1
15. }3}
16.
9. (Lesson 5.7)
2
#### > TH
####
GR
Given
1
Isosceles trapezoid
GTHR
3
##### > GT
####
GT
Same segment
Given
4
5
nRGT ! nHTG
SAS
6
##### > TR
####
GH
CPCTC
/RGT !/HTG
Isosceles Trapezoid
Conjecture
71
distance between the two points is twice the length
of the midsegment.
7. x 5 10°, y 5 40°
8. x 5 60 cm
9. a 5 116°, c 5 64° 10. 100
11. x 5 38 cm
12. y 5 34 cm, z 5 51 cm
14. a 5 72°, b 5 108°
15. a 5 120°, b 5 60°, c 5 60°, d 5 120°, e 5 60°,
f 5 30°, g 5 108°, m 5 24°, p 5 84°; Possible
explanation: Because c 5 60°, the angle that forms
a linear pair with e and its congruent adjacent
angle measures 60°. So 60° 1 2e 5 180°, and
e 5 60°. The triangle containing f has a 60° angle.
The other angle is a right angle because it forms a
linear pair with a right angle. So f 5 30° by the
Triangle Sum Conjecture. Because g is an interior
angle in an equiangular pentagon, divide 540° by 5
to get g 5 108°.
16. 15 stones
17. (1, 0)
18. When the swing is motionless, the seat, the bar
at the top, and the chains form a rectangle. When
you swing left to right, the rectangle changes to a
parallelogram. The opposite sides stay equal in
length, so they stay parallel. The seat and the bar
at the top are also parallel to the ground.
19. a = 60°, b = 120°
CHAPTER 5 REVIEW
1. 360° divided by the number of sides
2. Sample answers: Using an interior angle, set the
interior angle measure formula equal to the angle
and solve for n. Using an exterior angle, divide into
360°. Or find the interior angle measure and go
from there.
3. Trace both sides of the ruler as shown at right.
4. Make a rhombus using the double-edged
straightedge, and draw a diagonal connecting the
angle vertex to the opposite vertex.
5. Sample answer: Measure the diagonals with
string to see if they are congruent and bisect each
other.
6. Sample answer: Draw a third point and connect
it with each of the two points to form two sides of a
triangle. Find the midpoints of the two sides and
connect them to construct the midsegment. The
Opposite sides
are parallel
Opposite sides
are congruent
Opposite angles
are congruent
Isosceles
trapezoid
Parallelogram
Rhombus
Rectangle
Square
No
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
Diagonals bisect
each other
Diagonals are
perpendicular
No
No
Yes
Yes
Yes
Yes
Yes
No
No
Yes
No
Yes
Diagonals are
congruent
No
Yes
No
No
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
Yes
Yes
No
Exactly one line
of symmetry
Exactly two lines
of symmetry
72
Kite
20.
Resultant
vector
R
x
F
Y
900 km/h
x
F
z
R
Y
D
z
50 km/h
D
Speed: < 901.4 km/h. Direction: slightly west of
north. Figure is approximate.
E
21.
R
S
25.
26.
27.
28.
1x
2
1x
2
Q
20 sides
12 cm
possible answer:
A
B
4
R
R
x
y
C
Given: Parallelogram ABCD
# > CD
# and AD
# > CB
#
Show: AB
L
23.
N
F
L
2
D
Y
Y
F
1
3
E
x
L
z
P
2
1 DENI is
a rhombus
? Given
}
3
#
DI
?
DE > }
?
}
Definition of rhombus
#
NI
?
NE > }
?
Definition of }
rhombus
/2, /4
nDEN > nDIN
5 n ? >n ?
}
}
? SSS
}
##
DN
4
?
DN > }
? Same segment
}
6 /1 > / ?
}
?
/3 > / }
?
} CPCTC
7
DN bisects
/IDE and /INE
? Definition of
}
angle bisector
2
AD || CB
4 /1 > /3
5
1 ABCD is a parallelogram
Given
BD > BD
Same segment
3
AB || CD
Definition of
parallelogram
8
6 /2 > /4
AIA
AB > CD
CPCTC
AIA
Definition of
parallelogram
7 nABD > nCDB
ASA
9
AD > CB
CPCTC
73
6
LESSON 6.1
1.
3.
5.
6.
50°
30°
76 in.
possible answer:
2. 55°
4. 105°
11. Construct a line and label a point T on it. On
the line, mark off two points, L and M, each on
the same side of T at distances r and t from T,
respectively. Construct circle L with radius r.
Construct circle M with radius t.
r
L
M
T
t
Target
7. Possible answer: The perpendicular to the
tangent passes through the center of the circle. Use
the T-square to find two diameters of the Frisbee.
The intersection of these two lines is the center.
#$. Construct a line through point T
8. Construct OT
#$.
perpendicular to OT
r
T
O
#$, OY
#$, and OZ
#$. Construct tangents
9. Construct OX
through points X, Y, and Z.
Z
O
t
Y
X
10. Construct tangent circles M and N. Construct
equilateral nMNP. Construct circle P with radius s.
M
s
s
N
P
74
12. Start with the construction from Exercise 11.
@#$, mark off length TK so that TK 5 s
On line LM
and K is on the opposite side of T from L and M.
Construct circle K with radius s.
r
L
M
T
t
K
s
13. Sample answer: If the three points do not lie
on the same semicircle, the tangents form a circumscribed triangle. If two of the three points are
on opposite sides of a diameter, the tangent lines to
those two points are parallel. If the points lie on the
same semicircle, they form a triangle outside the
circle, with one side touching (called an exscribed
triangle).
14. sample answer: internally tangent: wheels on a
roller-coaster car in a loop, one bubble inside
another; externally tangent: touching coins, a
snowman, a computer mouse ball and its roller
balls
15. Constructions will vary.
16. 360° 2 30° 2 90° 2 90° 5 150° and
150°
#%
}} 5 41.6
360°
17. Angles A and B must be right angles, but this
would make the sum of the angle measures in the
quadrilateral shown greater than 360°.
18a. rhombus
18b. rectangle
18c. kite
18d. parallelogram
19. 78°
20. 9.7 km
21. x 1 55° 1 55° 5 180° and 40° 1 y 1
y 5 180°, so x 5 y 5 70°
11
22. }2}
1
LESSON 6.2
5 cm
5 cm
15. M(24, 3), N(24, 23), O(4, 23)
16. The center of the circle is the circumcenter of
the triangle. Possible construction:
17. possible construction:
O
18. < 13.8 cm
19. They can draw two chords and locate the
intersection of their perpendicular bisectors. The
radius is just over 5 km.
20. (Lesson 6.2)
1
AB > CD
?
} Given
2
AO > CO
All radii of a circle
are congruent
3
nAOB > nCOD
4
?
? >n }
n }
? SSS Congruence
}
Conjecture
5
/COD
?
/AOB > / }
? CPCTC
}
BO > DO
? All radii of a circle are congruent
}
75
1. 165°, definition of measure of an arc
2. 84°, Chord Arcs Conj.
3. 70°, Chord Central Angles Conj.
4. 8 cm, Chord Distance to Center Conj.
X 5 68°; m/B 5 34° (Because nOBC is
5. mAC
isosceles, m/B 5 m/C, m/B 1 m/C 5 68°,
and therefore m/B 5 34°.)
6. w 5 115°, x 5 115°, y 5 65°; Chord Arcs
Conjecture
7. 20 cm, Perpendicular to a Chord Conj.
8. w 5 110°, x 5 48°, y 5 82°, z 5 120°; definition
of arc measure
9. x 5 96°, Chord Arcs Conjecture; y 5 96°,
Chord Central Angles Conjecture; z 5 42°,
Isosceles Triangle Conjecture and Triangle Sum
Conjecture.
10. x 5 66°, y 5 48°, z 5 66°; Corresponding
Angles Conjecture, Isosceles Triangle Conjecture,
Linear Pair Conjecture
11. The length of the chord is greater than the
length of the diameter.
12. The perpendicular bisector of the segment
does not pass through the center of the circle.
13. The longer chord is closer to the center; the
longest chord, which is the diameter, passes
through the center.
14. The central angle of the smaller circle is larger,
because the chord is closer to the center.
1
21. y 5 }7} x; (0, 0) is a point on this line.
22a. true
C
A
X
B
D
# is the perpendicular
Possible explanation: If AB
##, then every point on AB
# is
bisector of CD
equidistant from endpoints C and D. Therefore
# > AD
## and BC
# > BD
#. Because CD
## is not the
AC
#, C is not equidistant
perpendicular bisector of AB
from A and B. Likewise, D is not equidistant from
# and BC
# are not congruent, and
A and B. So, AC
## and BD
# are not congruent. Thus ACBD
AD
has exactly two pairs of consecutive congruent
sides, so it is a kite.
22b. false, isosceles trapezoid
22c. false, rectangle
23. 140°, 160°, 60°; 180° minus the measure of
the intercepted arc
76
24. Using the Tangent Conjecture, /2 and /4
are right angles, so m/2 1 m/4 5 180°.
According to the Quadrilateral Sum Conjecture,
m/1 1 m/2 1 m/3 1 m/4 5 360°, so by the
Subtraction Property of Equality, m/1 1
m/3 5 180°, or m/1 5 180° 2 m/3. The
measure of a central angle equals the measure of
X.
its arc, so by substitution, m/1 5 180° 2 mAB
25. the circumcenter of the triangle formed by
the three light switches
26. 7
27. Station Beta is closer.
Downed aircraft
N
4.6 mi
488
7.2 mi
728
388
Alpha
28. D
Beta
3128
8.2 mi
LESSON 6.3
22. a 5 108°; b 5 72°; c 5 36°; d 5 108°;
e 5 108°; f 5 72°; g 5 108°; h 5 90°; l 5 36°;
m 5 18°; n 5 54°; p 5 36°
23. (22, 21); Possible method: Plot the
three points. Construct the midpoint and the
perpendicular bisector of the segments connecting
two different pairs of points. The center is the point
of intersection of the two lines. To check, construct
the circle through the three given points.
25. Start with an equilateral triangle whose vertices
are the centers of the three congruent circles. Then
locate the incenter/circumcenter/orthocenter/
centroid (all the same point because the triangle is
equilateral) to find the center of the larger circle. To
find the radius, construct a segment from the
incenter of the triangle through the vertex of the
triangle to a point on the circle.
It works on acute and right triangles.
20. The camera can be placed anywhere on the
major arc (measuring 268°) of a circle such that the
row of students is a chord intersecting the circle to
form a minor arc measuring 92°. This illustrates
the conjecture that inscribed angles that intercept
the same arc are congruent (Inscribed Angles
Intercepting Arcs Conjecture).
21. two congruent externally tangent circles with
half the diameter of the original circle
26. m/A 5 60°, m/B 5 36°, m/C 5 90°;
60° 1 36° 1 90° Þ 180°
24. (Lesson 6.3)
1 OR ' CD
2
/ORC and /ORD
are right angles
? Definition of
} perpendicular lines
3
6
nOCD is isosceles
7 /C > / ? D
}
?
} Given
5
OC > OD
?
}
All radii of
a circle are
congruent
?
} Definition of
isosceles triangle
m/ORC ! 908
m/ORD ! 908
? Definition of
} right angle
?
}
7. Isosceles Triangle
Conjecture
4
m/ORC ! m/ORD
(/ORC > /ORD)
Substitution
?
}
property
8 nOCR > n ?
}
?
}
nODR
SAA
9 CR > ? DR
} #
?
}
CPCTC
10 OR bisects CD
? Definition of bisect
}
77
1. 65°
2. 30°
3. 70°
4. 50°
5. 140°, 42° 6. 90°, 100° 7. 50°
8. 148°
9. 44°
10. 142°
11. 120°, 60°
12. 140°, 111° 13. 71°, 41°
14. 180°
15. 75°
16. The two inscribed angles intercept the same
arc, so they should be congruent.
17. /BFE > /DFA (Vertical Angles Conjecture).
/BGD > /FHD (all right angles congruent).
Therefore, /B > /D (Third Angle Conjecture).
X, m/D 5 }1} mEC
X, AC
X > EC
X
m/B 5 }21} mAC
2
18. Possible answer: Place the corner so that it is
an inscribed angle. Trace the inscribed angle. Use
the side of the paper to construct the hypotenuse
of the right triangle (which is the diameter). Repeat
the process. The place where the two diameters
intersect is the center.
LESSON 6.4
X5 m/ABD by the
1. Proof: m/ACD 5 }21}mAD
Inscribed Angle Conjecture. [ /ACD > /ABD.
2. Proof: By the Inscribed Angle Conjecture,
X
m/ACB 5 }21}mAD
B 5 }21}(180°) 5 90°. [ /ACB is a
right angle.
3. Proof: By the Inscribed Angle Conjecture,
X and m/L 5 }1}mYCI
X 5 }1}(360° 2
m/C 5 }21}mYLI
2
2
X
1 X
mYLI ) 5 180° 2 }2}mYLI 5 180° 2 m/C. [ /L
and /C are supplementary. (A similar proof can be
used to show that /I and /Y are supplementary.)
X 5 2m/2 5
4. Proof: /1 > /2 by AIA. mBC
X
2m/1 5 mAD by the Inscribed Angle Conjecture.
X > AD
X.
[ BC
5. True. Opposite angles of a parallelogram are
congruent. If it is inscribed in a circle, the opposite
angles are also supplementary. So they are right
angles, and the parallelogram is indeed
equiangular, or a rectangle.
# and DB
#. Diagonal RG
#
6. Construct diagonals RG
# by the Kite
is the perpendicular bisector of BD
Diagonal Bisector Conjecture. Because kite BRDG
# and RG
# are chords of
is inscribed in the circle, BD
the circle. By the Perpendicular Bisector of a Chord
#
Conjecture, the perpendicular bisector of BD
#
passes through the center of the circle. Because RG
is a chord of the circle that passes through the
# is a
center, by the definition of diameter, RG
diameter.
#, #
#, and AR
#. Because they
7. Draw in radii GR
ER , TR
#
#
# > TR
#.
>
ER
and AR
are radii of the same circle, GR
By the Parallel Lines Intercepted Arcs Conjecture,
X > ET
X. Their central angles must also be
GA
congruent, so /GRA > /ERT. Thus nGRA >
# > ET
# by CPCTC. [ GATE is
nERT by SAS, so GA
an isosceles trapezoid.
8. x 5 65°, y 5 40°, z 5 148°; half the measure of
the intercepted arc
9. Because the radii of a circle are congruent,
# > OB
#, so nOAB is isosceles. By the Isosceles
OA
Triangle Conjecture, /2 > /3, so m/2 5 m/3.
By the Triangle Sum Conjecture, m/1 1 m/2 1
m/3 5 180°, so by substitution, m/1 1
2(m/2) 5 180°, or m/1 5 180° 2 2(m/2).
78
@#$, so
# ' BC
By the Tangent Conjecture, OB
m/OBC 5 90°. By Angle Addition, m/2 1
m/4 5 m/OBC, so m/2 1 m/4 5 90°,
or m/2 5 90° 2 m/4. Substitute 90° 2 m/4
for m/2 in the earlier equation: m/1 5 180° 2
2(90° 2 m/4). Simplifying gives m/1 5 2(m/4).
X 5 m/1, substitution gives mAB
X5
Because mAB
X.
2(m/4), or m/4 5 }21}mAB
10a. S; An equilateral triangle is equiangular, but a
rhombus is not equiangular.
10b. A
10c. N; Only one diagonal is the perpendicular
bisector of the other.
10d. A
10e. S; An equilateral triangle has rotational
symmetry and three lines of symmetry; a
parallelogram has rotational symmetry but no line
of symmetry.
11. L
12. J
13. K
14. D
15. E
16. B
17. H
18. G
19. N
20. a 1 b 1 b 1 a 5 180°, so a 1 b 5 90°
2
21. }2}
1
# > RS
#. OP
# > OQ
## > OR
#>
22. It is given that PQ
#
because
all
radii
in
a
circle
are
congruent.
OS
Therefore, nOPQ > nORS by SSS and /2 > /1
by CPCTC. Now we move on to the smaller right
triangles inside nOPQ and nORS. It is given that
# ' PQ
# and OV
# ' RS
#. Therefore, /OTQ and
OT
/OVS are right angles by the definition of
perpendicular lines and /OTQ > /OVS because
all right angles are congruent. Thus, nOTQ >
# > OV
# by CPCTC.
nOVS by SAA and OT
LESSON 6.5
1. d 5 5 cm
2. C 5 10p cm
12
3. r 5 }p} m
11p
4. C 5 5.5p or }2}
18. (Lesson 6.5)
1
2
? MA
## > MT
##
}
Given
? SA
# > ST
#
}
Given
3
MAST is a kite
?
} Definition of kite
## is the perpendicular bisector of AT
#
MS
4 ?
}
? Kite Diagonal Bisector Conjecture
}
79
5. C 5 12p cm
6. d 5 46 m
7. C < 15.7 cm
8. C < 25.1 cm
9. r < 7.0 m
10. C < 84.8 in.
11. 565 ft
12. C 5 6p cm
13. 16 in.
14. Trees grow more in years with more rain;
244 yr
15. 1399 tiles
16. g 5 40°, n 5 30°, x 5 70°; y 5 142°; z 5 110°;
Conjecture: The measure of the angle formed by
two intersecting chords is one-half the sum of the
measures of the two intercepted arcs. In the
X1 mLG
X .
diagrams, m/AEN 5 }21}1mAN
2
X and m/LNG 5 }1}mLG
X
17. m/AGN 5 }21}mAN
2
by the Inscribed Angle Conjecture. m/AEN 5
m/AGN 1 m/LNG by the Triangle Exterior
X 1 mLG
X
Angle Conjecture. So, m/AEN 5 }21}1mAN
2
by substitution and the distributive property.
19. b 5 90°, c 5 42°, d 5 70°, e 5 48°, f 5 132°,
g 5 52°
150°
5
} }}
20. }
360° or 12
21. The base angles of the isosceles triangle have a
measure of 39°. Because the corresponding angles
are congruent, m is parallel to n.
22. 10x 1 2y
LESSON 6.6
1. < 4398 km/h
2. < 11 m/s
3. 37,000,000 revolutions
4. < 637 revolutions
5. Mama; C < 50 in.
6. < 168 cm
7. d < 7.6 ft. The table will fit, but the chairs may
be a little tight in a 12-by-14 ft room. 12 chairs 5
192 in., 12 spaces 5 96 in., C 5 288 in., d < 91.7 in.
< 7.6 ft.
8. 0.35 ft/s
9. a 5 37.5°, s 5 17.5°, x 5 20°; y 5 80°; z 5 61°;
Conjecture: The measure of an angle formed by
two secants that intersect outside a circle is
one-half the difference of the larger arc measure
and the smaller arc measure.
80
X and m/SAN 5 }1}mSN
X by
10. m/ESA 5 }21}mEA
2
the Inscribed Angle Conjecture. m/SAN 5
m/ESA 1 m/ECA by the Triangle Exterior Angle
X5 }1}mEA
X 1 m/ECA by
Conjecture. So, }21}mSN
2
substitution. With a little more algebra, m/ECA 5
X
X
1
}}1mSN 2 mEA 2.
2
11. Both triangles are isosceles, so the base angles
in each triangle are congruent. But one of each base
angle is part of a vertical pair. So, a 5 b by the
Vertical Angles Conjecture and transitivity.
12. C
13. 38°
14. 48°
15. 30 cm
400 rev 2p . 2 6 ft 1 min
} }} }}
16. }
1 min ? 1 rev ? 60 s < 1089 ft/s
17. 12 cm , third side , 60 cm. This is based on
the Triangle Inequality Conjecture.
1 2
1 2
50
40
Cost ($)
1
1. 2}2}, 3
2. (23, 23)
2
3. }5}, 3
4. (7, 22)
5. infinitely many solutions
6. no solution
7. 4. The lines intersect at the solution point, (7, 22).
The circumcenter is 1}27}2 , }17}8 2. Only two
perpendicular bisectors are needed because
the third bisector intersects at the same point.
9a. Plan A: y 5 4x 1 20; Plan B: y 5 7x;
x 5 6 h 40 min, y 5 $46.67
y
9b.
y
y 5 2x 2 16
5
30
Plan B
20 (0, 20)
10
2
5
( 7, 22)
x
10
4
6
Hours
x
The point of intersection shows when both plans
cost the same, the answer for 9a.
9c. Plan B; 7}21} hours, with Plan A
10. (23, 1), (0, 3), (3, 21)
y 5 2 12 x 1 32
–5
–10
y
y 5 2 13 x 1 2
x
6. The lines are parallel (the slopes are the same,
but the y-intercepts are different).
y
There is no solution.
y = –2x + 9
5
x
–5
–5
y = –2x – 1
–10
8. Possible solution using the equations
y 5 }21}x 1 1 and y 5 2}32}x 1 }13}4 :
1
y 5 }2}x 1 1
2
14 1
2}3}x 1 }3} 5 }2}x 1 1
24x 1 28 5 3x 1 6
22 5 7x
22
}} 5 x
7
1 22
y 5 }2} }7} 1 1
11a.
5
y 5 }43}x 1 }23}
y 5 2x 1 12
11b. (6, 6)
11c. (6, 6); (6, 6)
11d. The diagonals intersect at their midpoints,
which supports the conjecture that the diagonals of
a parallelogram bisect each other.
2
13
12. y 5 }3}x 2 }3}
13. (4, 27)
1 2
3
15. 13, 2}2}2
69 6
}
14. }1},
4 2}7
16. The circumcenter is the midpoint of the
hypotenuse. For a right triangle, the perpendicular
bisectors of the legs are two of the midsegments
of the triangle. As with any pair of triangle
midsegments, they intersect at the midpoint of
the third side.
1 2
18
y 5 }7}
81
5. The lines are the same. There are infinitely many
solutions.
4
16 _23 , 46 _232
Plan A
LESSON 6.7
4p
1. }3} in.
2. 8p m
3. 14p cm
4. 9 m
5. 6p ft
6. 4p m
7. 27 in.
8. 100 cm
9. 217 m/min
10. < 4200 mi
11. Desks are about 17 meters from the center.
About four desks will fit because an arc with
one-half the radius and the same central angle
will be one-half as long as the outer arc.
12. The measure of the central angle is 7.2°
because of the Corresponding Angles Conjecture.
Therefore, 500 5 }376.2}0 ? C, so C < 25,000 mi.
13. 18°/s. No, the angular velocity is measured in
degrees per second not in distance per second, so it
is the same at every point on the carousel.
14. Outer horse < 2.5 m/s, inner horse < 1.9 m/s.
One horse has traveled farther in the same amount of
time (tangential velocity), but both horses have
rotated the same number of times (angular velocity).
15. a 5 50°, b 5 75°, x 5 25°; y 5 45°; z 5 35°;
Conjecture: The measure of the angle formed by
an intersecting tangent and secant to a circle is
one-half the difference of the larger intercepted arc
measure and the smaller intercepted arc measure.
16. Let z represent the measure of the exterior
@#$.
# and tangent PB
angle of nPBA formed by AB
1 X
By the Tangent Chord Conjecture, z 5 }2}mAB . By
X.
the Inscribed Angle Conjecture, m/BAP 5 }21}mBC
By the Triangle Exterior Angle Conjecture, z 5
m/BPA 1 m/BAP, or m/BPA 5 z 2 m/BAP.
82
X 2 }1}mBC
X.
By substitution, m/BPA 5 }21}mAB
2
Using the distributive property, m/BPA 5
X
X
1
}}(mAB 2 mBC ).
2
17. a 5 70°, b 5 110°, c 5 110°, d 5 70°, e 5 20°,
f 5 20°, g 5 90°, h 5 70°, k 5 20°, m 5 20°,
n 5 20°, p 5 140°, r 5 80°, s 5 100°, t 5 80°,
u 5 120°
19. 170°
289
8
}}
20. y 5 2}1}x
5 1 15
21. 45°
22. The sum of the lengths is 8p cm for Case 1,
Case 2, Case 3, and Case 10.
23. 6
24. Yes, as long as the three points are noncollinear; possible answer: connect the points with
segments, then find the point of concurrency of
the perpendicular bisectors (same as circumcenter
construction).
CHAPTER 6 REVIEW
23. Sample answer: Construct a right angle and
# and RT
# with any
label the vertex R. Mark off RE
lengths. From point E, swing an arc with radius RT.
From point T, swing an arc with radius RE. Label
the intersection of the arcs as C. Construct the
# and RC
#. Their intersection is the
diagonals ET
center of the circumscribed circle. The circle’s
radius is the distance from the center to a vertex. It
is not possible to construct an inscribed circle in a
rectangle unless it is a square.
E
C
R
T
24. Sample answer: Construct acute angle R. Mark
off equal lengths RM and RH. From points M and
H, swing arcs of lengths equal to RM. Label the
intersection of the arcs as O. Construct RHOM. The
intersection of the diagonals is the center of the inscribed circle. Construct a perpendicular to a side to
find the radius. It is not possible to construct a circumscribed circle unless the rhombus is a square.
H
O
R
M
4
32
25. 4x 1 3y 5 32, or y 5 2}3}x 1 }3}
26. (23, 2)
27. d 5 0.318 m
28. Melanie: 151 m/min or 9 km/h; Melody:
94 m/min or 6 km/h.
2p(6357)
2p(6378)
?
?
29. } < 1.849 , 1.852 , 1.855 < }
360 60
360 60
30.
Possible
location
7700 ft
5280 ft
5500 ft
D
T
Possible
location
83
2. Draw two nonparallel chords. The intersection
of their perpendicular bisectors is the center of the
circle.
Fold the paper so that two semicircles coincide.
Repeat with two different semicircles. The center is
the intersection of the two folds.
Place the outside or inside corner of the L in the
circle so that it is an inscribed right angle. Trace the
sides of the corner. Draw the hypotenuse of the right
triangle (which is the diameter of the circle). Repeat.
The center is the intersection of the two diameters.
3. The velocity vector is always perpendicular to
the radius at the point of tangency to the object’s
circular path.
4. Sample answer: An arc measure is between 0° and
360°.An arc length is proportional to arc measure
and depends on the radius of the circle.
5. 55°
6. 65°
7. 128°
8. 118°
9. 91°
10. 66° 11. 125.7 cm 12. 42.0 cm
13. 15p cm
14. 14p ft
15. 2 ? 57° 1 2 ? 35° Þ 180°
16. 84° 1 56° 1 56° 1 158° Þ 360°
X 5 }1} (180° 2 108°) 5 36° 5
17. m/EKL 5 }21} mEL
2
# i YL
# by Converse of the Parallel
m/KLY. [ KE
Lines Conjecture.
X 5 360° 2 56° 2 152° 5 152° 5 mMI
X.
18. mJI
X 5 }1} mMI
X 5 m/MJI. By the
[ m/JMI 5 }21} mJI
2
Converse of the Isosceles Triangle Conjecture,
nJIM is isosceles.
X 5 2m/KEM 5 140°. [ mKI
X5
19. mKIM
X
140° 2 70° 5 70° 5 mMI . [ m/IKM 5
1 X
1 X
}} mMI 5 }} mKI 5 m/IMK. By the Converse of the
2
2
Isosceles Triangle Conjecture, nKIM is isosceles.
20. Ertha can trace the incomplete circle on paper.
She can lay the corner of the pad on the circle to
trace an inscribed right angle. Then Ertha should
mark the endpoints of the intercepted arc and use
the pad to construct the hypotenuse of the right
triangle, which is the diameter of the circle.
21. Sample answer: Construct
perpendicular bisectors of two
sides of the triangle. The point
at which they intersect (the
circumcenter) is the center of
the circle. The distance from
the circumcenter to each vertex
is the radius.
22. Sample answer: Construct the incenter (from
the angle bisectors) of the triangle. From the incenter,
which is the center of the circle, construct a perpendicular to a side. The distance from the incenter to
the foot of the perpendicular is the radius.
200
31. }p} ft < 63.7 ft
32. 8p m < 25.1 m
8
33. The circumference is }346}
0 ? 2p(45) 5 12p; the
diameter is 12 cm.
34. False. 20° 1 20° 1 140° 5 180°. An angle with
measure 140° is obtuse.
35. true
D
36. false
C
A
53. False. The ratio of the circumference to the
diameter is p.
24 cm
24 cm
54. false; 24 1 24 1
48 1 48 Þ 96
48 cm
55. true
56. This is a paradox.
57. a 5 58°, b 5 61°, c 5 58°, d 5 122°, e 5 58°,
f 5 64°, g 5 116°, h 5 52°, i 5 64°, k 5 64°,
l 5 105°, m 5 105°, n 5 105°, p 5 75°, q 5 116°,
r 5 90°, s 5 58°, t 5 122°, u 5 105°, v 5 75°,
w 5 61°, x 5 29°, y 5 151°
58. nTAR > nYRA by SAS, nTAE > nYR E
by SAA
59. nFTO > nYTO by SAA, SAS, or SSS;
nFLO > nYLO by SAA, SAS, or SSS;
nFTL > nYTL by SSS, SAS, or ASA
60. nPTR > nART by SSS or SAS;
nTPA > nRAP by SSS, SAS, SAA, or ASA;
nTLP > nRLA by SAA or ASA
61. ASA
X 5 84°, length of AC
X 5 11.2p < 35.2 in.
62. mAC
63. x 5 63°, y 5 27°, w 5 126°
64. sample answer:
B
37. true 38. true 39. true 40. true
41. False. (7 2 2) ? 180° 5 900°. It could have seven
sides.
42. False. The sum of the measures of any triangle
is 180°.
43. False. The sum of the measures of one set of
exterior angles for any polygon is 360°. The sum of
the measures of the interior angles of a triangle is
180° and of a quadrilateral is 360°. Neither is greater
than 360°, so these are two counterexamples.
44. False. The consecutive angles between the
bases are supplementary.
45. False. 48° 1 48° 1 132° Þ 180°
46. False. Inscribed angles that intercept the same
arc are congruent.
47. False. The measure of an inscribed angle is
half the measure of the arc.
48. true
# and BD
# bisect each other, but AC
# is
49. False. AC
#.
not perpendicular to BD
D
C
A
B
48 cm
308
1508
66a. The circle with its contents has 3-fold rotational symmetry, the entire tile does not.
66b. No, it does not have reflectional symmetry.
67.
68. 9.375 cm
50. False. It could be isosceles.
51. False. 100° 1 100° 1 100° 1 60° 5 360°
X À CD
X
52. false; AB
C
A
69.
908
908
70.
D
B
84
n
1
2
3
4
5
6
f (n)
5
1
23
27
211
215
...
n
...
20
. . . 9 2 4n . . .
271
7
11.
LESSON 7.1
1. Rigid; reflected, but the size and shape do not
change.
2. Nonrigid; the shape changes.
3. Nonrigid; the size changes.
4.
5.
P
P
,
6.
,
7. possible answer: a boat moving across the water
8. possible answer: a Ferris wheel
9a. Sample answer: Fold the paper so that the
images coincide, and crease.
or
18.
19. P(2a, b), Q(2a, 2b), R(a, 2b)
20. possible construction:
P
9b. Construct a segment that connects
two corresponding points. Construct the
perpendicular bisector of that segment.
10a. Extend the three horizontal segments onto
the other side of the reflection line. Use your
compass to measure lengths of segments and
distances from the reflection line.
10b.
21. 50th figure: 154 (50 shaded, 104 unshaded);
nth figure: 3n 1 4 (n shaded, 2(n 1 2) unshaded)
22. 46
23. It is given that /1 > /2, and /2 > /3
because of the Vertical Angles Conjecture, so
/1 > /3. Segment DC is congruent to itself.
/DCE and /DCB are both right angles, so they
are congruent. Therefore, nDCB > nDCE by
# > CE
# by CPCTC.
ASA, and BC
16. (Lesson 7.1)
Number of sides of
regular polygon
3
4
5
6
7
8
...
n
Number of reflectional
symmetries
3
4
5
6
7
8
...
n
Number of rotational
symmetries (# 360°)
3
4
5
6
7
8
...
n
85
13. 4-fold rotational and reflectional symmetry
15. 7-fold symmetry: possible answers are F or J.
9-fold symmetry: possible answers are E or H.
Basket K has 3-fold rotational symmetry but not
reflectional symmetry.
16. See table below; n, n
17.
LESSON 7.2
1.
y translation
5
x
5
6. Rules that involve x or y changing signs,
or switching places, produce reflections.
If both x and y change signs, the rule produces a
rotation. Rules that produce translations involve a
constant being added to the x and/or y terms.
k5, 0l is the translation vector for Exercise 1.
7. (x, y) → (x, 2y)
8. (x, y) → (2x, 2 y)
9.
N
2. reflection
y
Cue ball
8
W
E
8 ball
x
–5
S
10. There are two possible points, one on the N
wall and one on the W wall.
–8
3.
reflection
y
N
5
H
W
–4
x
7
T
y
S
11.
–5
4.
E
H''
reflection
N
5
W
E
T
x
5
H
12. by the Minimal Path Conjecture
Proposed freeway
5.
rotation
y
5
5
Mason
x
Perry
13.
,
14.
86
H'
S
15. possible answer: HIKED
16. one, unless it is equilateral, in which case it
has three
17. two, unless it is a square, in which case it has
four
18.
21. false; possible counterexample: trapezoid
with two right angles
22. false; possible counterexample: isosceles
trapezoid
87
LESSON 7.3
1. k10, 10l
2. A 180° rotation. If the centers of rotation differ,
rotate 180° and add a translation.
3a. 20 cm
3b. 20 cm, but in the opposite direction
4a. 80° counterclockwise
4b. 80° clockwise
5. 180°
6. 3 cm
7. possible answer:
11. Answers may vary. Possible answer: reflection
across the figure’s horizontal axis and 60°
clockwise rotation.
12.
13.
,
14. Sample answer: Draw a figure on an overhead
transparency and then project the image onto a
screen.
15. possible answers: rotational: playing card,
ceiling fan, propeller blade; reflectional: human
body, backpack
16. one: yes; two: no; three: yes
′O
′A
N ′′
A
H
′
A′
H ′′
O
N
′′
′H
O
′N
8. possible answer:
Center of rotation
A
O
B
9.
18a.
18b.
a b
d e
3
3
25
212
29
0
12
27
4 3
4 3
4
214 11
?
? }
11
1 }
5
0 13
?
?
20
} }
2a
4 3
c
a
1
f
2d
2b
d–e
4 3
3c
5
0
88
4c
4
0
10. Two reflections across intersecting lines yield
a rotation. The measure of the angle of rotation is
twice the measure of the angle between the lines of
reflection, or twice 90°, or 180°.
3b
?
? }
? }
}
?
?
?
} } }
d
f
LESSON 7.4
1. Answers will vary. 2. Answers will vary.
3. 33.42
4. 34.6
5. 32.4.3.4
6. 3.4.6.4/3.42.6
8. 36/32.4.12
7. 33.42/32.4.3.4
9a. The dual of a square tessellation is a square
tessellation.
9b. The dual of a hexagon tessellation is a triangle
tessellation.
9c. If tessellation A is the dual of tessellation B,
then tessellation B is the dual of tessellation A.
10. The dual is a 34/ 38 tessellation of isosceles
right triangles.
11.
1
16. y 5 2}2}x 1 4
y
4
–3
5
x
–6
17. possible answer: TOT
18.
12.
N
8-ball
W
E
Cue ball
13. A ring of ten pentagons fits around a decagon,
and another decagon can fit into any two of the
pentagons. But another ring of pentagons around
the second decagon doesn’t leave room for a third
decagon.
14.
S
89
LESSON 7.5
2. The dual is a 53/54 tessellation.
By the Triangle Sum Conjecture, a 1 b 1 c 5 180°.
Around each point, we have 2(a 1 b 1 c) 5
2 ? 180° 5 360°. Therefore, a triangle will fill the
plane edge to edge without gaps or overlaps. Thus, a
triangle can be used to create a monohedral tiling.
6. three ways
7.
8. y 5 22x 1 3
3.
y
8
4. Yes. The four angles of the quadrilateral
will be around each point of intersection in the
tessellation.
a
a
c
c
5.
b
b
b
a
c
c
a
b a
90
b a
c
b
c b a
a
c
b
c b a
c
a
c b
5
–2
x
LESSON 7.6
1.
2.
3.
4.
5.
6.
7.
Answers will vary.
Answers will vary.
Answers will vary.
regular hexagons
squares or parallelograms
squares or parallelograms
2
12. y 5 2}3}x 2 3; the slope is the opposite sign.
y
5
–10
10
x
13. 3.4.6.4 /4.6.12
440 rev 2p ? 28 ft 1 min < 1290 ft/s
}
}
14. }
?}
1 min ? }
60 s
1 rev
8.
11.
B
E
A
S
91
15. Possible explanations:
15a. true; The kite diagonal between vertex angles
is the perpendicular bisector of the other diagonal;
in a square, diagonals would bisect each other
15b. False; it could be an isosceles trapezoid.
15c. False; it could be a rectangle.
15d. true; Parallel lines cut off congruent arcs of a
circle, so inscribed angles (the base angles of the
trapezoid) are congruent.
LESSON 7.7
1. equilateral triangles.
2. regular hexagons.
3.
4.
9. true
10. true
11. False; it could be a kite or an isosceles
trapezoid.
12. The path would be }14} of Earth’s circumference,
approximately 6280 miles, which will take
126 hours, or around 5 }14} days.
13a. Using the Reflection Line Conjecture, the
line of reflection is the perpendicular bisector of
#9and BB
#9. Because these segments are both
AA
perpendicular to the reflection line, they are
# is parallel to
parallel to each other. Note that if AB
the reflection line, quadrilateral AA9B9B will be a
rectangle instead of a trapezoid.
13b. Yes; it has reflectional symmetry, so legs and
base angles are congruent.
13c. greatest: near each of the acute vertices;
least: at the intersection of the diagonals (where A,
C, and B9 become collinear and A9, C, and B
become collinear)
14a.
7. sample design:
331
43
5 26
0
4
4
92
9
3
4
228 15
0
14b.
3 128
4 3 }}? }}? 4 5 3 13}?
3
25
2
?
21 10
8. False; they must bisect each other in a
parallelogram.
108
8 7
?
? }
6 0 5 }
?
?
} }
29 2
29
4
30
250
LESSON 7.8
1. parallelograms
2. parallelograms
3.
7. Circumcenter is (3, 4); orthocenter is (10, 8).
8.
9.
4.
10.
93
1
1. y 5 2}6}x
2. y 5 22x 1 2
3. Centroid is 12, }23}2; orthocenter is (0, 5).
94
4. Centroid is (4, 0); orthocenter is (3, 0).
4
5. 1, }3}
6. (21, 21)
7. (5, 28)
1 2
22.
CHAPTER 7 REVIEW
T
H
23. Use a grid of squares. Tessellate by translation.
24. Use a grid of equilateral triangles. Tessellate by
rotation.
25. Use a grid of parallelograms. Tessellate by
glide reflection.
26. Yes. It is a glide reflection for one pair of sides
and midpoint rotation for the other two sides.
1. true
2. true
3. true
4. true
5. true
6. true
7. False; a regular pentagon does not create a
monohedral tessellation and a regular hexagon
does.
8. true
9. true
10. False; two counterexamples are given in
Lesson 7.5.
11. False; any hexagon with one pair of opposite
sides parallel and congruent will create a
monohedral tessellation.
12. This statement can be both true and false.
13. 6-fold rotational symmetry
14. translational symmetry
15. Reflectional; color arrangements will vary, but
the white candle must be in the middle.
16. The two towers are not the reflection (or
even the translation) of each other. Each tower
individually has bilateral symmetry. The center
portion has bilateral symmetry.
19. 36/32.4.3.4; 2-uniform
20. 4.82; semiregular
1
21. y 5 }2}x
27. No.Because the shape is suitable for glide
reflection,the rows of parallelograms should
alternate the direction in which they lean (row 1
leans right,row 2 leans left,row 3 leans right,and
so on).
28.
y
x
95
8
LESSON 8.1
1. 228 m 2
2. 41.85 cm 2
3. 8 yd
4. 21 cm
5. 91 ft 2
6. 182 m 2
7. 96 in2
8. 210 cm
9. A 5 42 ft 2
10. sample answer:
6 cm
12 cm
48 cm2
4 cm
8 cm
11.
12.
13.
14.
48 cm2
3 square units
10 square units
7}12} square units
sample answers:
8 cm
16 cm
64 cm2
23. 500 cm2
24a. smallest: 191.88 cm2; largest: 194.68 cm2
24b. Answers will vary. Sample answer: about
193 cm2.
24c. Answers will vary. The smallest and largest
area values differ at the ones place, so the digits
after the decimal point are insignificant compared
to the effect of the limit of precision in the
measurements.
25a. In one Ohio Star block, the sum of the red
patches is 36 in2, the sum of the blue patches is
72 in2, and the yellow patch is 36 in2.
25b. 42
25c. About 1814 in2 of red fabric, about 3629 in2
of blue fabric, and about 1814 in2 of yellow
fabric. The border requires 5580 in2 (if it does
not need the extra 20%).
26. 100; 36 1 64. The area of the square on the
longer side is the same as the sum of the areas on
the other two legs.
27. a 5 76°,b 5 52°,c 5 104°,d 5 52°,e 5 76°,
f 5 47°, g 5 90°, h 5 43°, k 5 104°, m 5 86°.
Explanations will vary.
64 cm2
4 cm
8 cm
16 cm
16 cm
4 cm
A
30°
M
30°
16 cm
16 cm
16. 23.1 m2
17. 2(4)(3) 1 2(5.5)(3) 5 57 m2
18. For a constant perimeter, area is maximized by
a square. 100 m 4 4 5 25 m per side; A 5 625 m 2.
1
}
19. }
530
20. 112
21. 96 square units
22. 32 square units
96
29a.
29c.
29b.
LESSON 8.2
1. 20 cm2
2. 49.5 m2
3. 300 square units
4. 60 cm2
5. 6 cm
6. 9 ft
7. 30 ft
8. 5 cm
9. 16 m
10. 168 cm
11. 12 cm
12. 3.6 ft; 10.8 ft
13. sample answer:
#.)
17. }12}(To see why, draw altitude PQ
18. more than half, because the top card
completely covers one corner of the bottom card
19a. 86 in. of balsa wood and 960 in2 of Mylar
19b. 56 in. (or less, if he tilts the kite)
20. 3600 shingles (to cover an area of 900 ft2)
21. The isosceles triangle is a right triangle
because the angles on either side of the right
angle are complementary. If you use the trapezoid
area formula, the area of the trapezoid is
1
}} (a 1 b)(a 1 b). If you add the areas of the
2
three triangles, the area of the trapezoid is
1
}} c 2 1 ab.
2
22. A b1 B
h
D
9 cm
9 cm
12 cm
14. sample answer:
4 cm
7 cm
8 cm
7 cm
10 cm
9 cm
15. sample answer:
46 cm
12 cm
45 cm
35 cm
12 cm
56 cm
16. The length of the base of the triangle equals
the sum of the lengths of both bases of the
trapezoid.
C
Given: trapezoid ABCD with height h. area
of nABD 5 }21} hb1; area of nBCD 5 }21}hb2; area
of trapezoid 5 sum of areas of two triangles
5 }21}h1b1 1 b22
23. 11}41} square units
24. 7 square units
25. 70 m
26. 144 cm2
27. 828 ft 2; 144 ft
28. 1440 cm2; 220 cm
29a. incenter
29b. orthocenter
29c. centroid
30. a 5 34°, b 5 68°, c 5 68°, d 5 56°, e 5 56°,
f 5 90°, g 5 34°, h 5 56°, m 5 56°, n 5 90°,
p 5 34°. Possible explanation: Let O be the center
X 5 112° by the Inscribed Angle
of the circle. mBC
X by the Central
Conjecture, and d 1 e 5 mBC
Angle Conjecture. nOBA is congruent to nOCA
X is a semicircle, so
by SSS, so d 5 e 5 56°. DEC
X
mDE 5 68°. By the Inscribed Angle Conjecture,
p 5 34°. Using nOEC and the Triangle Sum
Conjecture, n 5 90°.
31. 32.62/3.6.3.6
2 cm
3 cm
7 cm
3 cm
9 cm
97
12 cm
b2
LESSON 8.3
1a. 121,952 ft 2
1b. 244 gal of base paint and 488 gal of finishing
paint
2. He should buy at least four rolls of wallpaper.
(The area of each roll is 125 ft2. The total surface
area to be papered is 480 ft2.) If paper cut off at the
corners is wasted, he’ll need 5 rolls.
3. 1552 ft2; 776 ft 2 more surface area
4. 21
5. 336 ft2; $1780
98
6. $760
7. 220 terra cotta tiles, 1107 blue tiles; $1598.15
8. 72 cm 2
9. AB < 16.5 cm, BD < 15.3 cm
10. 60 cm2 by either method
11. Because nAOB is isosceles, m/A 5 20° and
X
X 5 82°.
m/AOB 5 140°. mA
A
B 5 140° and mCD
X5 mBD
X because parallel lines intercept
mAC
360° 2 140° 2 82°
} 5 69°.
congruent arcs on a circle. }
2
12. E
11. a2 1 2ab 1 b2
1. x 2 1 6x 1 5
a
b
a
a2
ab
a
b
ab
b2
2b
2. 2x 2 1 7x
x15
x
x
12. a2 2 b2
a
b
a2
ab
2ab
2b 2
5
x
x2
x11 x
13. (x 1 15)(x 1 4)
1
x
15
x
x2
15x
4
4x
60
2x 1 7 x
14. (x 2 12)(x 1 2)
x
212
x
x2
212x
2
2x
224
7
15. (x 1 5)(x 2 4)
3. 6x 2 1 19x 1 10
3x 1 2
x
x
x
x
5
x2
5x
24x
220
2
x
x
24
2x 1 5
16. (x 2 3)2 5 (x 2 3)(x 2 3)
x
x
4. (3)(2x 1 1)
5. (x 1 5)(x 1 3)
x
x2
23x
23x
9
17. (x 1 6)(x 2 6)
x15
3
23
23
5
x
5
x
6
x2
6x
26x
236
x
x2
x
2x 1 1
x
x13
3
x
26
1
18. (2x 1 7)(2x 2 7)
6. (2x 1 3)(x 1 4)
x
2x 1 3
x
3
2x
7
2x
4x 2
14x
27
214x
249
x
x14
4
7. x 2 2 26x 1 165
x
215
x2
215x
211x
165
x
211
9. x 2 1 8x 1 16
x
4
x
x2
4x
4
4x
16
8. 12x 2 2 13x 2 35
3x
27
4x
12x 2
228x
5
15x
235
10. 4x 2 2 25
2x
5
2x
4x 2
10x
25
210x
225
19. x 5 24 or x 5 21
20. x 5 210 or x 5 3
21. x 5 3 or x 5 8
1
22. x 5 2}2} or x 5 24
h
23a and b.
h
h14
1
23c. }2}[h 1 (h 1 4)]h 5 48
23d. h 5 28 or h 5 6. The height cannot be
negative, so the only valid solution is h 5 6.
The height is 6 feet, one base is 6 feet, and the
other base is 10 feet.
99
LESSON 8.4
2092 cm2
2. 74 cm
256 cm
4. 33 cm2
63 cm
6. 490 cm2
< 57.6 m
8. < 25 ft
2
< 42 cm
10. < 58 cm2
1
1
1 1
11. a 5 }2}s; A 5 }2}asn 5 }2} ? }2}s ? s ? 4 5 s 2
12. It is impossible to increase its area, because a
regular pentagon maximizes the area. Any
dragging of the vertices decreases the area.
(Subsequent dragging to space them out more
evenly can increase the area again, but never
beyond that of the regular pentagon.)
13. < 996 cm2
14. < 497 cm2
15. total surface area 5 13,680 in2 5 95 ft2;
cost 5 $8075
16. Area is 20 square units.
1.
3.
5.
7.
9.
y
y 5 – _4 x 1 12
3
y 5 – _1 x 1 6
3
(6, 4)
x
18. Conjecture: The three medians of a triangle
divide the triangle into six triangles of equal area.
Argument: Triangles 1 and 2 have equal area
because they have equal bases and the same
height. Because the centroid divides each median
into thirds, you can show that the height of
triangles 1 and 2 is }31} the height of the whole
triangle. Each has an area }61} the area of the whole
triangle. By the same argument, the other small
triangles also have areas }61} the area of the whole
triangle.
y
y 5 _12x 1 5
1
(2, 6)
y 5 22x 1 10
x
100
h
2
19. nw 1 ny 1 2x
20. 504 cm2
21. 840 cm2
LESSON 8.5
1. 9p in2
2. 49p cm2
3. 0.8 m2
4. 3 cm
5. Ïw
3 in.
6. 0.5 m
7. 36p in2
8. 7846 m2
9. 25p 2 48, or about 30.5 square units
10. 100p 2 128, or about 186 square units
11.
16. A < pr 2 because the 100-gon almost
completely fills the circle.
17. 456 cm2
18. 36 ft2
19. The triangles have equal area when the point
is at the intersection of the two diagonals. There is
no other location at which all four triangles have
equal area.
X 5 2 24° 5 48°
20. x 5 mDE
?
21. 90° 1 38° 1 28° 1 28° Þ 180°
22.
24 cm
r 5 18 cm
12 cm
804 m2
11,310 km2
154 m2
4 times
18 cm
12.
13.
14.
15.
6 cm
101
LESSON 8.6
1. 6p cm2
64p
2. }3} cm2
3. 192p cm2
4. (p 2 2) cm2
5. (48p 1 32) cm2
6. 33p cm2
7. 21p cm2
105p
8. }2} cm2
9. 6 cm
10. 7 cm
11. 75
12. 100
13. 42
14. $448
15a.
15b.
15c.
102
15d.
16. sample answer:
17a. (144 2 36p) cm2; 78.54%
17b. (144 2 36p) cm2; 78.54%
17c. (144 2 36p) cm2; 78.54%
17d. (144 2 36p) cm2; 78.54%
18. 480 m2
19. AB < 17.0 cm, AG < 6.6 cm
90
}
20. True. If 24p 5 }
360 ? 2pr, then r 5 48 cm.
360
21. True. If }n} 5 24, then n 5 15.
22. False. It could be a rhombus.
23. true; Triangle Inequality Conjecture
LESSON 8.7
1. 150 cm2
2. 4070 cm2
3. 216 cm2
4. 340 cm2
5. < 103.7 cm2
6. < 1187.5 cm2
7. < 1604.4 cm2
8. < 1040 cm2
2
9. < 414.7 cm
10. < 329.1 cm2
11. area of square 1 4 ? area of trapezoid 1 4 ?
area of triangle
12. $1570
13. sample answer:
15. a 5 75°, b 5 75°, c 5 30°, d 5 60°,
e 5 150°, f 5 30°
16. About 23 days. Each sector is about 1.767 km2.
17. a 5 50°, b 5 50°, c 5 80°, d 5 100°, e 5 80°,
f 5 100°, g 5 80°, h 5 80°, k 5 80°, m 5 20°,
n 5 80°. Explanations will vary. Sample
explanation: The angle with measure d corresponds
to the angle forming a linear pair with g. Because
d 5 100°, by the Parallel Lines Conjecture, the
angle adjacent to g measures 100°, and by the
Linear Pair Conjecture, g 5 80°. The angle with
measure f corresponds to the angle measuring
100°, so f 5 100°. The angles measuring g and k
are the base angles of an isosceles triangle, so by
the Isosceles Triangle Conjecture, k 5 80°.
18. 398 square units
14. sample tiling
33.42/32.4.3.4/44
103
CHAPTER 8 REVIEW
1. B (parallelogram)
3. C (trapezoid)
5. F (regular polygon)
7. J (sector)
9. G (cylinder)
11.
2. A (triangle)
4. E (kite)
6. D (circle)
8. I (annulus)
10. H (cone)
12.
20.
23.
26.
29.
32.
32 cm
21. 32 cm
81p cm2
24. 48p cm
2
153.9 cm
27. 72 cm2
300 cm2
30. 940 cm2
Area is 112 square units.
22.
25.
28.
31.
15 cm
40°
30.9 cm2
1356 cm2
y
Apothem
D (6, 8)
A (0, 0)
C (20, 8)
B (14, 0)
x
13.
y
R (4, 15)
U (9, 5)
F (0, 0)
14. Sample answer: Construct an altitude from
the vertex of an obtuse angle to the base. Cut off
the right triangle and move it to the opposite side,
forming a rectangle. Because the parallelogram’s
area hasn’t changed, its area equals the area of the
rectangle. Because the area of the rectangle is
given by the formula A 5 bh, the area of the
parallelogram is also given by A 5 bh.
b
O (4, –3)
34. 6 cm
35. 172.5 cm2
36. sample answers:
b
h
h
15. Sample answer: Make a copy of the trapezoid
and put the two copies together to form a
parallelogram with base 1b1 1 b22 and height h.
Thus the area of one trapezoid is given by the
formula A 5 }12} 1b1 1 b22h.
b2
b1
h
h
b1
b2
16. Sample answer:Cut a circular region into a large
number of wedges and arrange them into a shape
that resembles a rectangle.The base length of this
“rectangle”is pr and the height is r, so its area is pr 2.
ThustheareaofacircleisgivenbytheformulaA5pr 2.
r
pr
17. 800 cm2
18. 5990.4 cm2
19. 60p cm2 or about 188.5 cm2
104
x
37. 1250 m2
38. Circle. For the square, 100 5 4s, s 5 25,
A 5 252 5 625 ft2. For the circle, 100 5 2pr,
r < 15.9, A < p(15.9)2 < 794 ft2.
39. A round peg in a square hole is a better fit.
The round peg fills about 78.5% of area of the
square hole, whereas the square peg fills only about
63.7% of the area of the round hole.
40. giant
41. about 14 oz
42. One-eighth of a 12-inch diameter pie;
one-fourth of a 6-inch pie and one-eighth of a
12-inch pie both have the same length of crust,
which is longer than one-sixth of an 8-inch pie.
43a. 96 ft; 40 ft
43b. 3290 ft2
44. $3000
45. $4160
46. It’s a bad deal. 2pr1 5 44 cm. 2pr2 5 22 cm,
which implies 4pr2 5 44 cm. Therefore r1 5 2r2.
The area of the large bundle is 4p1r222 cm2. The
combined area of two small bundles is 2p1r222 cm2.
Thus he is getting half as much for the same price.
47. $2002
48. $384 (16 gal)
9
LESSON 9.1
b
a
b 2a
c
18. Sample answer: Yes, nABC > nXYZ by SSS.
Both triangles are right triangles, so you can
use the Pythagorean Theorem to find that
CB 5 ZY 5 3 cm.
m
120°
p
q
n
Or use the Exterior Angle Conjecture to get
q 1 n 5 120°. By AIA, q 5 m. Substituting,
m 1 n 5 120°.
22. a 5 122°, b 5 74°, c 5 106°, d 5 16°, e 5 90°,
f 5 74°, g 5 74°, h 5 74°, n 5 74°, r 5 32°,
s 5 74°, t 5 74°, u 5 32°, v 5 74°. Possible
explanation: By the Tangent Conjecture, the
quadrilateral containing g has two right angles
formed by radii intersecting tangent lines. Using
c 5 106° and the Quadrilateral Sum Conjecture,
g 1 90° 1 90° 1 106° 5 360°, so g 5 74°.
The angle measures e and f and the measure of the
inscribed angle that intercepts the arc with measure
u sum to 180°. Using e 5 90° and f 5 74°, the
inscribed angle measures 16°. Using the Inscribed
Angle Conjecture, u 5 32°.
T1
23 .
T2
105
1. c < 19.2 cm
2. a 5 12 cm
3. b < 5.3 cm
4. d 5 10 cm
5. s 5 26 cm
6. c < 8.5 cm
7. b 5 24 cm
8. x 5 3.6 cm
9. x 5 40 cm
10. s < 3.5 cm
11. r 5 13 cm
12. 127 ft
13. 512 m2
14. < 11.3 cm
15. 3, 4, 5
16. 28 m
17. The area of the large square is 4 ? area of
triangle 1 area of small square.
c 2 5 4 ? }21}ab 1 (b 2 a)2
c 2 5 2ab 1 b 2 2 2ab 1 a 2
c 2 5 a2 1 b2
19. 54 cm2
20. 36y32.4.3.4
21. Mark the unnamed angles as shown in the
figure below. By the Linear Pair Conjecture,
p 1 120° 5 180°, so p 5 60°. By AIA, m 5 q. By
the Triangle Sum Conjecture, q 1 p 1 n 5 180°.
Substitute m 5 q and p 5 60° to get
m 1 60° 1 n 5 180°. [ m 1 n 5 120°.
LESSON 9.2
1. yes
2. yes
3. no
4. no
5. no
6. no
7. no
8. No, the given lengths are not a Pythagorean
triple.
9. y 5 25 cm
10. y 5 24 units
11. y < 17.3 m
12. 6, 8, 10
13. 60 cm2
14. < 14.1 ft
15. < 17.9 cm2
16. 102 m
17a. < 1442 cm2
17b. < 74.8 cm
18. Sample answer: The numbers given satisfy the
Pythagorean Theorem, so the triangle is a right
triangle; but the right angle should be inscribed
in an arc of 180°. Thus the triangle is not a right
triangle.
106
19. Sample answer: (BD)2 5 62 2 32 5 27;
(BC)2 5 (BD)2 1 92 5 108; then (AB)2 1 (BC)2 5
(AC)2 (36 1 108 5 144), so nABC is a right
triangle by the Converse of the Pythagorean
Theorem.
20. centroid
3
21. }1}
0
22. Because m/DCF 5 90°, m/DCE 5 (90 2 x)°.
Because nDCE is isosceles, m/DEC 5 (90 2 x)°.
m/D 5 180° 2 2 ? (90 2 x)° 5 2x. Because /D is a
central angle, 2x 5 a. Therefore, x 5 1}21}2a.
23. The path from C to M to T lies on a straight
line and therefore must be shorter than the path
from C to A to T.
P
A
T
M
C
U
25.
26. 19
B
1. Ïw
6
2. 5
3. 18Ïw
2
4. 147
5. 8
6. 2Ïw
3
7. 3Ïw
2
8. 2Ïw
10
9. 5Ïw
3
10. Ïw
85
11. 4Ïw
6
12. 24
13. 12Ïw
5
14. 28
15. 6Ïw
23
16. Answers will vary. Possible answer: The length
of the hypotenuse of an isosceles right triangle
with legs of length 3 units is Ï18
w units. This is the
same as Ïw
2 1 Ïw
2 1 Ïw
2 , or 3Ïw
2.
18. Possible answer: A right triangle with legs of
lengths 1 and 3 units has a hypotenuse of length
w units.
Ï10
10
1
3
A right triangle with legs of lengths 6 and 2 units
has a hypotenuse of length Ï40
10 1
w 5 Ïw
5
2
10
.
10
Ïw
Ïw
40
2
6
19. Possible answer: A right triangle with legs of
lengths 2 and 3 units has a hypotenuse of length
w units.
Ï13
1
2
1
18
3
13
3
3
5
2
20. x 5 Ï3w, y 5 Ïw
12 . Because y is twice as
long as x, y 5 2x, so Ïw
12 5 2Ïw
3.
2
2
2
21. 2 1 1Ï3w2 5 1Ï7w2
7
3
1
2
2
20
2
4
107
17. The hypotenuse represents Ï5w units. In the
second right triangle, the legs have lengths 4 and 2,
so the hypotenuse has length Ïw
20 . The length of
the hypotenuse is twice the length of the hypotenuse
of the smaller triangle, so Ï20
5.
w 5 2Ïw
LESSON 9.3
72Ïw
2 cm
13 cm
10 cm, 5Ïw
3 cm
10Ïw
3 cm, 10 cm
34 cm, 17 cm
72 cm
12Ïw
3 cm
50 cm, 100 cm
16 cm2
1
1
10. } , }
2 Ïw
2
Ïw
1.
2.
3.
4.
5.
6.
7.
8.
9.
30°
x
3
2x
2
1
45°
11. A 30°-60°-90° triangle must have sides whose
lengths are multiples of 1, 2, and Ïw
3 . The triangle
shown does not reflect this rule.
12. Possible answer:
Use 12 1 1Ïw
3 22 5 22 and 42 1 1Ïw
48 22 5 82.
16. c 2 5 x 2 1 x 2 Start with the Pythagorean Theorem.
c 2 5 2x 2
Combine like terms.
c 5 xÏw
2
Take the square root of both sides.
17. 169Ïw
3 m2 < 292.7 m2
18. 390 m2
19.
3
Ïw
}
1
2
3
x
2
x
15°
45°
x
20. Construct an isosceles right triangle with legs
of length a, construct a 30°-60°-90° triangle with
legs of lengths a and a Ïw
3 , and construct a right
triangle with legs of lengths a Ïw
2 and a Ïw
3.
1
2
4
3
3
8
a
1
2
2
3
4
4
4
14. Possible answer:
Use 22 1 12 5 1Ïw
5 22 and 62 1 32 5 1Ïw
45 22.
5
a 3
1
45
5
3
4
3
6
15a. nCDA, nAEC, nAEB, nBFA, nBFC
15b. nMDB, nMEB, nMEC, nMFC, nMFA
108
a 5
a 2
21.
8p 5 12.5p; that is, the sum of the areas of the
semicircles on the two legs is equal to the area of
the semicircle on the hypotenuse.
73
22. }6} < 12.2 chih
23. Extend the rays that form the right angle.
m/4 1 m/5 5 180° by the Linear Pair
Conjecture, and it’s given that m/5 5 90°.
[ m/4 5 90°. m/2 1 m/3 1 m/4 5 m/2 1
m/3 1 90° 5 180°. [ m/2 1 m/3 5 90°.
m/3 5 m/1 by AIA. [ m/1 1 m/2 5 90°.
1
2
a 3
Areas: 4.5p cm2, 8p cm2, 12.5p cm2. 4.5p 1
1
2
2a
a
a
3
4
a 2
24. 80°
2
LESSON 9.4
1. No. The space diagonal of the box is about
33.5 in.
2. 10 m
3. 50 km/h
4. 8 ft
5. area: 60 m2; cost: $7200
6. surface area of prism 5 127Ïw
3 1 1802 cm2 <
2
226.8 cm ; surface area of cylinder 5 78p cm2 <
245.0 cm2
36 cm < 20.8 cm
7. }
3
Ïw
8. 48.2 ft; 16.6 lb
9a. 160 ft-lb
9b. 40 lb
9c. 20 lb
10. about 4.6 ft
2 2
2 2
11.
2
2
4
2
2
2
2
2
12.
13. 12 units
14. 18Ïw
2 cm
## and CB
#. /ABC 5 /ADC 5
15. Draw radii CD
90°. For quadrilateral ABCD, 54° 1 90° 1 m/C 1
X 5 126° but
90° 5 360°, so m/C 5 126°. BD
126° 1 226° Þ 360°.
16. 14, 4Ïw
32
17. SAA
18. orthocenter
19. 115°
#
20. PO
2
4
2
2
2
4
4
2
2
109
LESSON 9.5
1. 5 units
4. 354 m
7. rectangle
2. 45 units
5. 52.4 units
3. 34 units
6. isosceles
y
10
A
8
B
6
4
2
D
2
C
4
6
x
8 10
8. parallelogram
1
E2
4
x
F
H
28
6
G
210
9. kite
y
L
4
2
K
28
x
I 22
22
J
10. square
y
120°
8 P
M
2
Ï3w 1
19. 2}2}, }2}
6
20. k 5 Ïw
2 , m 5 Ïw
6
12
21. x 5 }, y 5 }
3
3
Ïw
Ïw
22. 96 cm
23. The angle of rotation is approximately 77°.
Connect two pairs of corresponding points.
Construct the perpendicular bisector of each
segment. The point where the perpendicular
bisectors meet is the center of rotation.
24. Any long diagonal of a regular hexagon
divides it into two congruent quadrilaterals. Each
(6 2 2) ? 180°
angle of a regular hexagon is }
} 5 120°, and
6
the diagonal divides two of the 120° angles into 60°
angles. Look at the diagonal as a transversal.
The alternate interior angles are congruent, thus
the opposite sides of a regular hexagon are parallel.
y
22
22
11b. Circle A: (x 2 1)2 1 ( y 1 2)2 5 64;
Circle B: x 2 1 (y 2 2)2 5 36
11c. (x 2 h)2 1 (y 2 k)2 5 (r)2
12. (x 2 2)2 1 y 2 5 25
13. Center is (0, 1), r 5 9.
14. (x 2 3)2 1 (y 1 1)2 5 18
15a. Ïw
14 units
15b. Ïw
176 5 4Ïw
11 units
2 1 (y 2 y )2 1 (z 2 z )2
15c. Ï(x
2
x
)
w
w
w
w
w
w
2
1
2
1
2
1
16. < 86.5 units
17. 14 units
18. n ? (n 1 2) 2 3 5 (n 1 3) ? (n 2 1)
60°
60°
120°
4
120°
O
N
24 22
11a.
2
60°
60°
x
4
y
y
(x, y)
(1, –2)
x
(x, y)
Circle A
110
(0, 2)
x
Circle B
120°
LESSON 9.6
18p cm2 < 56.5 cm2
(8p 2 16) m2 < 9.1 m2
456p cm2 < 1433 cm2
120p cm2 < 377.0 cm2
132p 2 32Ïw
3 2 m2 < 45.1 m2
(25p 2 48) cm2 < 30.5 cm2
64p
7. }3} 2 16Ïw
3 cm2 < 39.3 cm2
8. (240 2 18p) m2 < 183.5 m2
9. 102 cm
10. Possible proof:
## and AN
##
Given: Circle C with tangents AM
## > A
##
Show: AM
N
1.
2.
3.
4.
5.
6.
1
2
14. 112 1 6Ïw
3 2 cm < 22.4 cm
15. Ïw
77 cm < 8.8 cm
16. 76 cm
17. Inscribed circle: 3p cm2. Circumscribed circle:
12p cm2. The area of the circumscribed circle is four
times as great as the area of the inscribed circle.
18. 135°
19. (x 2 3)2 1 (y 2 3)2 5 36
20. The diameter is the transversal, and the
chords are parallel by the Converse of the Parallel
Lines Conjecture. The chords are congruent
because they can be shown to be the same distance
from the center. (Draw a perpendicular from each
chord to the center and use AAS and CPCTC.)
Sample construction:
M
C
A
##, NC
##, and AC
# to the diagram as shown.
Add MC
By the Tangent Conjecture, /M and /N are right
angles, so nAMC and nANC are right triangles.
Using the Pythagorean Theorem, (AM)2 1 (MC)2
5 (AC)2 and (AN)2 1 (NC)2 5 (AC)2. Solving for
AM and AN, AM 5 Ïw
(AC)2 2w
(MC)2 and
2
2
## and NC
## are
(AC) 2w
(NC) . Because MC
AN 5 Ïw
radii of the same circle, they have the same lengths.
So, substitute MC for NC in the second equation:
AN 5 Ïw
(AC)2 2w
(MC)2 5 AM. Therefore,
## > AN
##.
AM
11. 18 m
12. 324p cm2 < 1018 cm2
13. < 3931 cm2
21a. Because a carpenter’s square has a right angle
and both radii are perpendicular to the tangents, a
square is formed. The radius is 10 in., therefore the
diameter is 20 in.
10 in.
10
in.
10 in.
10 in.
d 5 20 in.
21b. Possible answer: Measure the circumference
with string and divide by p.
5
22. }6}
111
N
CHAPTER 9 REVIEW
1.
2.
3.
4.
20 cm
10 cm
obtuse
26 cm
3 1
Ïw
5. }2}, }2}
6. 2}1}, 2}1}
2
2
Ïw
Ïw
7. 200Ïw
3 cm2 < 346.4 cm2
8. d 5 12Ïw
2 cm < 17.0 cm2
9. 246 cm2
10. 72p in2 < 226.2 in2
11. 24p cm2 < 75.4 cm2
12. (2p 2 4) cm2 < 2.28 cm2
13. 222.8 cm2
14. isosceles right
15. No. The closest she can come to camp is 10 km.
16. No. The 15 cm diagonal is the longer diagonal.
17. 1.4 km; 8}21} min
18. yes
19. 29 ft
20. < 45 ft
21. 50 mi
22. < 707 m2
23. 6Ïw
3 and 18
24. 12 m
25. 42
26. No. If you reflect one of the right triangles into
the center piece, you’ll see that the area of the kite is
almost half again as large as the area of each of the
other triangles.
1
1
2
7
27. }9}
28. The quarter-circle gives the maximum area.
Triangle:
s
____
2
45°
s
2
45°
s
____
2
s
1
A 5 }2} ? }
2
Ïw
s
2
s
5 }4}
?}
2
Ïw
Square:
_1 s
2
_1 s
2
1 1
s2
A 5 }2}s ? }2}s 5 }4}
Quarter-circle:
s
2s
___
p
1
s 5 }4} ? 2pr
2s
r 5 }p}
1 2s 2 s 2
A 5 }4} p }p} 5 }p}
1 2
s 2 s2
}} . }}
p
4
Extra
30°
29. 1.6 m
30.
30°
30°
Or students might compare areas by assuming the
short leg of the 30°-60°-90° triangle is 1. The area
3
Ïw
of each triangle is then }
< 0.87 and the area of
2
the kite is 13 2 Ïw
3 2 < 1.27.
112
31. 4; 0; 10. The rule is }n2} if n is even, but 0 if n is
odd.
32. 4p in./s < 12.6 in./s
33. true
34. true
35. False. The hypotenuse is of length xÏw
2.
36. true
2
37. false; AB 5 Ïwx
(x 2 2 w
(y2 2 w
y1) 2
1) 1 w
38. False. A glide reflection is a combination of a
translation and a reflection.
39. False. Equilateral triangles, squares, and regular
hexagons can be used to create monohedral
tessellations.
40. true
41. D
42. B
43. A
44. C
45. C
46. D
48.
N
8-ball
W
B
E
Cue ball
A
S
49. 34 cm2; 22 1 4Ïw
2 < 27.7 cm
40p
50. }3} cm2 < 41.9 cm2
51. about 55.9 m
52. about 61.5 cm2
53. 48 cm
54. 322 ft2
1
ABCD is
a rectangle
Given
2
ABCD is
a parallelogram
Definition of
rectangle
4
/D > /B
Definition of
rectangle
3 #### ####
DA i CB
Definition of
parallelogram
5
/DAC > /BCA
AIA Conjecture
7
nABC > nCDA
SAA Congruence
Conjecture
6 #### ####
AC > AC
Same segment
113
10 CHAPTER 10 • CHAPTER
24.
LESSON 10.1
1. polyhedron; polygonal; triangles
2. nPQR, nTUS
3. PQUT, QRSU, RPTS
##, PT
#, RS
#
4. QU
5. 6 cm
6. GYPTAN
7. point E
#, YE
#, PE
#, TE
#, AE
#, NE
#
8. GE
9. 13 cm
10. D
11. L
12. C
13. G
14. B
15. H
16. E
17. A
18. J
19. J
20. M
21. H
22. I
23.
25.
x
2x
26.
x
2x
27. true
28. False. This statement is true only for a right
prism.
29. true
30. true
31. False. It is a sector of a circle.
32. true
33. false; counterexample:
34. true
35. true
114
39. 60
40. 30
41a. yes
Possible answers include that the number of lateral
faces of an antiprism is always twice the number
for the related prism; that the number of vertices is
the same for each related prism and antiprism; and
that the number of edges for a prism is three times
the number of faces, while for an antiprism, the
number of edges is twice the number of faces.
37. Answer should include the idea that the
painting “disappears” into the view out the window.
Students might also note the effect created by the
cone-shaped tower appearing similar to the road
disappearing into the distance.
38. 8
a
a
41b. yes
41c. no
41d. yes
3
3
36. (Lesson 10.1)
Triangular
prism
Rectangular
prism
Pentagonal
prism
Hexagonal
prism
Lateral
faces
3
4
5
6
...
n
Total
faces
5
6
7
8
...
n12
Edges
9
12
15
18
...
3n
Vertices
6
8
10
12
...
2n
Triangular
antiprism
Rectangular
antiprism
Pentagonal
antiprism
Hexagonal
antiprism
Lateral
faces
6
8
10
12
...
2n
Total
faces
8
10
12
14
...
2n 1 2
Edges
12
16
20
24
...
4n
Vertices
6
8
10
12
...
2n
n-gonal
prism
n-gonal
antiprism
115
20.
LESSON 10.2
1.
4.
5.
6.
7.
8.
72 cm3
2.
24 cm3
3.
108 cm3
r
r
160p cm3 < 502.65 cm3
r
36p cm3 < 113.10 cm3
324p cm3 < 1017.88 cm3
See table below.
960 in3
9. QTp cubic units
21. true
22. false
12 in.
Q
4 in.
T
23.
24 in.
8 in.
10. sample answer:
24. possible solution: prism
12
12
12
8
3
2
11. 2x 3
12. 3pr 3
13. 13x 3
14. Margaretta has room for 0.5625 cord. She
should order a half cord.
15. 170 yd3
16. 5100 lb
17. 11,140
18. The volume of the quilt in 1996 was 4000 ft3.
The quilt panels were stacked 2 ft 8 in. high.
19.
Salt crystal
25. approximately 1.89 m
26. 12 2 6Ïw
2
7. (Lesson 10.2)
Information about
base of solid
Height
of solid
Right triangular
prism
Right rectangular
prism
h
H
116
Right trapezoidal
prism
b2
h
b
Right
cylinder
b
H
r
h
H
b
H
b 5 6, b2 5 7,
h 5 8, r 5 3
H 5 20
a. V 5
480 cm3
d. V 5
960 cm3
g. V 5
1040 cm3
j. V 5
180p cm3
b 5 9, b2 5 12,
h 5 12, r 5 6
H 5 20
b. V 5
1080 cm3
e. V 5
2160 cm3
h. V 5
2520 cm3
k. V 5
720p cm3
b 5 8, b2 5 19,
h 5 18, r 5 8
H 5 23
c. V 5
1656 cm3
f. V 5
3312 cm3
i. V 5
5589 cm3
l. V 5
1472p cm3
14. 78,375 grams
15. 48 in3
16. 4p units3
17. 144x 3 cm3
18. 40,200 gal; 44 h 40 min
19. 71 ft 3
20. 403 barrels
21a. 16Ïw
3 cm2
21b. 96 cm2
21c. 80Ïw
3 cm2
21d. 124Ïw
13 1 1202 cm2
A
22.
D
LESSON 10.3
192 cm3
84p cm3 < 263.9 cm3
150 cm3
60 cm3
84p cm3 < 263.9 cm3
384p cm3 < 1206 cm3
m3
7. }3} cm3
2
8. }3} pb 3 cm3
9. 324px 3 cm3; 29.6%
1
11. V 5 }3}M 2H ft 3
1.
2.
3.
4.
5.
6.
D'
3 1
2
C
B
4
H
M
D''
Possible answer: From the properties of reflection,
/1 > /3 and /2 > /4. m/1 1 m/2 5 90°, so
m/3 1 m/4 5 90°, and m/1 1 m/2 1 m/3 1
m/4 5 180°. Therefore D9, C, and D0 are collinear.
23a. Y (a 1 c, d)
23b. Y (a 1 c, b 1 d)
23c. Y (a 1 c 2 e, b 1 d 2 f )
12. sample answer:
27
16
3
48
13. Mount Etna is larger. The volume for Mount
Etna is approximately 2193 km3, and the volume
for Mount Fuji is approximately 169 km3.
10. (Lesson 10.3)
Information about
base of solid
Height
of solid
Triangular
pyramid
Trapezoidal
pyramid
H
H
b
Rectangular
pyramid
H
h
b2
H
r
h
h
b
Cone
b
b 5 6, b2 5 7,
h 5 6, r 5 3
H 5 20
a. V 5
120 cm3
d. V 5
240 cm3
g. V 5
260 cm3
j. V 5
60p cm3
b 5 9, b2 5 22,
h 5 8, r 5 6
H 5 20
b. V 5
240 cm3
e. V 5
480 cm3
h. V 5
k. V 5
240p cm3
b 5 13, b2 5 29,
h 5 17, r 5 8
H 5 24
c. V 5
884 cm3
f. V 5
1768 cm3
i. V 5
2856 cm3
l. V 5
512p cm3
2480
}} cm3
3
117
M
LESSON 10.4
1.
2.
3.
4.
5.
6.
58.5 in3
32Ïw
3 cm3 < 55.43 cm3
15 cm
11 cm
5.0 cm
8.5 in. 2
3
}
p1}
2p 2 ? 11 in. < 63.24 in ;
11 in. 2
3
}
p1}
2p 2 ? 8.5 in. < 81.85 in
The short, fat cylinder has greater volume.
7. 257 ft 3
8. 4 cm
9. He must refute the statement.
10. 1502 lb
11. 192.4 gal
12. 13 min
13. Answers will vary, but pr 2H should equal
about 14.4 in3.
14. approximately 38 in3
15. 100,000p m 3, or about 314,159 m3;
16,528 loads
# > EC
# because the opposite sides of a
16. AB
# > BD
# because
parallelogram are congruent. EC
the diagonals of a rectangle are congruent. So,
# > BD
# because both are congruent to EC
#.
AB
Therefore, nABD is isosceles.
118
17. 8.2 cm
18. x 5 96°
19.
20.
D
45°
C
x
x
x
45°
45°
21a.
21b.
21c.
21d.
21e.
45°
x
A
A
S
N
S
A
x
3x
x
B
LESSON 10.5
1. 675 cm3
2. 36p cm3 < 113.1 cm3
3. < 47 in3
4. 1798.4 g
5. The gold has mass 2728.5 g, and the platinum
has mass 7529.8 g. The solid cone of platinum has
more mass.
6. 1.5 cm
7. 10.5 g/cm3; silver
8. 8000 cm3
9. The volume of the medallion is 160 cm3.Yes, it
is gold, and the Colonel is who he says he is.
10. 679 cm3
11. approximately 193 lb; 22 fish
9
12. }2} Ïw
3
13. flowchart proof:
#### i SP
###
QR
Given
/R > /S
#### > MR
####
SM
Given
/QMR > /PMS
Vertical Angles
AIA
nRMQ > nSMP
ASA
#### > MP
####
MQ
CPCTC
####
M is the midpoint of PQ
Definition of midpoint
14. 15 sides
15a. (1, 3)
15b. (x 2 1)2 1 (y 2 3)2 5 25
16. 58; 3n 2 2
119
LESSON 10.6
1. 36p cm3 < 113.1 cm3
p
2. }6} cm3 < 0.533 cm3
9p
3
3
3. }3}
2 cm < 0.884 cm
4. 720p cm3 < 2262 cm3
5. 30p cm3 < 94.3 cm3
6. 3456p cm3 < 10,857 cm3
7. 18p m3 < 56.6 cm3
8. No. The volume of the ice cream is 85.3#p cm3,
and the volume of the cone is 64p cm3.
9. only 20 scoops
148p
10. }3} m3, or about 155 m3
11. They have the same volume.
12. 9 in.
13. 3 cm
8192p
14. }3} cm3 < 8579 cm3
15. 18p in3, or about 57 in 3
16. No. The unused volume is 16p cm3, and the
volume of the golf ball is 10.6# p cm3.
120
17. approximately 15,704 gallons; 53 days
18. lithium
19. 31 ft
20. ABCD is a parallelogram because the slopes of
## and AB
# are both 0 and the slopes of BC
# and
CD
## are both }35}.
AD
y
C (3, 5) D (9, 5)
x
B (–3, –5)
A (3, –5)
21. They trace two similar shapes, except that the
one traced by C is smaller by a scale factor of 1:2.
22. The line traces an infinite hourglass shape. Or,
it traces the region between the two branches of a
hyperbola.
23. w 5 110°, x 5 115°, y 5 80°
1. V 5 972p cm3 < 3054 cm3
S 5 324p cm2 < 1018 cm2
2. V 5 0.972p cm3 < 3.054 cm3
S 5 3.24p cm2 < 10.18 cm2
3. V 5 1152p cm3 < 3619 cm 3
S 5 432p cm2 < 1357 cm2
4. S 5 160p cm2 < 502.7 cm2
256p
5. V 5 }3} cm3 < 268.1 cm3
6. S 5 144p cm2 < 452.4 cm2
7. Area of great circle 5 pr 2. Total surface area
of hemisphere 5 3pr 2. Total surface area of
hemisphere is three times that of area of great
circle.
8. 2 gal
9. V 5 }21} ? }34}p(1.8)3 1 }21}p(1.8) 2 (4.0) 5
10.368p m3 < 32.57 m3; S 5 }21} ? 4p(1.8) 2 1
1
}} ? 2p(1.8)(4.0) 5 13.68p m2 < 42.98 m 2
2
10a. approximately 3082 ft2
10b. 13 gal
10c. approximately 9568 bushels
11. 153,200,000 km2
12. The total cost is $131.95. He will stay under
budget.
13. 1.13%
14. 150p cm3 < 471.2 cm3
1
15. }4}
19a. (Lesson 10.7)
n
f(n)
1
2
3
4
5
6
...
n
...
200
22
1
4
7
10
13
. . . 3n 2 5 . . .
595
19b. (Lesson 10.7)
n
1
2
3
4
5
6
...
n
...
200
f(n)
0
1
}}
3
1
}}
2
3
}}
5
2
}}
3
5
}}
7
n21
}
... }
n11 . . .
199
}}
20 1
121
1
16. }2}
3
17. }4}
18. The ratio gets closer to 1.
19a and 19b. See tables below.
# > CB
# and AD
## > CD
## by the definition of
20. AB
#
#
rhombus, and BD > BD because it is the same
segment; therefore nABD > nCBD by SSS. By
CPCTC, /2 > /3 and /1 > /4, which shows
# bisects both /ABC and /ADC. Because
that BD
all four sides of the rhombus are congruent, a
similar proof can be used to show that nABC >
nADC and thus that /A and /C are both
#.
bisected by diagonal AC
# > CB
# by the definition of rhombus and
21a. AB
# > BE
# because it is the same segment. /1 > /2
BE
by the Rhombus Angles Conjecture. Therefore,
nAEB > nCEB by SAS.
# > CE
# by CPCTC, so BD
# bisects AC
#.
21b. AE
21c. /3 > /4 by CPCTC. Also, /3 and /4 form
a linear pair, so they are supplementary. Because
two angles that are congruent and supplementary
are right angles, /3 and /4 are right angles.
21d. Because /3 and /4 are right angles, the
diagonals are perpendicular.You still need to show
# bisects BD
#. Use a proof similar to that
that AC
given in 21a to show that nAEB > nAED. Then,
# > DE
#, which shows that AC
#
by CPCTC, BE
#.
bisects BD
LESSON 10.7
A
1. h 5 }b}
P 2 2h
P
2. b 5 }2} or b 5 }2} 2 h
3. r 5
3V
}
!}§
pH
2 2 a2
4. b 5 Ïcw
2 • SA
5. a 5 }} 2 l
P
6. y2 5 m1x 2 2 x12 1 y1 or y2 5 mx 2 2 mx 1 1 y1
d
7. v 5 }t}
The original formula gives distance in terms of
velocity and time.
9
8. F 5 }5}C 1 32
The original formula converts degrees Fahrenheit to
degrees Celsius.
T 2
9. L 5 g }2}
The original formula gives the period of a
pendulum (time of one complete swing) in terms
of length and acceleration due to gravity.
12
10a. V 5 2F 1 E 1 2
10b. See table below.
11a. The corresponding radii are approximately
3.63 cm, 3.30 cm, 3.02 cm, and 2.79 cm.
11b. The corresponding heights are approximately
9.32 cm,10.49 cm,11.61 cm,and 12.70 cm.
11c. The corresponding volumes are approximately
129 cm 3, 120 cm 3, 111 cm 3, and 104 cm 3.
11d. Answers will vary. Sample answer: The cone
with slant height 10 cm has the widest radius, so a
scoop of ice cream is least likely to fall off, and that
cone also has the greatest volume.
11e. Answers will vary. Sample answer: The cone
with slant height 13 cm has the greatest height, so
the cone appears bigger even though it has the same
surface area as the other cones; that cone also has
the smallest radius and smallest volume, so it could
hold less ice cream and still appear to be a bigger
cone.
m1 1 m2 1 m3 1 f 1 f
12a. A 5 }}}
5
12b. average of 60: 31; average of 70: 56; average
of 80: 81; average of 90: 106 (impossible)
10b. (Using Your Algebra Skills 10)
Pentahedron
Hexahedron
Octahedron
Decahedron
Dodecahedron
Number of faces
5
6
8
10
12
Number of edges
8
10
12
16
20
Number of vertices
5
6
6
8
10
122
CHAPTER 10 REVIEW
27.
3Ïw
3
m3 < 0.23 m3
1}p8} 2 }
32 2
28. 160p cubic units
1. They have the same formula for volume: V 5 BH.
2. They have the same formula for volume: V 5 }31}BH.
3. 6240 cm3
4. 1029p cm3 < 3233 cm3
5. 1200 cm3
6. 32 cm3
7. 100p cm3 < 314.2 cm3
8. 2250p cm3 < 7069 cm3
9. H 5 12.8 cm
10. h 5 7 cm
11. r 5 12 cm
12. r 5 8 cm
13. 960 cm3
14. 9 m
15. 851 cm3
16. four times as great
17a. Vextra large 5 54 in3
Vjumbo < 201.1 in3
Vcolossal < 785.4 in3
17b. 14.5 times as great
18. Cylinder B weighs }38} times as much as cylinder A.
19. 2129 kg; 9 loads
Vsphere }34} pr 3
20. H52r.}}5 }
(2r)3 < 0.524. Thus, 52.4% of
Vbox
the box is filled by the ball.
21. approximately 358 yd3
22. No. The unused volume is 98p in3, and the
volume of the meatballs is 32p in3.
23. platinum
24. No. The ball weighs 253 lb.
25. 256 lb
26. approximately 3 in.
123
3 3
1. }8}; }5}
AC 3 CD 5 BD
8
} }}, }} }}, }} }}
2. }
CD 5 5 BD 5 8 BC 5 13
3
3a. }1}
9
3b. }1}
4. a 5 6
5. b 5 16
6. c 5 39
7. x 5 5.6
8. y 5 68
9. x 5 12
10. z 5 6
11. d 5 1
12. y 5 5
13. 318 mi
14. 2.01
15. 12 ft by 15 ft
124
16. almost 80 years old
17a. true
3 1
17b. false; }6} 5 }2}, but 3 ? 1 Þ 6 ? 2
17c. true
17d. true
3 1
3 1
17e. false; }6} 5 }2}, but }2} Þ }6}
17f. true
18a. arithmetic: 10, 25, 40, 65; geometric: 10, 20,
40, 80 or 10, 220, 40, 280
18b. arithmetic: 2, 26, 50, 74; geometric: 2, 10, 50,
250 or 2, 210, 50, 2250
18c. arithmetic: 4, 20, 36, 52; geometric: 4, 12, 36,
108 or 4, 212, 36, 2108
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 5 Ïab
w; This formula holds for all positive
values of a and b.
19d. c 5 Ïw
2 • 50 5 Ï100
w 5 10;
c 5 Ïw
4 • 36 5 Ï144
w 5 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 5 6; RA 5 10; RG 5 4; KN 5 6
8. No; the corresponding angles are congruent,
but the corresponding sides are not proportional.
9. NY 5 21; YC 5 42; CM 5 27; MB 5 30
10. Yes; the corresponding angles are congruent,
and the corresponding sides are proportional.
11. x 5 6 cm, y 5 3.5 cm
2
12. z 5 10}3} cm
13. Yes, the corresponding angles are congruent.
Yes, the corresponding sides are proportional.Yes,
nAED , nABC.
9
9
14. m 5 }2} cm 5 4.5 cm, n 5 }4} cm 5 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: Ïw
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 $1,162.50 5 $2,987.50; and Omar,
$3,812.50.
23a.
23b.
19. 36
20. bc
60°
60°
24. approximately 92 gallons
125
5. possible answer:
x
LESSON 11.2
1.
2.
3.
4.
6 cm
40 cm; 40 cm
28 cm
54 cm; 42 cm
37 35
}}
5. No, }3}
0 Þ 28 .
6. Yes, nMOY , nNOT by SAS.
7. Yes, nPHY , nYHT because YH 5 12 and
20
16
12
}} 5 }} 5 }} (SSS).Yes, nPTY is a right triangle
15
12
9
because 202 1 152 5 25 2.
8. nTMR , nTHM , nMHR by AA.
x < 15.1 cm, y < 52.9 cm, h < 28.2 cm
9. Yes,/QTA > /TUR and /QAT > /ARU.
nQTA , nQUR by AA; 6}32} cm
10. 24 cm; 40 cm
11. Yes, /THU > /GDU and /HTU > /DGU;
52 cm; 42 cm
12. nSUN , nTAN by AA; 13 cm; 20 cm
13. Yes, /RGO > /FRG and /GOF > /RFO.
nGOS , nRFS by AA; 28 cm; 120 cm
14. 20 cm; 21 cm
15. x 5 50, y 5 9
126
2 21
Ïw
16. r 5 R } 5 R1Ïw
2 2 122
Ï2w 1 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 , A9B9C9D9. The ratio of the
perimeters is }12}. The ratio of the areas is }14}.
1
2
y
D' C'
A'
B'
x
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
nPRE , nPOC by AA.
8. 300 cm
6
13. nGHF , nFHK , nGFK by AA; h 5 18}1}
3,
9
4
}}
x 5 7}1}
3 , y 5 44 13 .
14. sample answer:
1
A 5 }2}(8.2)(1.7) 5 6.97 cm2
1
A 5 }2}(3)(4.6) 5 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
2
15. 5}3}
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} 5 }xy}. Finally, use the Pythagorean
Theorem to find the length of the guy
wire: Ïw
t 2 1 y 2w.
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. nMUN , nMSA by AA; x 5 31}3}.
12. nBDC , nAEC by AA; y 5 63.
C
Given: Parallelogram ABCD
# > CD
## and AD
## > BC
#
Show: AB
ABCD is a parallelogram
Given
#### i BC
####
AD
### i CD
####
AB
Definition of
parallelogram
Definition of
parallelogram
#### > BD
####
BD
/2 > /4
/1 > /3
Same segment
AIA Conjecture
AIA Conjecture
nABD > nCDB
ASA
#### > BC
####
AD
#### > CD
####
AB
CPCTC
CPCTC
17a. 4.6.12
17b. 3.12.12 or 3.12 2
127
1 + Ïw5
18. The golden ratio is }
}, or approximately
2
1.618. Here is one possible construction:
C
D
A
X
B
m/B 5 90°
1
Construct BC 5 }2}AB
Ï5
w
}
AC 5 }
2 AB
1
Ï5
w
}}AB
}
AX 5 AD 5 }
2 AB 2 2
Ï5
w21
} AB
AX 5 }
2
AB
1 + Ï5w
}} 5 }2}
}}
AX Ï5w 2 1 5 2
Therefore, X is the golden cut.
128
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.
c
a
17. }b} 1 1 5 }d} 1 1
a b c
d
}} 1 }} 5 }} 1 }}
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. 4}9} cm
p b
10. }q}; }q}
a1b c1d
}} 5 }}
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:
1 2
L
O
V
M
A
19. Yes, by the SSS or the SAS Similarity
Conjecture.
20a. 2a 5 b
20b. 2a 5 b
20c. all values of a and b
20d. no values of a and b
21. AB 5 3 cm, BC 5 7.5 cm
11. 6Ïw
3 cm
12. 6 cm
13
5Ïw
1
2
}
13. x 5 3}3} cm, y 5 }
3} cm, z 5 8}3 cm
3
k 7
14. B 5 (3, 5), R 5 1}4}, 7 ; }h} 5 }4}
2
Ïw
15. }1}
H
16.
E
T
Consider similar triangles nLVE and nMTH with
# and HA
##. To
corresponding angle bisectors EO
show that the corresponding angle bisectors are
proportional to corresponding sides, for example
EO
EL
}} 5 }}, show by AA that nLOE , nMAH,
HA
HM
EL
}
and then you can show that }HEO}A 5 }
HM . You
know that /L > /M. Use algebra to show that
/LEO > /MHA.
129
LESSON 11.5
1. 18 cm2
4. 27p cm2
2. 18
1
5. }4}
9
3. 5; 10
9
12
}},
16d. x 5 9, y 5 16, }1}
(9)(16)w 5
2 5 16 h 5 Ïw
5
12
144
Ïw
17.
7.1 m 14.2 m
14.2 m 6.5 m
6. 1: 3
1 2
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.
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) 5 2x 2
y
x
Area
in cm2
1
2
60
2
8
50
3
18
4
32
5
50
6
72
70
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
14. }3}, }3}
15. 8
16a. (i) 40°; 50°; 40°
(ii) 60°; 30°; 60°
(iii) 22°; 68°; 22°
Conjecture: similar; right triangle
h
s
h
b
16b. (i) }h}
(ii) }x}
(iii) }m} or }a}
p h
Conjecture: }} 5 }q}
h
16c. h 5 Ïpq
w
130
H
E
L
10
Area in cm2
x
Area
in cm2
Area in cm2
12a. a(x) 5 2x
O
V
M
A
T
Consider similar triangles nLVE and nMTH with
# and HA
##. Show
corresponding altitudes EO
EO
EL
}
that nLOE , nMAH, then }H}A 5 }
HM . You know
# and ##
HA are altitudes,
that /L > /M. Because EO
/LOE and /MAH are both right angles, and
/LOE > /MAH.
EL
}
So, nLOE , nMAH by AA. Thus }HEO}A 5 }
HM ,
which shows that the corresponding altitudes
are proportional to corresponding sides.
19. x 5 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 1}86}00 Þ }
135 2, 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 }61}4 .
LESSON 11.6
1. 1715 cm3
1 2
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 1 242 5 302 (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 5 3x 5 3 , or 2:3
y
800 Volume
600
Surface area
400
200
5
x
131
2. 16 cm, 4 cm; 768p cm3 < 2412.7 cm3;
64
12p cm3 < 37.7 cm3; }1}
3 125
3
3. }5}; }2};
7 1500 cm
H 3 64
}};
4. 1944p ft3 < 6107.3 ft3; }2}
4 5 27 32 ft
125
5. }2}
7
2
6. 2:5
7. }3}
8. $1,953.13
9. 2432 lb
10. Possible answer: No, a 4-foot chicken, similar
to a 14-inch 7-pound chicken, would weigh
3
approximately 282 pounds 1}4184}3 5 }7x}2. It is unlikely
that the legs of the giant chicken would be able to
support its weight.
11. Possible answer: Assuming the body types of
the African goliath frog and the Brazilian gold frog
are similar, the gold frog would weigh about
0.0001 kg, or 0.1 g.
12. surface area ratio 5 }11}6 , volume ratio 5 }61}4 . The
dolphin has the greater surface area to volume
ratio.
13. The ratio of the volumes is }21}7 . The ratio of the
surface areas is }91}.
14.
S(x) 5 22x 2 and V(x) 5 6x 3.
Possible answer: The surface area equation is
quadratic, so the graph is a parabola, and the
volume equation is cubic, so the graph is a cubic
graph.
15. Possible proof: The volume of the first
rectangular prism is lwh. The volume of the second
rectangular prism is rl ? rw ? rh, or r 3lwh. The ratio
r3lwh
3
}
of the volumes is }
lwh , or r .
3
16. 9,120p m or approximately 28,651 m3
s
s sÏ2w 2 s
17. }} 2 }4} or }4}
2Ïw
2
LESSON 11.7
1
2. 33}3} cm
3. 45 cm
4. 21 cm
5. 28 cm
6. no
7. José’s method is correct. Possible explanation:
Alex’s first ratio compares only part of a side of the
larger triangle to the entire corresponding side of the
smaller triangle, while the second ratio compares
entire corresponding sides of the triangles.
8. yes
9. 6 cm; 4.5 cm
10. 13.3# cm; 21.6 cm
11. yes; no; no
12. yes; yes; yes
13. 3Ïw
2 cm; 6Ïw
2 cm
14.
1. 5 cm
E
15.
F
I
J
So two pairs of corresponding sides of nXYZ and
nXAB are proportional. /X > /X, so
nXYZ , nXAB by the SAS Similarity Conjecture.
Because nXYZ , nXAB, /XAB > /XYZ.
# by the Converse of the Parallel
@#$ i 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 5 45 ft, y 5 40 ft, z 5 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. }3}pr
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.
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. 2064p cm3 < 6484 cm3
19. possible proof:
a b
}} 5 }}
d
c
ad 5 cb
ad 1 ab 5 cb 1 ab
a(d 1 b) 5 b(c 1 a)
a(d 1 b) b(c 1 a)
}} 5 }}
ab
ab
d1b c1a
}} 5 }}
b
a
132
P
Q
A
C
S
R
D
Given: Circumscribed quadrilateral ABCD, with
points of tangency P, Q, R, and S
Show: AB 1 DC 5 AD 1 BC
Use segment addition to show that each sum of
lengths of opposite sides is composed of four lengths:
AB 1 DC 5 (AP 1 BP) 1 (DR 1 CR) and AD 1
BC 5 (AS 1 DS) 1 (BQ 1 CQ). Using the Tangent
Segments Conjecture, AP 5 AS, BP 5 BQ,
CR 5 CQ, and DR 5 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 1 DC 5 (AP 1 BP) 1 (DR 1 CR) Segment
addition.
5 (AS 1 BQ) 1 (DS 1 CQ) Substitute AS
for AP, BQ
for BP, DS for
DR, and CQ
for CR.
5 (AS 1 DS) 1 (BQ 1 CQ) Regroup the
measurements
by common
points of
tangency.
AB 1 DC 5 AD 1 BC
Use segment
addition to
rewrite the
right side as
the other sum
of opposite
sides.
133
CHAPTER 11 REVIEW
1. x 5 24
2. x 5 66
3. x 5 66
4. x 5 17
5. 6 cm; 4.5 cm; 7.5 cm; 3 cm
1
1
6. 4}6} cm; 7}2} cm
7. 13 ft 2 in.
8. It would still be a 20° angle.
9.
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
}} 5 }}
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 5 Hsculpture’s photo
1
134
2
14. 9:49
5 125
15. }4}; }6}
4
16. $266.67
17. 640p cm3
18. 32; 24; 40; 126
19. 841 coconuts
p 1
20a. 1 to }4} to }2}, or 4 to p 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.
LESSON 12.1
1.
2.
3.
4.
5.
6.
0.6018
0.8746
0.1405
11.57
30.86
62.08
s
s
r
7. sin A 5 }t}; cos A 5 }t}; tan A 5 }r}
4
3
4
8. sin u 5 }5}; cos u 5 }5}; tan u 5 }3}
24
7
7
}};
}};
9. sin A 5 }2};
5 cos A 5 25 tan A 5 24
10. 30°
11. 53°
30°
24°
a < 35 cm
b < 15 cm
c < 105 yd
d < 40°
e < 50 cm
f < 33°
g < 18 in.
approximately 237 m
x < 121 ft
6.375
2.2
16-inch pizza
box of ice cream
8Ïw
3 cm < 13.9 cm
V 5 288p ft3 < 905 ft 3, S 5 144p ft 2 < 452 ft 3
135
7
24
24
}};
}}
sin B 5 }2};
5 cos B 5 25 tan B 5 7
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
LESSON 12.2
1. 9 cm
2. 64°
3. 7 cm
4. 24 m
5. 22 in.
6. 49°
7. 127°
8. 30°
9. 64 cm
10. approximately 655 m
16a. approximately 974 m
16b. approximately 1007 m
16c. Yes, the height of the balloon would be 1 m
less because you don’t have to account for the
tripod. The distance to a point under the balloon
would not change.
136
17. 45°
18. 60°
19. 60°
20a. 15.04
20b. 2.05
20c. 5.2304
20d. < 2.644
1
21. x 5 3.5, y 5 9}7} < 9.14
22. The block has a volume of 90 cm3 but it
displaces only 62.8 cm3 of water. So not all of the
block is under water, which means it floats.
23. CZ , AX , BY
24a. decreases, approaching 0
24b. decreases, approaching 0
24c. remains 90°
24d. increases, approaching (AO)2
24e. decreases, approaching 1
24f. decreases, approaching 1
25. R(8, 7) and E(3, 9), or R(4, 23) and
E(21, 21)
LESSON 12.3
1. 329 cm2
2. 4 cm2
3. 11,839 cm2
4. 407 cm2
5. 35 cm
6. 17 cm
7. 30 cm
8. 56°
9. 45°
10. 66°
11a. approximately 2200 m
11b. approximately 1600 m
11c. approximately 1400 m
12. The other two walls were 300 ft and
approximately 413 ft. The area was approximately
45,000 ft2.
14. 2366 cm; approximately 41°
15. approximately 5°
9pÏ3w 3
3
16. }}
8 cm , or approximately 6 cm
# i CD
##, /A > /D by the AIA
17. Because AB
Conjecture. Because /D and /B intercept the
same arc, /D > /B. Therefore /A > /B by the
transitive property. So nABE is isosceles by the
Converse of the Isosceles Triangle Conjecture.
18. Draw line through the two points. Then fold
the paper so that the two points coincide. Draw
another line along the fold. These two lines contain
the diagonals of the square. Now fold the paper so
that the two lines coincide. Mark the vertices on
the other line.
19. AC 5 60 cm, AE 5 93.75 cm, AF < 117 cm
20. Box 1; about 1.2 in. longer
137
LESSON 12.4
1. 32 cm
2. 47 cm
3. 341 cm
4. 74°
5. 64°
6. 85°
7. approximately 43 cm
8. approximately 30°
9. approximately 116° and 64°
10. approximately 6 min
11. approximately 87.8 ft
14. The ladder at approximately a 76° angle is not
safe. The ladder should be at least 6.5 ft and no
more than 14.3 ft from the base of the wall.
15.
c
A
a
b
a
sin A 5 }c}
b
cos A 5 }c}
a
tan A 5 }b}
a
}}
sin A
a c a
c
}} 5 } 5 }} ? }} 5 }} 5 tan A
cos A
c b b
b
}}
c
16a. PR 5 x Ïw
3
16b. Since x2 1 1x Ïw
3 22 5 (2x)2, nTPR is a right
triangle by the Converse of the Pythagorean
Theorem. Therefore m/TPR 5 90°.
17a. increases
17b. decreases
17c. increases then decreases
18a. The base perimeters are equal.
18b. The cone has the greater volume.
18c. The cone has the greater surface area.
19a. a translation right 10 units;
(x, y) → (x 1 10, y)
19b. a rotation 180° about the origin;
(x, y) → (2x, 2y)
138
x
y
x
2y
x
120°
y
120°
120°
x
x
60°
60°
120°
y
y
60°
60°
2y
2x
22. Answers will vary from simple (circles and
segments) to quite complex. See below for a more
detailed answer.
Students might begin with the simplest case where
all factors are equal. With circles of the same radius
and with endpoints of the segment traveling in the
same direction, at the same speed, and starting
from the same relative position, the midpoint will
trace a circle equal in size to those constructed.
Changing the size of both circles will change the
size of the circle traced, as will changing the
starting positions of the endpoints. The maximum
radius of the traced circle is limited by the radius of
the constructed circles; the smallest observable
trace is a single point.
With all other factors equal and the endpoints of
the segment traveling in opposite directions
around the circles, the midpoint will trace a
segment. Changing the relative speeds of the
endpoints, or the relative radii of the circles, will
produce more complex polar curves, which
students might describe as flowers, radii, or
“spirograph-like.” Changing the distance between
the centers has no effect on the shape of the trace,
only its position.
LESSON 12.5
2. approximately 142 mi/h
3. approximately 51 km
5a. approximately 9.1 km
5b. approximately 255°
7. approximately 42 ft
8. They must dig at an angle of approximately
71.6° and dig for approximately 25.3 m.
9. approximately 168 km/h
6 cm
10.
For simplicity, let u 5 }326n}0 . Now use the tangent
ratio to find the length of the apothem.
s
}}
2
tan u 5 }
a
s
}
a5}
2 tan u
Now use the area formula for a regular polygon.
1
A 5 }2}aP
1
s
}
A 5 }2} ? }
2 tan u ? ns
2
360
ns
}
}}
A5}
4 tan u where u 5 2n
13. approximately 12.8 m at an angle of
approximately 48°
14. 1536Ïw
3 cm3 < 2660 cm3
15. The area increases by a factor of 9.
y
(0, 2)
10a. apothem < 9.23 cm; area < 277 cm2
10b. leg < 9.7 cm; area of triangle < 27.7 cm2;
area of decagon < 277 cm2
10c. They are the same.
11. approximately 108 cm3
12. Sample answer: Any regular polygon can be
divided into n congruent isosceles triangles, each
360°
}
with a vertex angle of }
n .
360°
___
n
The altitude from the vertex angle of an isosceles
triangle bisects the angle and the opposite side.
This altitude is also an apothem of the regular
polygon.
–6
_1
A5 2
(1, 0)
_1
A5 2
? 1 ? 2 5 1x
6
?3?659
–6 (0, –6)
5 5 1 15
}
16. }4} Þ }12
17a. decreases then increases
17b. does not change
18a. a reflection across the x-axis;
(x, y) → (x, 2y)
18b. a reflection across the y-axis; (x, y) → (2x, y)
360°
___
___ 360°
2n 2n
a
_s
2
_s
2
139
(–3, 0)
4.
y
2
1.
y
x
–5
6
–2
4
–4
f(x) 5 x
2
–6 –4 –2
2
4
6
–6
x
p(x) 5 2(x 1 3) 2 5
–2
The graph of y 5 2x 1 1 is the same as the
graph of p(x). The slope of a line is equal to
the vertical stretch of the linear function. The
y-intercept is equal to 2ah 1 k.
5. a vertical stretch by a factor of 2 and a vertical
translation down 5 units
–4
–6
y
y
10
6
8
4
6
2
4
–2
2
–2
g(x) 5 x 2
x
2
–2
h(x) 5 x 3
x
2
y
2
–3
3
–4
–2
–6
–4
x
f(x) 5 2x3 2 5
–6
2.
y
–8
4
f(x) 5 |x|
2
–5
5
g(x) 5 |x 2 5|
x
10
–2
6. a horizontal translation right 5 units and a
vertical translation up 2 units
y
g(x) 5 (x 2 5)3 1 2
6
h(x) 5 |x | 2 5
–4
4
2
All three graphs are V-shaped, consisting of two
parts that have slopes of 21 and 1. Graph g is a
translation of graph f right 5 units. Graph h is a
translation of graph f down 5 units.
y
3.
2
–4 –2
–2
2
8
x
7. a vertical stretch by a factor of }21} (a vertical
shrink), a horizontal translation right 2 units,
and a vertical translation down 5 units
x
2 4
f(x) 5 2x2
y
1
–4 –2
–4
6
–2
2
4
6
8
x
–2
–6
–6
y
h(x) 5 12 |x 2 2| 2 5
2
–4 –2
2
4
x
–2
–4
g(x) 5 2|x|
–6
y
2
–4 –2
2
–2
–4
4
x
h(x) 5 2x 3
A vertical stretch by a factor of 21 reflects the
graph across the x-axis.
140
8. f(x) 5 |x 2 3| 2 1; a horizontal translation
right 3 units and a vertical translation down 1 unit
9. f(x) 5 3x2; a vertical stretch by a factor of 3
10. f(x) 5 2(x 2 3)2; a vertical stretch by a factor
of 21 (a reflection across the x-axis) and a horizontal translation right 3 units
11. f(x) 5 22|x| 1 3; A vertical stretch by a factor
of 22 and a vertical translation up 3 units
12. f(x) 5 3(x 1 1)2 2 3; A vertical stretch by a factor of 3, a horizontal translation left 1 unit, and a
vertical translation down 3 units
13. f(x) 5 }21}(x 2 3)2 1 2; A vertical stretch by a factor of }21} (a vertical shrink), a horizontal translation
right 3 units, and a vertical translation up 2 units
14. Both graphs have the same “hills and valleys”
shape, but one is a horizontal translation of the
other.
17. Possible answer: f(x) 5 2 sin(x 2 90°) 21;
a vertical stretch by a factor of 2, a horizontal
translation right 90°, and a vertical translation
down 1 unit
y
p(x) 5 sin(x)
2
x
360
–360
–2
q(x) 5 cos(x)
15. a horizontal translation; a 5 1, h 5 90°, k 5 0
16. The dashed function in each graph is the parent
function f(x) 5 sin(x).
y
p(x) 5 2sin(x)
y
2
2
–360
360
x
360
–360
x
–2
q (x) 5 _12 sin(x)
y
r(x) = –sin(x)
2
360
–360
x
–2
Varying a in the equation of the sine function
causes the same types of vertical stretches as with
other functions.
141
CHAPTER 12 REVIEW
1. 0.8387
2. 0.9877
3. 28.6363
a c a
4. }b}; }b}; }c}
8 15 8
}}; }}
5. }1};
7 17 15
s
6. s; t; }t}
7. 33°
8. 86°
9. 71°
10. 1823 cm2
11. 15,116 cm3
12. Yes, the plan meets the act’s requirements. The
angle of ascent is approximately 4.3°.
13. approximately 52 km
14. approximately 7.3°
15. approximately 22 ft
17. approximately 2973 km/h
18. 393 cm2
19. 30 cm
20. 78°
21. 105 cm
22. 51°
23. 759 cm2
25. 72 cm2
26. approximately 15.7 cm
27. approximately 33.5 cm2
28. approximately 10.1 km/h at an approximate
bearing of 24.5°
29. False; an octahedron is a polyhedron with
eight faces.
30. false
33.
34.
35.
36.
37.
38.
39.
40.
41.
2
False; the ratio of their areas is }mn}2 .
true
length of leg opposite /T
false; tangent of /T 5 }}}}
length of leg adjacent to /T
true
true
true
true
False; the slope of line , 2 is 2}m1}.
false
42. B
43. C
44. A
45. D
46. B
47. B
48. A
49. B
50. C
51. D
52. C
53. A
100p
54. }3} cm3
55. 28p cm3
56. 30.5p cm3
57. 33
58. (x, y) → (x 1 1, y 2 3)
59. w 5 48 cm, x 5 24 cm, y 5 28.5 cm
61. (x 2 5)2 1 (y 2 1)2 5 9
62. Sample answer: Each interior angle in a regular
pentagon is 108°. Three angles would have a sum of
324°, or 36° short of 360, which would leave a gap.
Four angles would have a sum exceeding 360° and
hence create an overlap.
64. 30 ft
65. 4 cm
B
66.
A
31. true
32. true
142
D
C
66a. m/ABC 5 2 ? m/ABD
# is the perpendicular
66b. Possible answers: BD
#. It is the angle bisector of /ABC, it
bisector of AC
is a median, and it divides nABC into two
congruent right triangles.
LESSON 13.1
1. A postulate is a statement accepted as true
without proof. A theorem is deduced from other
theorems or postulates.
2. Subtraction: Equals minus equals are equal.
Multiplication: Equals times equals are equal.
Division: Equals divided by nonzero equals are equal.
3. Reflexive: Any figure is congruent to itself.
C
nABC > nABC
B
A
Transitive: If Figure A is congruent to Figure B and
Figure B is congruent to Figure C, then Figure A is
congruent to Figure C.
P
Q
>
R
S
>
X
, then
Y
P
Q
>
X
Y
.
# > RS
# and RS
# > XY
#, then PQ
# > XY
#.
If PQ
Symmetric: If Figure A is congruent to Figure B,
then Figure B is congruent to Figure A.
X
L
>
If
,
Y
Z
L
then
M
N
X
>
.
M
N
Y
Z
If /XYZ > /LMN, then /LMN > /XYZ.
4. reflexive property of equality; reflexive
property of congruence
5. transitive property of congruence
19. (Lesson 13.1)
# and BO
# are radii
AO
1
2
?
Given
AO > BO
nAOB is isosceles
3 ?
Definition of ?
isosceles
Definition of circle
20. (Lesson 13.1)
1
/1 > ? /2
? Given
2
? min
Corresponding
Angles Postulate
3
/3 > ? /4
? Corresponding
Angles Postulate
143
If
6. subtraction property of equality
7. division property of equality
8. Distributive; Subtraction; Addition; Division
9. Given; Addition property of equality;
Multiplication property of equality; Commutative
property of addition.
10. true, definition of midpoint
11. true, Midpoint Postulate
12. true, definition of angle bisector
13. true, Angle Bisector Postulate
14. false, Line Intersection Postulate
15. false, Line Postulate
16. true, Angle Addition Postulate
17. true, Segment Addition Postulate
18.
• That all men are created equal.
• That they are endowed by their creator with
certain inalienable rights, that among these are life,
liberty, and the pursuit of happiness.
• That to secure these rights, governments are
instituted among men, deriving their just powers
from the consent of the governed.
• That whenever any form of government becomes
destructive to these ends, it is the right of the
people to alter or to abolish it, and to institute new
government, laying its foundation on such
principles and organizing its powers in such
form as to them shall seem most likely to effect
their safety and happiness.
28. 30 ft
29. FG 5 2Ïw
6 and DG 5 2Ïw
3 because ABGF
and BCDG are parallelograms. Triangle FGD is
right (m/FGD 5 90°) by the Converse of the
Pythagorean Theorem because 12Ïw
6 22 1 12Ïw
3 22
5 62. But m/FGB 5 128° and m/DGB 5 140° by
conjectures regarding angles in a parallelogram.
So, m/FGD 5 92° because the sum of the angles
around G is 360°. So, m/FGD is both 90° and 92°.
30. x 5 54°, y 5 126°, a 5 7.3 m
p
31a. }4}
p
31b. 1 2 }4}
p
31c. }2} 2 1
23. (2n 2 1) 1 (2m 2 1) 5 2n 1 2m 2 2 5
2(n 1 m 2 1)
24. (2n 2 1)(2m 2 1) 5 4nm 2 2n 2 2m 1 1 5
4nm 2 2n 2 2m 1 2 2 1 5
2(2nm 2 n 2 m 1 1) 2 1
25. Let n be any integer. Then the next two
consecutive integers are n 1 1 and n 1 2. The sum
of these three integers is (n) 1 (n 1 1) 1 (n 1 2) 5
n 1 n 1 1 1 n 1 2. Combining like terms: 3n 1 3 5
3(n 1 1), which is divisible by 3.
26. <299 m
27. sphere, cylinder, cone; cylinder, cone, sphere;
cone, cylinder, sphere
21. (Lesson 13.1)
1
AC > BD
? Given
/D > /C
nBAD
2
4
## > BC
#
? AD
? Given
3
5
nABC > ?
? SSS Congruence
Postulate
?
CPCTC
AB > BA
Reflexive property
of congruence
22. (Lesson 13.1)
2 ? AB
# > CB
#
Given
BD
bisector ####
Angle Bisector
Postulate
3
/ABD > ?
5
nBAD > ?
? Reflexive property
of congruence
6
?
? SAS Congruence ? CPCTC
Definition
of ? angle bisector Postulate
4 #### ####
BD > BD
144
/A > /C
nBCD
/CBD
1 Construct angle
LESSON 13.2
10.
,3
1
1. Linear Pair Postulate
2. Parallel Postulate, Angle Addition Postulate,
Linear Pair Postulate, Corresponding Angles
Postulate
3. Parallel Postulate
4. Perpendicular Postulate
5.
4
Use the VA Theorem and the transitive property to
get /3 > /4. Therefore the lines are parallel by the
Converse of the AIA Theorem.
,3
11.
2
Use the definition of supplementary angles and the
substitution property to get m/1 1 m/1 5 180°.
Then use the division property and the definition
of a right angle.
6.
1 2
3 4
1
2
Use the definition of a right angle and the
transitive property to get m/1 5 m/2. Then use
the definition of congruence to get /1 > /2.
,3
8.
,1
1
2
,2
Use the Linear Pair Postulate and the definition of
supplementary angles to get m/1 1 m/3 5 180°.
Then use the CA Postulate and the substitution
property to get m/1 1 m/2 5 180°.
,3
12.
,1
31
2
,2
supplementary angles to get m/1 1 m/3 5
m/1 1 m/2. Then use the subtraction property
and the Converse of the AIA Theorem to get
,1 i ,2.
13.
,1
A
1
B 2
,2
3
,3
,2
3
Use the VA Theorem and the transitive property to
get /1 > /3. Therefore the lines are parallel by the
CA Postulate.
,3
9.
1
,1
,1
3
,2
Construct a transversal across lines ,1 and ,2; it
will intersect line ,3 by the Parallel Postulate. Use
the Interior Supplements Theorem and the
definition of supplementary angles to get m/1 1
m/2 5 180°. Use the CA Postulate and the
substitution property to get m/1 1 m/3 5 180°.
Therefore ,1 i ,3 by the Converse of the Interior
Supplements Theorem.
2
Use the VA Theorem and the CA Postulate to get
/1 > /3 and /2 > /3. Then use the transitive
property to get /1 > /2.
145
Use the definition of supplementary angles and the
transitive property to get m/1 1 m/2 5 m/3 1
m/4. Then use the substitution property and the
subtraction property to get m/1 5 m/4.
7.
,2
2
3
1
21
,1
3
14.
,3
1
2
,1
,2
Use the definition of perpendicular lines and the
transitive property to get m/1 5 m/2. Therefore
lines ,1 and ,2 are parallel by the Converse of the
AIA Theorem.
15. 2
1
3
By the Triangle Sum Theorem, m/1 1 m/2 1
m/3 5 180°. By the definition of a right triangle,
/1 is a right angle. By the definition of right
angle, m/1 5 90°. Using the subtraction property,
m/2 1 m/3 5 90°, so by the definition of
complementary angles, /2 and /3 are
complementary.
16.
Linear Pair
Post.
VA Thm.
CA Post.
Converse of
AIA Thm.
Converse of
AEA Thm.
146
17. 1066 cm3
18.
Area (ft2)
Bottom
6
2 sides
9
Back and front
6
2 rooftops
< 8}12}
2 gable ends
2
Total
31}12}
Plywood (4 by 8)
32
The area is less than that of one sheet of plywood.
However, it is impossible to cut the correct size
pieces from one piece. This answer assumes that
the bottom of the dog house is included. If the
bottom is not included, and the gables can be cut
separately from the rectangular part of the front
and back, then the pieces can be cut from a single
sheet of plywood.
19. A9(6, 210), B9(22, 26), C9(0, 0); mapping
rule: (x, y) → (2x 2 8, 2y 2 2)
Isosceles Triangle Theorem (CB 5 CA), the reflexive
property, and the SSS Congruence Postulate to get
nACP > nBCP. Therefore, /ACP > /BCP by
CPCTC.
6. n C
m
LESSON 13.3
1. Case 1: P is collinear with A and B. Use the Line
Intersection Postulate to show that P and E are the
same point and use the definitions of bisector and
midpoint to get AP 5 BP.
Case 2: P is not collinear with A
and B. Use the SAS Congruence Postulate to get
# > BP
#.
nAEP > nBEP. Then use CPCTC to get AP
P
A
Use the Line Intersection Postulate and the
Perpendicular Bisector Theorem to get AP 5 BP
and BP 5 CP. Then use the transitive property and
the Converse of the Perpendicular Bisector
Theorem to prove that point P is on line n.
C
7.
C
P
A
B
,
B
E
m
D
,
n
Q
A
Use the Line Intersection Postulate and the Angle
Bisector Theorem to prove that Q is equally distant
# and AC
# and from AB
# and BC
#. Then use the
from AB
transitive property and the Converse of the Angle
Bisector Theorem to prove that point Q is on line n.
B
8.
P
A
B
E
2
Case 2: P is not collinear with A and B. Draw
# . Use the SSS Congruence
midpoint E and PE
Postulate to get nAEP > nBEP. Then use CPCTC
and the Congruent and Supplementary Theorem
to prove that /AEP and /BEP are both right
# is the perpendicular bisector
angles. Therefore PE
# by the definitions of midpoint and
of AB
perpendicular.
C
3. Use the reflexive property and
the SSS Congruence Postulate to get
nABC > nBAC. Therefore,
/A > /B by CPCTC.
1
5.
3 4
A
C
B
P
C
A
Draw line BA.
Use the Isosceles Triangle Theorem to get
/PAB > /PBA. Use the Angle Addition
Postulate and the subtraction property to get
/BAC > /ABC. Then use the Converse of the
4
C
supplementary angles to get m/3 1 m/4 5 180°.
Then use the Triangle Sum Theorem and the
transitive property to get m/1 1 m/2 1
m/3 5 m/3 1 m/4. Therefore, m/1 1
m/2 5 m/4 by the subtraction property.
D
9.
B
A
3
A
A
4. Use the reflexive property and
the ASA Congruence Postulate to
get nABC > nBAC. Then use
CPCTC and the definition of
isosceles triangle.
B
1
C
2
B
B
Use the Triangle Sum Theorem and the addition
property to get m/A 1 m/1 1 m/3 1 m/C 1
m/4 1 m/2 5 360°. Then use the Angle Addition
Postulate and the substitution property to get
m/A 1 m/ABC 1 m/C 1 m/CDA 5 360°.
C
10. Use the definitions of median
and midpoint to get BM 5 }21}BC and
N
M
AN 5 }21}AC. Then use the multiplication property and the substitution
B
A
## > BM
##. By the
property to get AN
reflexive property, the Isosceles Triangle Theorem,
and the SAS Congruence Postulate, nABN >
# > AM
## by CPCTC.
nBAM. Therefore, BN
147
2. Case 1: P is collinear with A and B. Use the
definitions of congruence and midpoint to show
#. Then use the
that P is the midpoint of AB
definition of perpendicular bisector.
11.
median by CPCTC and the definitions of midpoint
and median.
C
Q
P
C
B
A
Use the Angle Addition Postulate and the
definition of angle bisector to get
m/PAB 5 }21} m/CAB and
m/QBA 5 }21} m/CBA. Then use the Isosceles
Triangle Theorem, the multiplication property, and
the substitution property to get /PAB > /QBA.
By the reflexive property and the ASA Congruence
# > BQ
#
Postulate, nABP > nBAQ. Therefore, AP
by CPCTC.
C
12.
T
S
B
A
Use the Isosceles Triangle Theorem, the Right
Angles Are Congruent Theorem, and the SAA
Theorem to get nABT > nBAS. Therefore,
# > BT
# by CPCTC.
AS
C
13.
A
B
D
median → angle bisector
Use the definitions of median, midpoint, and
isosceles triangle, the reflexive property, and the
SSS Congruence Postulate to prove that
nADC > nBDC. Then use CPCTC and
the definition of angle bisector.
A
D
B
angle bisector → altitude
Use the definitions of angle bisector and isosceles
triangle, the reflexive property, and the SAS
Congruence Postulate to get nADC > nBDC.
Then use CPCTC, the Linear Pair Postulate, and
the Congruent and Supplementary Theorem to
prove that /ADC and /BDC are both right
## is the altitude by the
angles. Therefore CD
definitions of perpendicular and altitude.
14. x 5 6, y 5 3
15. <21.9 m
16. first image: (0, 3); second image: (6, 1)
17. BC 5 FC makes ABCF a rhombus, so its
diagonals are perpendicular. m/FGC 5 90°, so
# ' GE
#, so by
m/CFG 1 m/FCG 5 90°. FD
subtraction and transitivity m/2 5 m/FCG.
/1 > /FCG by AIA, so /1 > /2.
p Ï3w
w
18a. Ï3
18b. }p}
18c. }} 2 }
}
4
3
4
6
19. One possible sequence:
1. Fold A onto B and crease. Label as ,1. Label the
midpoint of the arc M.
2. Fold line ,1 onto itself so that M is on the crease.
Label as ,2.
X, and , is the desired
M is the midpoint of AB
2
tangent.
,1
C
M
,2
A
D
B
altitude → median
Use the Right Angles Are Congruent Theorem, the
Isosceles Triangle Theorem, and the SAA Theorem
## is the
to get nADC > nBDC. Therefore, CD
148
A
B
20. m/BAC ø 13°
21a. B
21b. B
1
1
1
22a. x 5 }2}(a 1 c), y 5 }2}(a 1 b), z 5 }2}(b 1 c)
1
1
1
22b. w 5 }2}(a 1 b), x 5 }2}(b 1 e), y 5 }2}(e 1 d),
1
z 5 }2}(d 1 a)
LESSON 13.4
1.
D
B
Use x to represent the measures of one pair of
congruent angles and y for the other pair. Use the
Quadrilateral Sum Theorem and the division
property to get x 1 y 5 180°. Therefore, the
opposite sides are parallel by the Converse of the
Interior Supplements Theorem.
C
2. D
3
2
4
B
A
8
2
C
A
B
5
3
4
B
A
Use the definition of rhombus, the reflexive
property, and the SSS Congruence Postulate to get
nABC > nADC. Then use CPCTC and the
# bisects
definition of angle bisector to prove that AC
/DAB and /BCD. Repeat the steps above using
#.
diagonal DB
C
4. D
Use the Converse of the Opposite Angles Theorem
to prove that ABCD is a parallelogram. Then use
the definition of rectangle.
C
7. D
A
B
Use the definition of rectangle to prove that ABCD
is a parallelogram and /DAB > /CBA. Then use
the Parallelogram Opposite Sides Theorem, the
reflexive property, and the SAS Congruence Postulate
to get nDAB > nCBA. Finish with CPCTC.
C
8. D
A
B
Use the Parallelogram Opposite Sides Theorem,
the reflexive property, and the SSS Congruence
Postulate to get nDAB > nCBA. Repeat the above
steps to get nADC > nCBA and nDAB > nBCD.
Then use CPCTC and the transitive property to get
/DAB > /ABC > /BCD > /ADC. Finish with
the Four Congruent Angles Rectangle Theorem.
D
C
9.
A
B
A
## i BC
#
Use the definition of parallelogram to get AD
# i DC
##. Then use the Interior Supplements
and AB
Theorem.
C
5. D 2
3
1
A
E
B
# i CB
#.
Use the Parallel Postulate to construct DE
Then use the Parallelogram Opposite Sides
Theorem and the transitive property to prove that
nAED is isosceles. Therefore, /A > /B by the
Isosceles Triangle Theorem, the CA Postulate, and
substitution.
D
C
10.
4
B
Use the reflexive property and the SSS Congruence
Postulate to get nABD > nCDB. Then use
CPCTC and the Converse of the AIA Theorem to
# i CD
## and AD
## i CB
#. Therefore,ABCD is a
get AB
rhombus by the definitions of parallelogram and
rhombus.
A
B
Use the Isosceles Trapezoid Theorem, the reflexive
property, and the SAS Congruence Postulate to get
# > BD
# by CPCTC.
nDAB > nCBA. Then AC
149
Use the AIA Theorem, the reflexive property, and
the SAS Congruence Postulate to get nADC >
nCBA. Then use CPCTC and the Converse of the
# i DC
##.
AIA Theorem to get AB
3. D 7 6 C
1
D
C
A
1
6.
11.
D
2
A
4
1
14.
C
3
Parallelogram
Diagonal Lemma
B
Use the Parallelogram Opposite Angles Theorem,
the multiplication property, and the definition of
angle bisector to get /1 > /3. Then use the
Converse of the Isosceles Triangle Theorem,
the definition of isosceles triangle, and the
Parallelogram Opposite Sides Theorem to get
# > BC
# > DC
## > AD
##.
AB
12.
W
Z
Q
X
P
ASA Congruence
Postulate
Y
Use the Converse of the Angle Bisector Theorem
## is the angle bisector of /Y. In
to prove that WY
##
like manner, WY is the angle bisector of /W.
Therefore, WXYZ is a rhombus by the Converse
of the Rhombus Angles Theorem.
CA Postulate
Interior Supplements Theorem
Parallelogram Consec. Angle Theorem
Opposite Sides
Theorem
Line Postulate
Converse of the
IT Theorem
2 diagonals
2 ' bisectors of sides
isosceles
trapezoid
150
Converse of the Angle
Bisector Theorem
15a. A
15b. S
15c. S
15d. N
16. 2386 ft2
V 2 has length 12.8 and bearing 72.6°.
18. Y
V1 1 Y
19a. B
19b. A
20a. 19°
20b. 52°
20c. 52°
20d. 232°
20e. 19°
Lines of symmetry
Rotational symmetry
none
2-fold
trapezoid
none
none
kite
1 diagonal
none
parallelogram
IT Theorem
Double-Edged
Straightedge Theorem
17. (Lesson 13.4)
Name
SSS Congruence
Postulate
Opposite Angles
Theorem
Converse of the Rhombus
Angles Theorem
Angle Addition
Postulate
4-fold
square
rectangle
2 ' bisectors of sides
2-fold
rhombus
2 diagonals
2-fold
1 ' bisector of sides
none
LESSON 13.5
B
C
D
O
E
A
10. a 5 75°, b 5 47°, c 5 58°
11. 42p ft3 < 132 ft 3
3
Ïw
p
12a. }
4 2 }1}
2
p
12b. 1 2 Ïw
3 1 }3}
13a. A
13b. N
13c. S
13d. S
13e. A
1. D: Paris is in France; Tucson is in the U.S.;
London is in England. Bamako must be the capital
of Mali.
2. C: The “Sir” in part A shows that Halley was
English; Julius Caesar was an emperor, not a
scientist; Madonna is a singer. Galileo Galilei must
be the answer.
3. No, the proof is claiming only that if two
particular angles are not congruent, then the two
particular sides opposite them are not congruent. It
still might be the case that a different pair of angles
are congruent and that therefore a different pair of
sides are congruent.
4. Yes, this statement is the contrapositive of the
conjecture proved in Example B, so they are
logically equivalent.
5. 1. Assume the opposite of the conclusion;
2. Triangle Sum Theorem; 3. Substitution property
of equality; 4. 0°; Subtraction property of equality
6. Assume ZOID is equiangular. Use the definition
of equiangular and the Four Congruent Angles
Rectangle Theorem to prove that ZOID is a
rectangle. Therefore ZOID is a parallelogram, which
creates a contradiction.
## is the altitude to AB
#. Use the
7. Assume CD
definitions of altitude, median, and midpoint, the
Right Angles Are Congruent Theorem, and the
SAS Congruence Postulate to get nADC > nBDC.
# > BC
#, which creates a contradiction.
Therefore AC
8. Assume ZO 5 ID. Use the Opposite Sides
Parallel and Congruent Theorem to prove that
ZOID is a parallelogram, which creates a
contradiction.
# and perpendicu9. Given: Circle O with chord AB
@#$
lar bisector CD
@#$ passes through O
Show: CD
@#$ does not pass through O. Use the Line
Assume CD
# and OA
# and the
Postulate to construct OB
#. Then use
Perpendicular Postulate to construct OE
the Isosceles Triangle Theorem, the Right Angles
Are Congruent Theorem, and the SAA Theorem to
get nOEA > nOEB. From CPCTC and the
definition of midpoint, prove that E is the midpoint
#, which creates a contradiction.
of AB
151
LESSON 13.6
1.
A
X
Use the Cyclic Quadrilateral Theorem, the
Opposite Angles Theorem, and the Congruent and
Supplementary Theorem to get /R, /C, /E, and
/T are right angles.
P
5.
1
S
B
2
W
Case 1 The same: Use the Inscribed Angle Theorem
and the transitive property to get /A > /B.
A
T
Z
Y
B
X
W
O
Case 2 Congruent: Use the multiplication
X 5 }1} mWX
X. Then follow
property to get }12} mYZ
2
the steps in Case 1.
2. D
#, OT
#, and
Use the Line Postulate to construct OS
#. Then use the Tangent Theorem, the Converse
OP
of the Angle Bisector Theorem, and the SAA
# > PT
# by
Theorem to get nOSP > nOTP. PS
CPCTC.
6.
A
D
3
1
2
E
C
C
B
B
A
Use the Inscribed Angle Theorem, the addition
property, and the distributive property to get
X 1 mDAB
X . Then use
m/A 1 m/C 5 }12} 1mBCD
2
the definition of degrees in a circle, the substitution
property, and the definition of supplementary
angles to get /A and /C are supplementary.
Repeat the steps using m/B and m/D to get /B
and /D are supplementary.
3.
B
##. Then use
Use the Line Postulate to construct AD
the Inscribed Angle Theorem, the addition
X 1 mBD
X . Therefore,
m/2 1 m/3 5 }12} 1mAC
2
X 1 mBD
X by the Triangle Exterior
m/1 5 }12} 1mAC
2
Angle Theorem and the transitive property.
P
7.
x
A
b
B
C
O
a
A
D
D
C
##. Then use
Use the Line Postulate to construct AD
the AIA Theorem, the Inscribed Angle Theorem,
X > BD
X.
and substitution to get AC
4.
T
C
R
E
152
Intersecting Secants Theorem: The measure of an
angle formed by two secants intersecting outside a
circle is half the difference of the measure of the
larger intercepted arc and the measure of the
smaller intercepted arc. Use the Triangle Exterior
Angle Theorem and the subtraction property to get
x 5 b 2 a. Then use the Inscribed Angle Theorem,
the substitution property, and the distributive
X 2 mAC
X.
property to get x 5 }12} 1mBD
2
8.
D
C
A
B
E
Use the Inscribed Angle Theorem to get
1 X
m/BDC 5 }2} mBEC
. By the definition of
X 5 mBDC
X 5 180°, so by the
semicircle, mBEC
division property, m/BDC 5 90°. By the
definition of right angle, /BDC is a right angle.
Because only definitions, properties, and the
Inscribed Angle Theorem are needed to prove this
conjecture, it is a corollary of the Inscribed Angle
Theorem.
9.
Angle Addition
Postulate
Line
Postulate
ASA Congruence
Postulate
SAA
Theorem
IT
Theorem
SAS Congruence
Postulate
Right Angles Are
Congruent Theorem
Converse of the Angle
Bisector Theorem
Midpoint
Postulate
Tangent
Theorem
h
c
a
b
P
16a. 27°
16b. 47°
16c. 67°
16d. 133°
16e. cannot be determined
16f. cannot be determined
17.
Perpendicular
Postulate
Tangent Segments
Theorem
x
A
O
10.
Linear Pair
Postulate
CA
Postulate
Parallel
Postulate
Angle Addition
Postulate
Converse of the
IT Theorem
SSS Congruence
Postulate
11. A(4.0, 2.9), P(21.1, 22.8)
12. 3; 1; 9
#, AB
#, AC
#, CD
##, AD
##
13. BC
14. 32.5
15. As long as point P is inside the triangle,
a 1 b 1 c 5 h.
Proof: Let x be the length of a side. The areas of the
three small triangles are }21} xa, }21} xb, and }21} xc. The area
of the large triangle is }21} xh. So, }21} xa 1 }21} xb 1 }21} xc 5
1
1
}} xh. Divide both sides by }} x. So, a 1 b 1 c 5 h.
2
2
M
P
Arc Addition
Postulate
B
SSS
Congruence
Postulate
VA
Theorem
IT
Theorem
AIA
Theorem
Triangle Sum
Theorem
ASA Congruence
Postulate
Exterior Angle
Sum Theorem
Inscribed Angle
Theorem
Cyclic Quadrilateral
Theorem
Parallelogram
Diagonal Lemma
Opposite Angles
Theorem
Congruent and
Supplementary Theorem
Right Angles
Are Congruent
Theorem
Steps:
#.
1. Construct OP
#. Label midpoint M.
2. Bisect OP
3. Construct circle with center M and radius PM.
4. Label the intersection of the two circles A and B.
# and PB
#. PA
# and PB
# are the
5. Construct PA
required tangents.
/OAP and /OBP are right angles because they are
inscribed in semicircles.
Parallelogram Inscribed
in a Circle Theorem
153
6.
LESSON 13.7
1.
C
V
T
1
A
B
H
A
G
L
E
Use the Right Angles Are Congruent Theorem,
CASTC, and the AA Similarity Postulate to get
nBHT , nGEV. Therefore, }GBVT} 5 }TVH}E by CSSTP. See
the Solutions Manual for the family tree.
A
G
2.
2
B
D
Use the Right Angles Are Congruent Theorem, the
reflexive property, and the AA Similarity Postulate
to get nADC , nACB and nACB , nCDB.
Therefore, nADC , nACB , nCDB by the
transitive property of similarity.
C
7.
h
S
L
M
B
Y
Use the definitions of median and midpoint, the
Segment Addition Postulate, the substitution
property, and the multiplication property to get
BY 5 }12}BI and SL 5 }12}SM. Then use CASTC,
CSSTP, and the SAS Similarity Theorem to get
nBYG , nSLA. Therefore, }BSAG} 5 }GA}YL by CSSTP. See
the Solutions Manual for the family tree.
3.
F
C
2 4
13
D
A
Q
E
B
P
Use the definition of angle bisector, the Angle
Addition Postulate, and the substitution property to
get m/ACB 5 2m/1 and m/DFE 5 2m/2. Then
use CASTC and the AA Similarity Postulate to get
nAPC , nDQF. Therefore, }DAC}F 5 }CFQ}P by CSSTP.
C
4.
D 1
A
E
2
A
I
B
Use the CA Postulate and the AA Similarity
Postulate to get nCDE , nCAB. Then use CSSTP
CD 1 DA
}5
and the Segment Addition Postulate to get }
CD
CE 1 EB
DA
EB
}}. Therefore, }} 5 }} by algebra.
CE
CD
CE
C
5.
x
Use the Three Similar Right Triangles Theorem to
get nADC , nCDB. Then use CSSTP to get
x
h
}} 5 }}.
h
y
8.
b
A
2
E
B
A
Use the addition property to get }DCD}
1 1 5 }CEB}E 1 1.
Then use algebra and the Segment Addition
Postulate to get }CCA}D 5 }CC}EB . Therefore nABC ,
nDEC by the SAS Similarity Theorem, /1 > /2
# by the CA Postulate.
@#$ i AB
by CASTC, and DE
154
a
d
c–d
c
Draw the altitude to the hypotenuse, then use the
ratios given by the Three Similar Right Triangles
a
c
2
2
}
}}
Theorem. In particular, }
c 2 d 5 a yields a 5 c 2 cd.
Now look at the other small triangle and use }bd} 5 }bc}
to get b 2 5 cd.
C
E
9.
a
B
b
a
A
c
b
D
x
F
Begin by constructing a second triangle, right
triangle DEF (with /E a right angle), with legs of
lengths a and b and hypotenuse of length x. The
plan is to show that x 5 c, so that the triangles
are congruent. Then show that /C and /E are
congruent. Once you show that /C is a right angle,
then nABC is a right triangle.
L
10. H
Y
D 1
B
y
D
P
E
G
Use the Pythagorean Theorem to write expressions
for the lengths of the unknown legs. Show that the
expressions are equivalent. The triangles are
congruent by SSS or SAS.
11.
CA
Postulate
Segment Addition
Postulate
AA Similarity
Postulate
Parallel/Proportionality
Theorem
12.
Segment Duplication
Postulate
Parallel
Postulate
CA
Postulate
AA Similarity
Postulate
SAS
Congruence
Postulate
SSS Similarity
Theorem
13.
Perpendicular
Postulate
AA Similarity
Postulate
Right Angles
Are Congruent
Theorem
Three Similar
Right Triangles
Theorem
Pythagorean
Theorem
14. approximately 1.5 cm or 6.5 cm
15a. C
15b. A
15c. A
15d. A
15e. C
16a. The vectors are diagonals of your quadrilateral.
16b. A 180° rotation about the midpoint of the
common side; the entire tessellation maps onto
itself.
20a. nCDG > nCFG by SAA; nGEA > nGEB
by SAS; nDGA > nFGB by the HL Theorem
## > CF
# and DA
## > FB
# by CPCTC; CD 1
20b. CD
DA 5 CF 1 FB (addition property of equality).
# > CB
#, and nABC is isosceles.
Therefore, CA
20c. The figure is inaccurate.
20d. The angle bisector does not intersect the
perpendicular bisector inside the triangle as
shown, except in the special case of an isosceles
triangle, when they coincide.
21. 173 cm, 346 cm, 20 stones. Draw the trapezoid
and extend the legs until they meet to form a
triangle. Use parallel proportionality to find the
rise. The span is twice the rise. Use inverse tangent
to find the central angle measure. Divide into 180°
to find the number of voussoirs.
155
SAS Similarity
Theorem
SSS
Congruence
Postulate
17a. a 180° rotation about the midpoint of any
side
17b. possible answer: a vector running from each
vertex of the quadrilateral to the opposite vertex
(or any multiple of that vector)
18a. 33°
18b. 66°
18c. 57°
18d. 62°
18e. /PSQ and /PTQ are 90° by the Tangent
Theorem and so are supplementary. So, /SPT
and /SQT must also be supplementary by the
Quadrilateral Sum Theorem. Therefore, opposite
angles are supplementary, so it’s cyclic.
# is a tangent, m/PSQ 5 90°
18f. Because PS
#
(Tangent Theorem). Because m/PSQ 5 90°, SQ
must be a tangent (Converse of the Tangent
Theorem).
3
Ïw
p
}
19a. }1}
1
21
2
2
3
p Ïw
19b. 2}6} 2 }
4 11
6. possible answer:
y
1. B(a, 0)
2. C(a 1 b, c)
P (c, d )
2 2 b2
3. C 1a 1 b, Ïw
a 2 2 b 2w2, D1b, Ïaw
w2
4. possible answer:
T (0, 0)
A (a – c, d )
R (a, 0)
x
y
C (a, b)
T (0, b)
R (0, 0)
020 0
#5}
} }}
slope of TR
a20 5 a 5 0
E (a, 0)
d20
d
#5}
} }}
slope of RA
a 2 c 2 a 5 2c
d2d
0
#5}
} }}
slope of AP
a 2 2c 5 a 2 2c 5 0
x
d20 d
#5}
} }}
slope of PT
c20 5 c
020 0
#5}
} }}
slope of RE
a20 5 a 5 0
b20 b
#5}
} }}
slope of CE
a2a 5 0
(undefined)
b2b 0
#5}
} }}
slope of TC
a20 5 a 5 0
b20 b
#5}
} }}
slope of TR
020 5 0
(undefined)
# and AP
# have the same slope and are parallel by
TR
the parallel slope property. So TRAP has only one
pair of parallel sides and is a trapezoid by definition.
Opposite sides have the same slope and are
therefore parallel by the parallel slope property.
Two sides are horizontal and two sides are vertical,
so the angles are all congruent right angles. RECT
is an equiangular parallelogram and is a rectangle
by definition.
5. possible answer:
PT 5 Ïw
(c 2 0w
)2 1 (dw
2 0)2 5 Ïw
c 2 1 d 2w
RA 5 Ïw
(a 2 cw
2 a)2w
1 (d 2w
0)2 5 Ïw
(2c)2 w
1 d2
2
2
5 Ïw
c 1 dw
The nonparallel sides of the trapezoid have the
same length. So trapezoid TRAP is isosceles by
definition.
y
7.
U (0, a 3)
E
(–a, 0)
Q
(a, 0)
x
y
I (b, c)
Z b, c
2 2
(
T (0, 0)
Y a+b, c
2
2
)
(
X a,0
2
(
)
R (a, 0)
)
x
#.
Let X be the midpoint of TR
a10 010
a
X 5 }2}, }2} 5 }2}, 0
#.
Let Y be the midpoint of RI
1
2 1 2
1
2 1
1
2 1 2
2
a1b 01c
a1b c
Y 5 }2}, }2} 5 }2}, }2}
#.
Let Z be the midpoint of TI
b10 c10
b c
Z 5 }2}, }2} 5 }2}, }2}
#, RI
#, and TI
#,
X, Y, and Z are the midpoints of TR
respectively, by the coordinate midpoint property.
#, YZ
#, and ZX
# are midsegments by definition.
So XY
156
Show: nEQU is equilateral
EQ 5 2a
EU 5 Ï(2a
2w
w
0)2 1wa
(0 2w
Ïw
3 )2
2 1 3a 2
5 Ïaw
w
2
5 Ï4a
w
5 2a
21 02a
UQ 5 Ïw
(a 2 0)w
w22
1 w
Ï3w
2 1 3a 2
5 Ïaw
w
5 Ïw
4a 2
5 2a
[ EQ 5 EU 5 UQ
[ nEQU is equilateral
8. Task 1: Given: A rectangle with both diagonals
Show: The diagonals are congruent
y
Task 2:
C (a, b)
T (0, b)
R (0, 0)
E (a, 0)
Task 2:
x
#
Task 3: Given: Rectangle RECT with diagonals RC
#
and TE
# > TE
#
Show: RC
Task 4: To show that two segments are congruent,
you use the distance formula to show that they
have the same length.
Task 5: RC 5 Ïw
(a 2 0w
)2 1 (bw
2 0)2 5 Ïw
a2 1 bw2
TE 5 Ïw
(a 2 0w
)2 1 (0w
2 b)2 5 Ïw
a2 1 b2w
I (c, d)
(
T (0, 0)
)
Y a+c, d
2
2
(
R (a, 0)
y
P (b, c)
M
T (0, 0)
A (d, c)
N
R (a, 0)
One possible set of the coordinates for TRAP is
shown in the figure. By the coordinate midpoint
property, the coordinates of M are 1}2b}, }2c}2 and of N
a1d c
} }}
are 1}
2 , 2 2.
Task 3: Given: Trapezoid TRAP with midsegment
##
MN
## i TR
#
Show: MN
Task 4: To show that the midsegment and bases
are parallel, you need to find their slopes.
Task 5:
## 5
slope of MN
)
c c
}} 2 }}
0
2
2 5 }}
50
}}
a1d b
a1d2b
}}
}} 2 }}
2
2
2
x
020 0
#5}
} }}
slope of TR
a20 5 a 5 0
#
Task 3: Given: nTRI and midsegment YZ
1
# i TR
# and YZ 5 }}TR
Show: YZ
2
Task 4: To show that two segments are parallel, use
the parallel slope property. The segments are
horizontal, so to compare lengths, subtract their
x-coordinates.
Task 5:
020 0
#5}
} }}
Slope of TR
a20 5 a 5 0
d d
}} 2 }}
0 50
2 2 5}
#
Slope of YZ 5 }}
a1c c
a
}} 2 }}
}}
2
2
2
The slopes are the same, so the segments are parallel
by the parallel slope property.
TR 5 a 2 0 5 a
a1c c a 1
1
YZ 5 }2} 2 }2} 5 }2} 5 }2}a 5 }2}TR
x
c2c
0
# is }
}
}}
The slope of PA
d 2 b 5 d 2 b 5 0. The slopes are
equal, therefore the lines are parallel.
11. Task 1: Given: A quadrilateral in which only
one diagonal is the perpendicular bisector of the
other
Show: The quadrilateral is a kite
y
Task 2:
I (0 , b)
T (–a, 0)
K (a, 0)
M(0, 0)
x
E (0, c)
Task 3: Given: Quadrilateral KITE with diagonal
#, which is the perpendicular bisector of diagonal
IE
#
TK
157
# > TE
# because both segments have the same
So RC
length. Therefore the diagonals of a rectangle are
congruent.
9. Task 1: Given: A triangle with one midsegment
Show: The midsegment is parallel to and half the
length of the third side
y
Task 2:
Z c,d
2 2
So the midsegment is half the length of the third
side. Therefore the midsegment of a triangle is
parallel to the third side and half the length of the
third side.
10. Task 1: Given: A trapezoid
Show: The midsegment is parallel to the bases
Show: KITE is a kite
Task 4: To show that a quadrilateral is a kite, you
use the distance formula to show that only two
pairs of adjacent sides have the same length.
Task 5:
2 1 ( b 2 0) 2 5
2 1 b2
KI 5 Ï(0
2 a)w
w
ww Ïaw
w
IT 5 Ïw
(0 2 (w
2a))2 1
(b 2w
0)2 5 Ïw
a2 1 b2w
w
TE 5 Ïw
(0 2 (w
2a))2 w
1 (c 2w
0)2 5 Ïw
a2 1 c2w
EK 5 Ï(a
2 0w
)2 1 (0w
2 c)2 5 Ïw
a 2 1 c 2w
w
# and IT
# have the same length and
Adjacent sides KI
#
#
adjacent sides TE and EK have the same length,
and because b Þ c the pairs are not equal in
length to each other. Therefore, KITE is a kite by
definition. Therefore, if only one diagonal of a
quadrilateral is the perpendicular bisector of the
other diagonal, then the quadrilateral is a kite.
12. Task 1: Given: A quadrilateral with midpoints
connected to form a second quadrilateral
Show: The second quadrilateral is a parallelogram
Task 2: y
D (d, e)
(
d, e
2 2
G
)
L b+d, c+e
2
2
A (b, c)
(
)
R a+b, c
2
2
(
Q (0, 0) P a , 0
2
( )
U (a, 0)
x
c
c
}} 2 0
}}
2
2
# 5 }} 5 } 5 }c}
Task 5: slope of PR
a1b a
b
b
}} 2 }}
}}
2
2
2
c1e c
e
}} 2 }}
}}
2
2
2
# 5 }} 5 } 5 }e}
slope of RL
b1d a1b
d2a
d2a
}} 2 }}
}}
2
2
2
e c1e
2c
}} 2 }}
}}
2
2
2
c
# 5 }} 5 }
slope of LG
5 }b}
d b1d
2b
}} 2 }}
}}
2
2
2
e
e
}} 2 0
}}
2
2
e
#5}
} 5 }}
slope of GP
5
d a
d2a
d
2
a
}} 2 }}
}}
2 2
2
D a2 , h2
(
)
A (a, h)
F 32a, h2
(
)
x
B (0, 0) E (a, 0) C (2a, 0)
Task 3: Given: Isosceles triangle ABC with
midpoint of base, E, and midpoints of legs, D and
F, connected to form quadrilateral ADEF.
Show: ADEF is a rhombus
Task 4: You need to show that all the sides of
ADEF have the same length.
Task 5: By the distance formula,
)
Task 3: Given: Quadrilateral QUAD with
midpoints P, R, L, and G
Show: PRLG is a parallelogram
Task 4: To show that a quadrilateral is a
parallelogram, we need to show that opposite
sides have the same slope.
158
# and LG
# have the same slope, and
Opposite sides PR
# and GP
# have the same slope. So
opposite sides RL
they are parallel by the parallel slope property, and
PRLG is a parallelogram by definition. Therefore
the figure formed by connecting the midpoints of
the sides of a quadrilateral is a parallelogram.
13. Task 1: Given: An isosceles triangle with the
midpoint of the base connected to the midpoint of
each leg, to form a quadrilateral
Show: The quadrilateral is a rhombus
Task 2: y
AD 5
1 2 §
1 2 !§
12 12
!§
2
a
a 2 }2}
h
1 h 2 }2}
2
5
a
}}
2
2
h
1 }2}
2
a 2 1 h w2
Ïw
5 }}
2
AF 5
1 2 §
1 2 !§
12 §
1 2§
!§
3a
}} 2 a
2
2
h
1 }2} 2 h
2
a
}}
2
5
2
h
1 2}2}
a 2 1 h w2
Ïw
5 }}
2
DE 5
2 1§2§ !§
12 §
1 2§
1 §
!§
a
a 2 }2}
2
h
1 0 2 }2}
2
5
a
}}
2
2
h
1 2}2}
2
a 2 1 h w2
Ïw
5 }}
2
EF 5
1 2 §
1 2 !§
12 12
!§
3a
}} 2 a
2
2
h
1 }2} 2 0
2
5
a
}}
2
2
h
1 }2}
a 2 1 h w2
Ïw
5 }}
2
AD 5 AF 5 DE 5 EF by the transitive property
of equality.
Therefore, ADEF is a rhombus by the definition of
a rhombus.
2
2
CHAPTER 13 REVIEW
1408
408
208
408
1408
208
208
208
20. False; x 5 360° 2 2b so x 5 90° only if
b 5 135°.
x
a
b
a
b
x 5 5408 2 (1808 1 2b)
x 5 3608 2 2b
21. True (except in the special case of an isosceles
right triangle, in which the segment is not defined
because the feet coincide).
C
/EAB > /DBA by the Isosceles
Triangle Theorem. /AEB >
E
D
/BDA by the definition of
X
altitude and by the Right Angles A
B
Are Congruent Theorem.
# > AB
# by the reflexive property of congruence,
AB
# > BD
# by
so nAEB > nBDA by SAA. [ AE
CPCTC. By the definition of congruence, AC 5 BC
and AE 5 BD, so EC 5 DC by the subtraction
property of equality and the Segment Addition
Postulate. By the division property of equality,
DC
EC
# divides the sides of nABC
}} 5 }}, so ED
BD
AE
# i AB
# by the
proportionally. Therefore ED
Converse of the Parallel/Proportionality Theorem.
22. true
## and
Given: Rhombus ROME with diagonals RM
#
EO intersecting at B
E
4 B
3
2
R
M
1
O
## ' EO
#
Show: RM
Because diagonals bisect the angles in a rhombus,
the diagonals are angle bisectors.
Statement
Reason
1. /1 > /2
1. Rhombus Angles
Theorem
2. RO 5 RE
2. Definition of rhombus
#
#
3. RO > RE
3. Definition of
congruence
# > RB
#
4. RB
4. Reflexive property of
congruence
5. nROB > nREB
5. SAS Congruence
Postulate
6. /3 > /4
6. CPCTC
7. /3 and /4 are
7. Definition of linear
a linear pair
pair
8. /3 and /4 are
supplementary
9. /3 and /4 are right 9. Congruent and
angles
Supplementary
Theorem
## ' EO
#
10. RM
10. Definition of
perpendicular
159
1. False. The quadrilateral could be an isosceles
trapezoid.
2. true
3. False. The figure could be an isosceles trapezoid
or a kite.
4. true
5. False. The angles are supplementary but not
necessarily congruent.
6. False. See Lesson 13.5, Example B.
7. true
8. perpendicular
9. congruent
10. the center of the circle
11. four congruent triangles that are similar to the
original triangle
12. an auxiliary theorem proven specifically to
help prove other theorems
13. If a segment joins the midpoints of the
diagonals of a trapezoid, then it is parallel to the
bases.
14. Angle Bisector Postulate
15. Perpendicular Postulate
16. Assume the opposite of what you want to
prove, then use valid reasoning to derive a contradiction.
17a. Smoking is not glamorous.
17b. If smoking were glamorous, then this smoker
would look glamorous. This smoker does not look
glamorous, therefore smoking is not glamorous.
18. False. The parallelogram is a rhombus.
19. False. Possible counterexample:
Given: Isosceles nABC with
## and BE
#; AC
# > BC
#; ED
#
altitudes AD
# i AB
#
Show: ED
23. true
D
E
C
2
1
3
A
F
B
/BAD > /DCB by the Opposite Angles Theorem.
[ m/1 5 }21} m/BAD 5 }21} m/DCB 5 m/2 by
# i DC
## by the
the definition of angle bisector. AB
definition of parallelogram. [ /2 and /3 are
supplementary by the Interior Supplements
Theorem.
[ /1 and /3 are supplementary by the
substitution property.
# i FC
# by the Converse of the Interior
[ AE
Supplements Theorem.
24. Use the Inscribed Angle Theorem, the addition
m/P 1 m/E 1 m/N 1 m/T 1 m/A 5
X
X
X
X
X
1
}} 1mTN 1 mAT 1 mPA 1 mEP 1 mNE 2. Because
2
there are 360° in a circle, m/P 1 m/E 1 m/N 1
m/T 1 m/A 5 180°.
25. T
# > OP
# by CPCTC. Use
SAA Theorem; thus DY
the Triangle Midsegment Theorem and the
substitution property to get TR 5 }21} (ZO 1 DY).
Also, use the Triangle Midsegment Theorem to
# i ZO
#.
get TR
28a. The quadrilateral formed when the
midpoints of the sides of a rectangle are connected
is a rhombus.
G
C
28b. D
H
F
A
B
E
Use the Right Angles Are Congruent Theorem and
the SAS Congruence Postulate to get nAEH >
nBEF > nDGH > nCGF. Then use CPCTC to
prove that EFGH is a rhombus.
29a. The quadrilateral formed when the
midpoints of the sides of a rhombus are connected
is a rectangle.
I
29b.
P
L
R
Assume m/H . 45° and m/T . 45°. Use the
Triangle Sum Theorem, the substitution
property, and the subtraction property to get
m/H 1 m/T . 90°, which creates a
contradiction.Therefore m/H # 45° or m/T # 45°.
Y
26.
S
T
R
Use the definition of midpoint, the Segment
Addition Postulate, the substitution property, and
MY
1
YS
1
}
}}
}}
}}
the division property to get }
TY 5 2 and YR 5 2 .
Then use the reflexive property and the SAS
Similarity Theorem to get nMSY , nTRY.
Therefore, MS 5 }21}TR by CSSTP and the
## i TR
# by
multiplication property and MS
CASTC and the CA Postulate.
D
Y
27.
1
3
T
R
42
Z
O
P
# and DR
#. Then
Use the Line Postulate to extend ZO
use the Line Intersection Postulate to label P as the
@#$ and DR
@#$. nDYR > nPOR by the
intersection of ZO
160
X
J
N
O
H
M
M
Y
K
##
By the Triangle Midsegment Theorem both PM
##
#
#
##
and ON are parallel to LJ , and both PO and MN
#
#
#
are parallel to IK . Because LJ and IK are
perpendicular, we can use corresponding angles
on parallel lines to prove that the lines that are
adjacent sides of the quadrilateral are also
perpendicular.
30a. The quadrilateral formed when the midpoints
of the sides of a kite are connected is a rectangle.
I
30b. By the Triangle Midsegment
P
M
##
##
Theorem both PM and ON are
L
J
#, and both PO
# and
parallel to LJ
## are parallel to IK
#. Because
MN
N
#
#
LJ and IK are perpendicular, we can O
use the CA Postulate to prove that
K
the lines that are adjacent sides of
the quadrilateral are also perpendicular.
C
31. Use the Line Postulate
# and
to construct chords DB
#. Then use the Inscribed
AC
B
Angles Intercepting Arcs
P
O
A
Theorem and the AA
Similarity Postulate to get
nAPC , nDPB. Therefore,
D
by CSSTP and the multiplication
property, AP ? PB 5 DP ? PC.

Answers to Exercises

Transcription

Similar documents

If ABC is obtuse, what are the restrictions on x? ∠

Stress: How Does Your Body Respond? Poster

Pilates exercises for non

Wkst- Law of Sines-Area of Triangle

Leon Russianoff

Original Green™ Diamond - Popp Dental Supply, LLC

Core Training with Jenny Meadows

Section 4

w w w . b o d y s o l i d . c o m