Chapter 11 Answers

Chapter 11 Answers
Practice 11-1
1. 32 2. 50 3. 72 4. 15 5. "91 6. 6
7. "634 8. "901 9. 4"3
10. circumscribed
11. inscribed 12. circumscribed 13. 24 cm
14. 28 in. 15. 52 ft
Practice 11-2
0
0
1. r = 13; mAB
134.8 2. r = 3 "5;mAB
53.1
#41 0
3. r =
2 ; mAB
102.7 4. 3 5. 4.5
6. 3 7. 20.8 8. 13.7 9. 8.5 10. Q T;
0 0
PR SU 11. A J; BC KL 12. Construct
radii OD and OB. FD = EB because a diameter perpendicular
to a chord bisects it, and chords CD and AB are congruent
(given). Then, by HL, OEB OFD, and by CPCTC,
OE = OF. 13. Because congruent arcs have congruent
chords, AB = BC = CA. Then, because an equilateral
triangle is equiangular, mABC = mBCA = mCAB.
17.
18.
5
4
3
2
1
y
(0, 5)
(0, 0) (5, 0)
x
5
y
3
(3, 5)
1 2 3 4 5 x
Practice 11-3
1. A and D; B and C 2. ADB and CDB
3. ADB and CAD 4. 55 5. x = 45; y = 50;
z = 85 6. x = 90; y = 70 7. 180 8. 70
9. x = 120; y = 60; z = 60 10. x = 120; y = 100;
z = 140 11. x = 63; y = 63; z = 54
0
12. x = 50; y = 80; z = 80 13a. m AE = 170
13b. mC = 85 13c. mBEC = 10
13d. mD = 85 14a. mA = 90
14b. mB = 80 14c. mC = 90
14d. mD = 100
19.
4
2
2
2
(2, 4)
4
6
y
2 4
x
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Practice 11-4
1. 87 2. 35 3. 45 4. 120 5. 72 6. 186
7. x = 58; y = 59; z = 63 8. x = 30; y = 66
9. x = 30; y = 30; z = 120 10. x = 16; y = 52
11. x = 138; y = 111; z = 111 12. x = 30; y = 60
13. 10 14. 14.8 15. 4.7 16. 4 17. 3.2
18. 6
Practice 11-5
1. C(0, 0), r = 6 2. C(2, 7), r = 7 3. C(-1, -6), r = 4
4. C(-3, 11), r = 2"3
5. x 2 + y 2 = 49
6. (x - 4) 2 + (y - 3) 2 = 64 7. (x - 5) 2 + (y - 3) 2 = 4
8. (x + 5) 2 + (y - 4) 2 1
= 9. (x + 2) 2 + (y + 5) 2 = 2
10. (x + 1) 2 + (y - 6) 2 4
= 5 11. x 2 + y 2 = 4
12. (x + 3) 2 + (y - 3) 2 = 1 13. x 2 + (y - 3) 2 = 16
14. (x - 7) 2 + (y + 2) 2 = 4 15. x 2 + (y + 20) 2 = 100
16. (x + 4) 2 + (y + 6) 2 = 25
20.
(1, 7)
(1, 1)
4
6
4
2
2
2
4
21. x 2 + y 2 = 25 22. (x - 5) 2 + (y - 9) 2 = 9
23. (x + 4) 2 + (y + 3) 2 = 61
24. (x - 7) 2 + (y + 2) 2 = 80
25. (x + 4) 2 + (y + 3) 2 = 4
26. (x - 4) 2 + (y - 1) 2 = 16
y
6
2 4
(5, 1)
x
Geometry Chapter 11 Answers 33

Chapter 11 Answers (continued)
Practice 11-6
1.
11.
2 cm
Q
T
2.
1.5 cm
1 in.
12.
R
0.75 in.
S
0.75 in.
3.
P
Q
13.
T
U
0.5 cm
14.
A
B
5 mm
6 mm
4.
5.
0.5 in.
R
0.5 in.
X
Y
S
0.75 in.
0.75 in.
Reteaching 11-1
1. "157 2. "741 3. "2 4. 7.5
Reteaching 11-2
1. 5"5
2. 12"2
3. 2"14
4. 11.83 cm
5. 2.54 cm 6. 8.66 in. 7. 2.78 in.
6.
7.
F
H
P
G
A
E
C
J
8. a third line parallel to the two given lines and midway
between them 9. a sphere whose center is the given
point and whose radius is the given distance 10. the
points within, but not on, a circle of radius 1 in. centered
at the given point
B
Reteaching 11-3
1. 87 2. 40 3. 60 4. 55 5. x = 94, y = 80
6. 120 7. 40 8. 20 9. 70
Reteaching 11-4
1. 93 2. 156 3. 42 4. 35 5. 60 6. 55
7. x = 36; y = 60; z = 48 8. x = 64; y = 64; z = 52
9. x = 46; y = 90; z = 44
Reteaching 11-5
1. (x - 3) 2 + (y - 11) 2 = 4 2. (x + 5) 2 + y 2 = 225
3. (x - 6) 2 + (y + 6) 2 = 7 4. x 2 + y 2 = 20
5. (x + 2) 2 + (y + 2) 2 = 4
6. (x - 3) 2 + (y - 1) 2 = 50
7. (x - 5) 2 + (y - 2) 2 = 36
8. (x - 4) 2 + (y - 2) 2 = 25
9. (x + 2) 2 + (y - 3) 2 = 25 10. C(-3, -5); r = 5
11. C(0, 0); r = 0.2 12. C(4, 0); r = "6
13. C(3, 5); r = 4
Reteaching 11-6
1.–5. Check students’ work. 6. the line perpendicular to
AB at its midpoint 7. one, the midpoint of AB
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34
Answers Geometry Chapter 11

Chapter 11 Answers (continued)
© Pearson Education, Inc. All rights reserved.
Enrichment 11-1
1. Given 2. Two points determine a line segment.
3. Two tangents drawn to a circle from an external point
are congruent. 4. Radii of a circle are congruent.
5. A radius and a tangent drawn to the same point of contact
form a right angle. 6. Definition of a square 7. Sides
of a square are congruent. 8. Addition Property
9. Segment Addition Postulate 10. Substitution Property
11. Addition Property 12. Substitution Property
13. Subtraction Property
Enrichment 11-2
1a. 1 1b. 3 1c. 6 1d. 10 1e. 15
2. 1, 3, 6, 10, 15 3. triangular numbers 4.
2
5. From each point, (N - 1) chords may be drawn. So, a total
of N(N - 1) chords may be drawn for N points. Because each
chord has been drawn twice, divide by 2. 6. N(N - 1)
Enrichment 11-3
1. Given 2. Two points determine a line segment.
3. The measures of exterior angles of a triangle equal the
sum of the measures of the remote interior angles.
4. Radii of a circle are congruent. 5. Definition of
isosceles triangle 6. Base angles of an isosceles triangle
are congruent. 7. Definition of congruent angles
8. Base angles of an isosceles triangle are congruent.
9. Definition of congruent angles 10. The measures of
exterior angles of a triangle equal the sum of the measures
of the remote interior angles. 11. Substitution Property
12. Division Property 13. drawing diameter DE passing
through A so that radius OB AB
Enrichment 11-4
1. 2x + 2x + 8 + x + x - 32 = 360; 6x - 24 = 360;
6x = 384 2. x = 64 3. 128 4. 136 5. 64
6. 32 7. chords ) AB, BD; 16 8. secants EB, EC ; 52
9. tangent FB and chord AB; 64 10. chords
)
BD, AC; 84
11. chords AC, BC; 64 12. tangent FB and secant FD;
36 13. chords AC, BD; 96 14.
)
AF, AE; 100
15. DE, DA; 48 16. tangent FB and chord BC; 68
Enrichment 11-5
1a. the edge of the fountain pool 1b. plaza
N(N 2 1)
1c. fountain pool 2. a circle with center (-3, 4) and a
radius of 3 3. all the interior points of a circle with center
(4, 0) and a radius of "12 4. all points outside a circle
with center (0, 0) and a radius of "18 5. a circle and all
its interior points with center (0, -2) and a radius of 5
6. x 2 + y 2 = 100 7. x 2 + y 2 36 8. The sergeant
can draw a circle with center (0, 0) and a radius of 2. The team
will search in the circle.
Enrichment 11-6
1.–9. Check students’ drawings. 1. P with a radius
of 4 in. 2. two lines parallel to line l, one 2 in. above l
and the other 2 in. below l 3. 4 points 4. two lines
parallel to line l, one 4 in. above l and the other 4 in. below l
5. 2 points 6. two lines parallel to line l, one 6 in. above l
and the other 6 in. below l 7. empty set 8. 6 points
9. empty set
Chapter Project
Activity 1: Doing
Check students’ work.
Activity 2: Exploring
Both figures are made from 9 circles with the same center. In
Figure A, the smallest circle and alternate rings are black. In
Figure B, vertical and horizontal tangents to the smallest circle
are drawn. Then the tangents at 45° angles to the first ones are
drawn. Alternating sections of the 6 outer rings are black, as is
the smallest ring.
Check students’ work.
Activity 3: Constructing
Check students’ work.
✔ Checkpoint Quiz 1
1. 52 in. 2. 44 cm 3. 40 m 4. 6 5. 12
6. 11.5 7. x = 37, y = 100 8. x = 116, y = 88,
z = 79 9. x = 120
✔ Checkpoint Quiz 2
1. x = 140 2. x = 75, y = 105 3. x = 20, y = 70
4. x 2 + (y - 1) 2 = 6.76 5. (x + 3) 2 + (y - 2) 2 = 100
6. 12 7.
51
7
8. 20
Chapter Test, Form A
1. C(0, 0); r = 10 2. C(11, -6); r = 9
3. C(-1, -4); r = "7 4. (x + 3) 2 + (y - 2) 2 = 1
5. (x - 2) 2 + (y - 1) 2 = 16 6. x 2 + (y - 2) 2 = 9
7. C = 31.4; A = 78.5 8. (x + 3) 2 + (y - 4) 2 = 41
9.
y
10. 90
C(1, 2)
r 2
x
Geometry Chapter 11 Answers 35

Chapter 11 Answers (continued)
11.
y
x
Chapter Test, Form B
1. C(0, 0); r = 9 2. C(-5, 2); r = 4
3. C(4, -8); r = "2 4. (x - 1) 2 + (y - 3) 2 = 4
5. (x - 1) 2 + y 2 = 1 6. (x + 3) 2 + (y - 1) 2 = 16
7. C = 25.1; A = 50.3 8. (x - 5) 2 + (y - 1) 2 = 89
9.
y
C(2, 1)
r 4
12.
y
x
x
10. 90
11.
y
13.
x 5
y
x
x
12.
y
14.
y
y 4
y 6
15. 120 16. 105 17. 118.0 18. 40 19. 70
20. 115 21. 6 22. 11.8 23. 5.5 24. 45
25. 35 26. 80 27. 5.8 28. 5.7 29. x = 63;
y = 71 30. x = 87; y = 95; w = 85; z = 93
31. x = 120; y = 64; z = 176 32. 44
x
13. y
y 2
x
x
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36
Answers Geometry Chapter 11

Chapter 11 Answers (continued)
© Pearson Education, Inc. All rights reserved.
14.
x 7 x 3
y
15. 110.3 16. 77.4 17. 120 18. 42 19. 80
20. x = 125; y = 90 21. 8.4 22. 20 23. 2
24. 40 25. 145 26. 74 27. 4.5 28. 3.5
29. x = 60; y = 80 30. x = 88; y = 93; z = 92;
w = 87 31. x = 116; y = 80; z = 164 32. 45.5
Alternative Assessment, Form C
TASK 1: Scoring Guide
(a) center (2, 5), radius = 4 (b) The distance from point
(x, y) on a circle to the center (2, 5) is given by the formula
D = "(x 2 2) 2 1 (y 2 5) 2 .In any circle, the distance, or
radius, is the same for all points on the circle. Therefore the
distance formula, when applied to the circle in this problem,
becomes 4 = "(x 2 2) 2 1 (y 2 5) 2 for all points (x, y) on
the circle. Squaring both sides of this equation yields the
equation of the circle, 16 = (x - 2) 2 + (y - 5) 2 .
3 Student gives accurate answers and a correct explanation.
2 Student gives answers or an explanation that may contain
minor errors.
1 Student gives wrong answers or an incomplete or inaccurate
explanation.
0 Student makes little or no effort.
TASK 2: Scoring Guide
(a) Construct the perpendicular bisector of two chords. The
point at which they meet is the center of the circle. (b)
Because the pentagon is regular, all chords are congruent.
Therefore the corresponding arcs are congruent. Because
there are 5
0
congruent arcs in the circle, each arc, and in
particular AB , has measure 72. To find
0
AB, call the center O,
and consider triangle AOB. Because AB has measure 72, so
does angle AOB. Because OA = OB, AOB is isosceles.
Therefore, the bisector of angle AOB is perpendicular to
(and bisects) chord AB, forming a 36°-54°-90° triangle. Then
1
sin 36
2AB AB
= = , yielding AB = 7.05.
OA 12
3 Student devises a correct method and gives a correct answer
and a valid explanation.
2 Student devises a method, gives an answer, and gives an
explanation that may contain some errors.
1 Student gives a method, an answer, and an explanation that
may contain major errors or omissions.
0 Student makes little or no effort.
x
TASK 3: Scoring Guide
(a) Angles A and C are inscribed angles. Each is
inscribed in a different arc, but together the arc in which
they are inscribed comprise the entire circle. Therefore,
1
mA + mC = 2(360) = 180. And, because ABCD is a
parallelogram, angles A and C are congruent. Finally, if angles
are congruent and supplementary, then they are right. (b) The
diagonals of ABCD bisect each other because ABCD is a
parallelogram. Suppose that they intersect at point X. Then
AX = CX and BX = DX. Also, because ABCD is
inscribed in a circle, the diagonals are chords, and therefore
AX ? CX = BX ? DX. Substituting yields AX 2 = BX 2 ,
and therefore AX = BX = CX = DX. Then AC = BD.
(c) Either (a) or (b) allows you to conclude that ABCD must
be a rectangle.
3 Student gives valid and accurate arguments and
explanations.
2 Student gives arguments that, although basically valid,
may contain minor flaws.
1 Student gives arguments containing major flaws.
0 Student makes little or no attempt.
TASK 4: Scoring Guide
(a) Because the circumscribing lines are tangent to the circle,
AB = AX, CB = CY, and DX = DY. Then, because it is
given that AB = BC, we have AX = CY. Adding segments
together yields DA
0
= DC, showing that
1
ACD is isosceles.
Secondly, because XY has measure 100, XBY has measure
1
260. Then mD = 2(260 - 100) = 80, and because
ACD is isosceles, mA = mC = 50. Finally, AC = 20
AB
because AB = BC = 10, and, by trigonometry, cos 50 = AD,
AB 10
AD = cos 50 = 0.6428 = 15.56. And again, AD = CD.
(b) Because angle A has measure 50 and line AB is tangent
at B, triangle AOB is a 25°-65°-90° triangle. Therefore, by
OB
trigonometry, tan 25 = AB,OB= 10 ? tan 25 = 4.66.
3 Student gives accurate answers and explanations.
2 Student gives answers and explanations that may contain
minor errors.
1 Student gives answers and explanations that contain
significant errors.
0 Student makes little or no attempt.
Cumulative Review
1. B 2. D 3. E 4. A 5. D 6. B 7. A
8. A 9. D 10. B 11. C 12. C 13. D
14. C 15. B 16. "89 17. Check students’
work. 18. A tangent intersects a circle at exactly one
point. A secant intersects a circle at two points. 19. Check
students’ work. 20. This follows from the Pythagorean
Theorem (c 2 = a 2 + b 2 ) or the fact that the longest side of
a triangle is opposite the largest angle. 21. If a triangle
is not a right triangle, then it is acute. 22. 135
Geometry Chapter 11 Answers 37