Chapter 10
The diameter of the smaller circle centered at A is 3 inches, and the diameter of the larger circle
centered at A is 9 inches. The diameter of D is 11 inches. Find each measure.
1. BC
SOLUTION: BC = the radius of the large circle at A – the radius of the small circle at A.
4.5 – 1.5 = 3
2. CD
SOLUTION: CD = the radius of circle D – BC.,
BC = the radius of the large circle at A – the radius of the small circle at A.
4.5 – 1.5 = 3
CD = 5.5 – 3 = 2.5
3. DECORATIONS To decorate for homecoming, Brittany estimates that she will need to purchase enough
streamers to go around the school's circular fountain twice. If the diameter of the fountain is 88 inches, about how
many feet of streamers should she buy?
SOLUTION: d = 88 in.
r = 44 in.
We want to cover the circumference twice.
Use
C to find the length of each arc. Round to the nearest hundredth.
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4. Arc XY if the radius is 5 feet
SOLUTION: Page 1
Chapter 10
Use
C to find the length of each arc. Round to the nearest hundredth.
4. Arc XY if the radius is 5 feet
SOLUTION: 5. Arc YZ if the diameter is 8 meters
SOLUTION: 6. TRANSPORTATION The graph shows the results of a survey in which students at a high school were asked how
they get to school.
a. Find the measure of arc CD.
b. Find the measure of arc BC. SOLUTION: a.
b.
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Chapter 10
6. TRANSPORTATION The graph shows the results of a survey in which students at a high school were asked how
they get to school.
a. Find the measure of arc CD.
b. Find the measure of arc BC. SOLUTION: a.
b.
Find the value of x.
7. SOLUTION: arc AB is congruent to arc CD.
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Chapter 10
8. SOLUTION: arc PQ is congruent to arc SR, so chord QS is congruent to chord PR.
In
M, MZ = 12 and WY = 20. Find each measure. Round to the nearest hundredth.
9. CM
SOLUTION: MZ = 12
WY = 20, so CY = 10
If we connect a line from Y to M , we form a right triangle. YM = MZ = 12
Use the Pythagorean theorem to find CM.
10. XC
SOLUTION: XC + CM = MZ. Find CM.
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MZ = 12
WY = 20, so CY = 10
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Chapter 10
10. XC
SOLUTION: XC + CM = MZ. Find CM.
MZ = 12
WY = 20, so CY = 10
If we connect a line from Y to M , we form a right triangle. YM = MZ = 12
Use the Pythagorean theorem to find CM.
XC = 12 – 6.63 = 5.37
Find each measure. 11. m∠N
SOLUTION: arc MP + arc PN + arc MN = 360
arc MP = 360 – (128 + 114) = 118
∠N = 0.5(arc MP) = 0.5(118) = 59
12. m∠B
SOLUTION: arc AC = arc AD + arc DC = 180
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arc DC = 180 –
112by=Cognero
68
∠B = 0.5(arc DC) = 0.5(68) = 34
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arc MP = 360 – (128 + 114) = 118
Chapter
10
∠N = 0.5(arc MP) = 0.5(118) = 59
12. m∠B
SOLUTION: arc AC = arc AD + arc DC = 180
arc DC = 180 – 112 = 68
∠B = 0.5(arc DC) = 0.5(68) = 34
13. Find x. Assume that segments that appear to be tangent are tangent.
SOLUTION: 14. Quadrilateral ABCD is circumscribed about
H. Find m.
SOLUTION: DZ = DY, so DY = 54
YC = DC – DY = 87 – 54 = 33
CX = YC = 33
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m = 33
Find each measure. Assume that segments that appear to be tangent are tangent.
Page 6
Chapter 10
14. Quadrilateral ABCD is circumscribed about
H. Find m.
SOLUTION: DZ = DY, so DY = 54
YC = DC – DY = 87 – 54 = 33
CX = YC = 33
m = 33
Find each measure. Assume that segments that appear to be tangent are tangent.
15. Measure of arc XYZ SOLUTION: arc XYZ = arc XY + arc YZ
arc XYZ = arc XY + arc YZ = 80 + 60 = 140
16. Measure of arc MP
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SOLUTION: Page 7
Chapter
10 = arc XY + arc YZ = 80 + 60 = 140
arc XYZ
16. Measure of arc MP
SOLUTION: Find x to the nearest tenth. Assume that segments that appear to be tangent are tangent.
17. SOLUTION: x cannot equal 0 because length cannot be negative.
18. SOLUTION: PHONES
phone
19. CELL
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tower covers a circular area with a radius of 15 miles.
a. If the tower is located at the origin, write an equation for this circular area of coverage.
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Chapter 10
19. CELL PHONES A cell phone tower covers a circular area with a radius of 15 miles.
a. If the tower is located at the origin, write an equation for this circular area of coverage.
b. Will a person 11 miles west and 12 miles south of the tower have coverage? Explain. SOLUTION: 2
a. It is centered at the origin, so h = 0 and k = 0. The radius is 15, so r = 225.
2
2
x + y = 225
b. The magnitude of 11 miles west and 12 miles south is about 16.27 miles, which is greater than the radius, so the
tower will not have coverage.
20. Write an equation of a circle that contains points A(–1, 5), B(–5, 9), and C(–9, 5). Then graph the equation.
SOLUTION: Graph the three points and form a triangle. Identify the perpendicular bisectors of two of the segments. Find the
intersection. This is the radius of the circle.
In the graph above, the red point is the center of the circle. One perpendicular bisector went through (–5, 9) and the
other went through (–7, 7). The center is located at (–5, 5).
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2
The other vertices are 3 units away from the center, so r = 3 and r = 9.
Chapter
10
The magnitude
of 11 miles west and 12 miles south is about 16.27 miles, which is greater than the radius, so the
tower will not have coverage.
20. Write an equation of a circle that contains points A(–1, 5), B(–5, 9), and C(–9, 5). Then graph the equation.
SOLUTION: Graph the three points and form a triangle. Identify the perpendicular bisectors of two of the segments. Find the
intersection. This is the radius of the circle.
In the graph above, the red point is the center of the circle. One perpendicular bisector went through (–5, 9) and the
other went through (–7, 7). The center is located at (–5, 5).
2
The other vertices are 3 units away from the center, so r = 3 and r = 9.
h = –5 and k = 5
2
2
The equation is y = (x + 5) + (y – 5) = 9.
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