Name: ________________________ Class: ___________________ Date: __________
ID: A
Unit 3 Practice Test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
The radius, diameter, or circumference of a circle
is given. Find the missing measures. Round to the
nearest hundredth if necessary.
Use the diagram to find the measure of the given
angle.
1. d = 22.3 km, r = ? , C = ?
a. r = 44.6 km, C = 35.03 km
b. r = 11.15 km, C = 35.03 km
c. r = 11.15 km, C = 70.06 km
d. r = 44.6 km, C = 70.06 km
2. Find the exact circumference of the circle.
3. ∠PRS
a. 95°
b. 20°
c. 50°
d. 85°
a.
b.
c.
d.
4. ∠PRQ
a. 85°
b. 20°
c. 50°
d. 95°
7π cm
5π cm
10π cm
4π cm
1
Name: ________________________
ID: A
Find x. Assume that any segment that appears to be
tangent is tangent.
5. Find x. Assume that segments that appear tangent
are tangent.
7.
a.
b.
c.
d.
7
6
14
5
Find the measure of the numbered angle.
6.
a.
b.
c.
d.
65
66
68
62
a.
b.
c.
d.
35
20
25
30
8.
a.
b.
c.
d.
115
125
120
130
2
Name: ________________________
ID: A
Find x. Round to the nearest tenth if necessary.
11. Find the area of a circle having a circumference of
79.3. Round to the nearest tenth. Use 3.14 for π.
a. 430.2 units 2
b. 1001.4 units 2
c. 500.7 units 2
d. 469.9 units 2
9.
Find the area of the shaded region. Round answers
to the nearest tenth. Assume all inscribed polygons
are regular.
12.
a.
b.
c.
d.
6
5
4
3
Find x. Round to the nearest tenth if necessary.
Assume that segments that appear to be tangent are
tangent.
10.
a.
b.
c.
d.
10.3 units2
9.9 units2
9.2 units2
13.8 units2
Find the area of the figure. Round to the nearest
tenth if necessary.
13.
a.
b.
c.
d.
4
5
6
7
a.
b.
c.
d.
3
764.7 units2
366.7 units2
249.7 units2
300.3 units2
Name: ________________________
ID: A
14. Find the shape resulting from the cross-section of
the cylinder.
Find the volume of the cylinder. Use 3.14 for π.
Round to the nearest tenth.
16.
a.
b.
c.
d.
rectangle
square
circle
ellipse
Find the lateral area of each prism. Round to the
nearest tenth if necessary.
15.
a.
b.
c.
d.
7877.6 in 3
627.2 in 3
281.3 in 3
1969.4 in 3
a.
b.
c.
d.
888.3 in 3
12,258.6 in 3
49,034.5 in 3
3904.0 in 3
17.
a.
b.
c.
d.
456 units2
225 units2
184 units2
429 units2
4
Name: ________________________
ID: A
Find the volume of the pyramid. Round to the
nearest tenth if necessary.
20. Find the volume of a sphere that has a radius of 9.5
meters. Use 3.14 for π. Round to the nearest tenth.
a. 3589.5 m 3
b. 2692.2 m 3
c. 377.8 m 3
d. 897.4 m 3
18.
a.
b.
c.
d.
21. Suppose a snow cone has a paper cone that is 8
centimeters deep and has a diameter of 5
centimeters. The flavored ice comes in a spherical
scoop with a diameter of 5 centimeters and rests on
top of the cone. If all the ice melts into the cone,
will the cone overflow? Explain.
a. No. The volume of the ice is less than the
volume of the cone.
b. No. The volume of the ice is exactly the same
as the volume of the cone.
c. Yes. The volume of the ice is greater than the
volume of the cone.
d. There is not enough information given to solve
this problem.
11,520 m 3
14,400 m 3
4800 m 3
3840 m 3
19. Find the surface area of a sphere if the
circumference of a great circle is 43.96 centimeters.
Use 3.14 for π. Round to the nearest tenth.
a. 4308.1 cm2
b. 196 cm2
c. 153.9 cm2
d. 615.4 cm2
Short Answer
22. A CD has a diameter of about 116 millimeters.
Find the circumference of the CD.
5
Name: ________________________
ID: A
Use the information to answer the questions that
follow.
26. Find the values of y and z.
Annie wants to grow flowers all around a circular
garden with a radius of 6 feet.
27. Using the properties of tangents, find the value of
x. Also mention the property of tangent applied to
find x.
23. If the outermost circle is 2 to 3 feet farther from the
center than the inner circle, find the minimum and
maximum circumference of the inner circle to the
nearest foot.
28. In C , mRQ = 150, mTQ = 22, mSR = 92. What is
m∠P ?
24. In C , CL = CM , XY = 5x + 8, and ST = 15x − 32.
Find XL.
29. Find the volume of the solid. Round to the nearest
tenth.
25. In C , the diameter is 42 units long, and,
m∠CRT = 30°. Find x.
6
ID: A
Unit 3 Practice Test
Answer Section
MULTIPLE CHOICE
1. ANS: C
radius = diameter ÷ 2
Circumference = (2 × radius × π) or (diameter × π)
Feedback
A
B
C
D
Check both your radius and circumference calculations.
Check your circumference calculation.
Correct!
Check your radius calculation.
PTS: 1
DIF: Basic
REF: Lesson 10-1 OBJ: 10-1.1 Identify and use parts of circles.
NAT: NCTM ME.2
TOP: Identify and use parts of circles.
KEY: Circles | Parts of Circles
2. ANS: B
The circumference formula is diameter × π. The diameter shown also happens to be the hypotenuse of the right
triangle inscribed in the circle, so it can be found by using the Pythagorean Theorem.
Feedback
A
B
C
D
Use the Pythagorean Theorem.
Correct!
How did you find the diameter?
Use the Pythagorean Theorem.
PTS:
OBJ:
NAT:
TOP:
1
DIF: Average
REF: Lesson 10-1
10-1.2 Solve problems involving the circumference of a circle.
NCTM GM.1 | NCTM GM.1a | NCTM ME.2
Solve problems involving the circumference of a circle. KEY: Circles | Circumference
1
ID: A
3. ANS: A
∠PRS and ∠QRT are vertical angles and are therefore congruent. So, set the two expressions equal and solve for
x.
Feedback
A
B
C
D
Correct!
That's the answer for x, you need the measure of ∠PRS .
Did you use vertical angles?
How are vertical angles related?
PTS: 1
DIF: Average
REF: Lesson 10-2
OBJ: 10-2.1 Recognize major arcs, minor arcs, semicircles, and central angles, and their measures.
NAT: NCTM ME.2
TOP: Recognize major arcs, minor arcs, semicircles, and central angles, and their measures.
KEY: Major Arcs | Minor Arcs | Semicircles | Central Angles
4. ANS: A
∠PRQ and ∠QRT are a linear pair and are therefore supplementary. First find m∠QRT . ∠PRS and ∠QRT are
vertical angles and are therefore congruent. So, set the two expressions equal and solve for x.
Feedback
A
B
C
D
Correct!
That's the answer for x, you need the measure of ∠PRQ.
Check over your work.
How many degrees are in a linear pair?
PTS: 1
DIF: Average
REF: Lesson 10-2
OBJ: 10-2.1 Recognize major arcs, minor arcs, semicircles, and central angles, and their measures.
NAT: NCTM ME.2
TOP: Recognize major arcs, minor arcs, semicircles, and central angles, and their measures.
KEY: Major Arcs | Minor Arcs | Semicircles | Central Angles
5. ANS: B
The triangle shown is a right triangle since the tangent segment, FE, intersects a radius, DE, which always results
in a right angle. So to solve for x, use the Pythagorean Theorem. Note that mDE = x since they are both radii of
the same circle.
Feedback
A
B
C
D
Use the Pythagorean Theorem and mDE = x.
Correct!
Is the triangle a right?
Use the Pythagorean Theorem and mDE = x.
PTS: 1
DIF: Average
NAT: NCTM GM.1 | NCTM GM.1a
KEY: Tangents
REF: Lesson 10-5 OBJ: 10-5.1 Use properties of tangents.
TOP: Use properties of tangents.
2
ID: A
6. ANS: B
When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is
equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this
diagram, the measures of the intercepted arcs for ∠6 are not given, but they have a sum of 250 since the arcs
shown have a sum of 110 (360 – 110 = 250).
Feedback
A
B
C
D
Add the intercepted arcs and divide by 2.
Correct!
Did you divide correctly?
How do you find the measures of the other two arcs?
PTS: 1
DIF: Average
REF: Lesson 10-6
OBJ: 10-6.1 Find measures of angles formed by lines intersecting on or inside a circle.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P1.c
TOP: Find measures of angles formed by lines intersecting on or inside a circle.
KEY: Measure of Angles | Circles
7. ANS: A
When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the
positive difference of the measures of the intercepted arcs.
Feedback
A
B
C
D
Correct!
What is the measure of the other intercepted arc?
Check your subtraction.
Did you find the positive difference of the measures of the intercepted arcs?
PTS: 1
DIF: Average
REF: Lesson 10-6
OBJ: 10-6.2 Find measures of angles formed by lines intersecting outside the circle.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P1.c
TOP: Find measures of angles formed by lines intersecting outside the circle.
KEY: Measure of Angles | Circles
8. ANS: C
When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to
one-half the positive difference of the measures of the intercepted arcs.
Feedback
A
B
C
D
Check your subtraction.
Did you subtract carefully?
Correct.
Check your subtraction.
PTS:
OBJ:
NAT:
TOP:
KEY:
1
DIF: Average
REF: Lesson 10-6
10-6.2 Find measures of angles formed by lines intersecting outside the circle.
NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P1.c
Find measures of angles formed by lines intersecting outside the circle.
Measure of Angles | Circles
3
ID: A
9. ANS: D
The products of the segments for each intersecting chord are equal.
Feedback
A
B
C
D
Multiply the segments and set them equal to each other.
Did you factor correctly?
Use multiplication, not addition.
Correct!
PTS: 1
DIF: Average
REF: Lesson 10-7
OBJ: 10-7.1 Find measures of segments that intersect in the interior of a circle.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P1.c
TOP: Find measures of segments that intersect in the interior of a circle.
KEY: Circles | Interior of Circles
10. ANS: C
When two secant segments intersect in the exterior of a circle, set an equality between the product of each external
segment and the entire segment.
Feedback
A
B
C
D
Check your multiplication.
Check the segments in your multiplication.
Correct!
Check your multiplication.
PTS: 1
DIF: Average
REF: Lesson 10-7
OBJ: 10-7.2 Find measures of segments that intersect in the exterior of a circle.
NAT: NCTM AL.2 | NCTM AL.2c | NCTM RE.2
STA: MC2P1.c
TOP: Find measures of segments that intersect in the exterior of a circle.
KEY: Circles | Exterior of Circles
11. ANS: C
The area formula for a circle is π × radius 2 . To find the radius, use the formula for the circumference, which is
circumference
.
circumference = 2 × π × radius or radius =
2×π
Feedback
A
B
C
D
How do you find the area of a circle?
What is the formula for the area of a circle?
Correct!
The area of a circle is pi times the square of the radius.
PTS: 1
DIF: Average
NAT: NCTM ME.2 | NCTM ME.2b
KEY: Area | Circles | Area of Circles
REF: Lesson 11-3
STA: MC2P5.b
4
OBJ: 11-3.1 Find areas of circles.
TOP: Find areas of circles.
ID: A
12. ANS: C
The shaded area here represents two-fifths of the area of the circle after subtracting the area of the regular
pentagon.
Feedback
A
B
C
D
Re-check all of your calculations.
Re-check all of your calculations.
Correct!
It’s two-fifths of the leftover area.
PTS: 1
DIF: Average
REF: Lesson 11-3
OBJ: 11-3.3 Solve problems involving segments of circles.
NAT: NCTM PS.1 | NCTM PS.2 | NCTM PS.3
TOP: Solve problems involving segments of circles.
KEY: Circles | Segments of Circles
13. ANS: B
This figure consists of a rectangle and two semicircles which combine to make one circle with radius 6.5.
Feedback
A
B
C
D
Use the radius, not the diameter in the area formula for the circle.
Correct!
The area of the rectangle is 18 × 13.
There are two semicircles that make one full circle.
PTS: 1
DIF: Average
REF: Lesson 11-4 OBJ: 11-4.1 Find areas of composite figures.
NAT: NCTM ME.2 | NCTM ME.2b
STA: MC2P5.b
TOP: Find areas of composite figures.
KEY: Area | Composite Figures | Area of Composite Figures
14. ANS: A
A cross-section is the intersection of a three-dimensional body with a plane.
Feedback
A
B
C
D
Correct!
Check your answer.
It will be the vertical cross-section of the cylinder shown.
Refer to the hint and try again.
PTS:
OBJ:
NAT:
TOP:
KEY:
1
DIF: Basic
REF: Lesson 12-1
12-1.2 Investigate cross sections of three-dimensional figures.
NCTM GM.4 | NCTM GM.4a | NCTM GM.4b
STA: MC2P4.a
Investigate cross sections of three-dimensional figures.
Cross Sections | Three-Dimensional Figures
5
ID: A
15. ANS: C
The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L
square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of
all of the sides of the base.
Feedback
A
B
C
D
Lateral area is perimeter of the base times the height. The height of this prism is 4.
Lateral area is perimeter of the base times the height. What is the height of this prism?
Correct!
Lateral area is perimeter of the base times the height. The height of this prism is 4.
PTS: 1
DIF: Average
REF: Lesson 12-2 OBJ: 12-2.1 Find lateral areas of prisms.
NAT: NCTM ME.2 | NCTM ME.2b
STA: MC2P5.b
TOP: Find lateral areas of prisms.
KEY: Lateral Area | Prisms | Lateral Area of Prisms
16. ANS: D
The volume of a cylinder is found by the formula π × radius 2 × height. In this figure, the height and diameter are
given. To find the radius, divide the diameter by 2.
Feedback
A
B
C
D
What is the formula for the volume of a cylinder?
How do you find the volume of a cylinder?
What is the radius of the base?
Correct!
PTS: 1
DIF: Average
REF: Lesson 12-4 OBJ: 12-4.2 Find volumes of cylinders.
NAT: NCTM ME.2 | NCTM ME.2b
STA: MC2P5.b
TOP: Find volumes of cylinders.
KEY: Volume | Cylinders | Volume of Cylinders
17. ANS: B
The volume of a cylinder is found by the formula π × radius 2 × height. In this figure, the height and radius are
given.
Feedback
A
B
C
D
Did you square the radius?
Correct!
Did you use the correct formula?
Did you include pi in the formula?
PTS: 1
DIF: Average
REF: Lesson 12-4
NAT: NCTM ME.2 | NCTM ME.2b
STA: MC2P5.b
KEY: Volume | Cylinders | Volume of Cylinders
6
OBJ: 12-4.2 Find volumes of cylinders.
TOP: Find volumes of cylinders.
ID: A
18. ANS: D
The volume formula for a pyramid is
1
3
Bh, where B is the area of the base and h is the height of the pyramid. In
the figure, the length and width of the base are given. Since the base is a rectangle, the area is found by
multiplying the length and width. To find the height of the pyramid, it is necessary to use the Pythagorean
Theorem. As shown in the figure, there is a right triangle formed with legs that are the height and half the width.
The hypotenuse is the slant height of the pyramid’s face. So the height is equal to
slant 2 − (width ÷ 2) 2 .
Feedback
A
You need to divide your answer by three.
B
The formula you need is V =
C
You need to find the height of the pyramid.
Correct!
D
1
3
Bh.
PTS: 1
DIF: Average
REF: Lesson 12-5 OBJ: 12-5.1 Find volumes of pyramids.
NAT: NCTM ME.2 | NCTM ME.2b
STA: MC2P5.b
TOP: Find volumes of pyramids.
KEY: Volume | Pyramids | Volume of Pyramids
19. ANS: D
The surface area of a sphere is found by the formula 4π r 2 , where r is the radius of the sphere. This problem gives
c
the circumference, so begin by computing the radius which can be found by the formula
. For example, if the
2π
12.56
circumference is given as 12.56, then the radius is 2 × 3.14 = 2 and the surface area would be 4 × 3.14 × 2 2 = 50.2. If
the solid is a hemisphere, then the formula for the surface area is 3π r 2 , since a hemisphere has half the surface
Ê1
ˆ
Ê
ˆ
area of the sphere ÁÁÁ 2 4π r 2 = 2π r 2 ˜˜˜ plus the area of the great circle on its base ÁÁ 2π r 2 ˜˜ .
Ë
¯
Ë
¯
Feedback
A
B
C
D
Does the formula indicate cubing the radius?
Did you leave pi out of the formula?
Did you use the correct formula?
Correct!
PTS: 1
DIF: Average
REF: Lesson 12-6 OBJ: 12-6.2 Find surface areas of spheres.
NAT: NCTM PS.1 | NCTM PS.2 | NCTM PS.3
STA: MC2P5.b
TOP: Find surface area of spheres.
KEY: Surface Area | Spheres | Surface Area of Spheres
7
ID: A
20. ANS: A
The volume of a sphere is found by the formula
4
3
π × radius 3 . In this problem, the radius or diameter is given. To
find the radius from the diameter, simply divide by 2. If the solid is a hemisphere, then it has one-half the volume
of a sphere. So, divide the volume of the sphere by 2 to get the volume of a hemisphere.
Feedback
A
Correct!
B
You need to multiply your answer by 3 .
C
You need to cube the radius, not square it.
D
The volume formula has
4
4
3
1
in it, not 3 .
PTS: 1
DIF: Average
REF: Lesson 12-6 OBJ: 12-6.3 Find volumes of spheres.
NAT: NCTM ME.2 | NCTM ME.2b
STA: MC2P5.b
TOP: Find volumes of spheres.
KEY: Volume | Spheres | Volume of Spheres
21. ANS: C
1
4
The volume of the cone is 3 × 3.14 × 2.5 2 × 8 = 52.3 cm3. The volume of the ice (a sphere) is 3 × 3.14 × 2.5 3 = 65.4
cm3. So the volume of the ice is greater than the volume of the cone, causing it to overflow the cone.
Feedback
A
B
C
D
Check your volume calculations.
Check your volume calculations.
Correct!
This problem can be solved with the information given.
PTS:
OBJ:
STA:
KEY:
1
DIF: Average
REF: Lesson 12-6
12-6.4 Solve problems involving volumes of spheres.
NAT: NCTM GM.1 | NCTM GM.1b
MC2P5.b
TOP: Solve problems involving volumes of spheres.
Volume | Spheres | Volume of Spheres
SHORT ANSWER
22. ANS:
about 364.4 mm
Circumference = (2 × radius × π) or (diameter × π)
PTS: 1
DIF: Basic
REF: Lesson 10-1
NAT: NCTM GM.1 | NCTM GM.1a | NCTM ME.2
KEY: Solve Multi-Step Problems
8
OBJ: 10-1.3 Solve multi-step problems.
TOP: Solve multi-step problems.
ID: A
23. ANS:
18.8 ft; 25.1 ft
Find the range of values for the radius of the inner circle.
The minimum radius will give the minimum circumference and the maximum radius will give the maximum
circumference.
PTS: 1
DIF: Advanced
REF: Lesson 10-1
NAT: NCTM GM.1 | NCTM GM.1a | NCTM ME.2
KEY: Solve Multi-Step Problems
24. ANS:
14
OBJ: 10-1.3 Solve multi-step problems.
TOP: Solve multi-step problems.
In a circle, two chords are congruent if they are equidistant from the center.
In a circle, if a diameter is perpendicular to a chord then it bisects the chord.
PTS: 1
DIF: Average
REF: Lesson 10-3 OBJ: 10-3.2 Solve multi-step problems.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P4.a
TOP: Solve multi-step problems.
KEY: Solve Multi-Step Problems
25. ANS:
1.5
sin ∠CRT =
CT
CR
PTS: 1
DIF: Average
REF: Lesson 10-3 OBJ: 10-3.2 Solve multi-step problems.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P4.a
TOP: Solve multi-step problems.
KEY: Solve Multi-Step Problems
26. ANS:
70°; 105°
If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.
PTS: 1
DIF: Average
REF: Lesson 10-4 OBJ: 10-4.3 Solve multi-step problems.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2 | NCTM GM.1a
STA: MC2P1.c
TOP: Solve multi-step problems.
KEY: Solve Multi-Step Problems
27. ANS:
16; Two segments from the same exterior point are tangent to the circle.
If two segments from the same exterior point are tangent to a circle, then they are congruent.
AD ≅ AB
PTS: 1
DIF: Basic
REF: Lesson 10-5 OBJ: 10-5.3 Solve multi-step problems.
NAT: NCTM GM.1 | NCTM GM.1a | NCTM GM.1b | NCTM ME.2
STA: MC2P1.c
TOP: Solve multi-step problems.
KEY: Solve Multi-Step Problems
9
ID: A
28. ANS:
35
When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the
positive difference of the measures of the intercepted arcs.
ˆ
1Ê
m∠P = ÁÁÁ mSR − mTQ ˜˜˜
2Ë
¯
PTS: 1
DIF: Basic
REF: Lesson 10-6 OBJ: 10-6.3 Solve multi-step problems.
NAT: NCTM GM.1 | NCTM GM.1b | NCTM ME.2
STA: MC2P1.c
TOP: Solve multi-step problems.
KEY: Solve Multi-Step Problems
29. ANS:
385.1 in 3
V = π × radius 2 × height −
2
π × radius 3
3
PTS: 1
DIF: Advanced
REF: Lesson 12-6 OBJ: 12-6.5 Solve multi-step problems.
NAT: NCTM ME.2 | NCTM ME.2b | NCTM PS.1 | NCTM PS.2 | NCTM PS.3 | NCTM GM.1 | NCTM GM.1b
STA: MC2P5.b
TOP: Solve multi-step problems.
KEY: Solve Multi-Step Problems
10