Circles, Arc Lengths, and Sector Areas Practice Questions — GRE General Test

Question 1Question →

Sector $S_1$ belongs to a circle with radius $r$ and has a central angle of measure $\theta^\circ$ , where $0 < \theta < 360$ . Sector $S_2$ belongs to a circle with radius $2r$ and has a central angle of measure $\left(\frac{\theta}{2}\right)^\circ$ . Which of the following statements must be true? Select all such statements.

Select all that apply

The area of sector

S_2

is twice the area of sector

S_1

.

The arc length of sector

S_2

is equal to the arc length of sector

S_1

.

The perimeter of sector

S_2

is twice the perimeter of sector

S_1

.

The ratio of the area of sector

S_1

to its arc length is half the ratio of the area of sector

S_2

to its arc length.

The perimeter of sector

S_2

is equal to the perimeter of sector

S_1

.

Show answer & explanation

Answer: The area of sector $S_2$ is twice the area of sector $S_1$ .; The arc length of sector $S_2$ is equal to the arc length of sector $S_1$ .; The ratio of the area of sector $S_1$ to its arc length is half the ratio of the area of sector $S_2$ to its arc length.

Answer

The true statements are that the area of sector

S_2

is twice the area of sector

S_1

, the arc length of sector

S_2

is equal to the arc length of sector

S_1

, and the ratio of area to arc length for sector

S_1

is half that of sector

S_2

.

The area of sector

S_2

is twice that of

S_1

because quadrupling

r^2

combined with halving the central angle results in a factor of 2. The arc length of sector

S_2

equals that of

S_1

because doubling

r

and halving the angle cancel each other out. The area-to-arc-length ratio of any sector reduces to

\frac{R}{2}

, so sector

S_1

with radius

r

has ratio

\frac{r}{2}

, which is half the ratio

r

of sector

S_2

.

Step-by-Step Solution

1

Calculate and compare the sector areas.

\text{Area}(S_1) = \frac{\theta}{360}\pi r^2

and

\text{Area}(S_2) = \frac{\theta/2}{360}\pi (2r)^2 = \frac{\theta}{720}\pi (4r^2) = \frac{2\theta}{360}\pi r^2 = 2 \cdot \text{Area}(S_1)

.

The sector area formula is

\text{Area} = \frac{\text{angle}}{360^\circ} \pi R^2

. Doubling the radius quadruples

R^2

, while halving the angle reduces the fraction by half, producing a net doubling of area.

2

Calculate and compare the arc lengths.

\text{Arc}(S_1) = \frac{\theta}{360}(2\pi r)

and

\text{Arc}(S_2) = \frac{\theta/2}{360}(2\pi \cdot 2r) = \frac{\theta}{360}(2\pi r) = \text{Arc}(S_1)

.

The arc length formula is

\text{Arc} = \frac{\text{angle}}{360^\circ} (2\pi R)

. Doubling the radius doubles

R

, while halving the angle halves the fraction, leaving the product unchanged.

3

Calculate and compare the sector perimeters.

\text{Perimeter}(S_1) = 2r + \text{Arc}(S_1)

and

\text{Perimeter}(S_2) = 4r + \text{Arc}(S_2) = 4r + \text{Arc}(S_1)

.

The perimeter of a sector consists of two straight radii and the curved arc length.

2 \cdot \text{Perimeter}(S_1) = 4r + 2\text{Arc}(S_1) \neq \text{Perimeter}(S_2)

.

4

Evaluate the area-to-arc-length ratios for both sectors.

For

S_1

,

\frac{\text{Area}(S_1)}{\text{Arc}(S_1)} = \frac{\frac{\theta}{360}\pi r^2}{\frac{\theta}{360}2\pi r} = \frac{r}{2}

. For

S_2

,

\frac{\text{Area}(S_2)}{\text{Arc}(S_2)} = \frac{2r}{2} = r

.

The ratio of area to arc length simplifies to

\frac{R}{2}

for any sector, so doubling the radius doubles this ratio.

Key Concept

Geometric properties of circle sectors, including proportional relationships between radius, central angle, arc length, sector area, and total sector perimeter.

Estimated Time:2m 0s

Question 2Question →

In circle $O$ , sector $AOB$ has a central angle of $60^\circ$ and a radius of $12$ . A smaller circle $C$ is inscribed within sector $AOB$ such that it is tangent to radii $OA$ and $OB$ , as well as to arc $AB$ . What is the area of the region inside sector $AOB$ that lies outside circle $C$ ?

4\pi

8\pi

12\pi

16\pi

18\pi

Show answer & explanation

Answer: $8\pi$

Answer

The area of the region inside sector

AOB

outside circle

C

is

8\pi

.

The correct answer is derived by first finding the area of sector

AOB

using

\frac{60}{360} \pi (12^2) = 24\pi

. Then, analyzing the geometry of the inscribed circle reveals that the line from

O

to the center of circle

C

bisects the

60^\circ

angle. In the resulting

30^\circ-60^\circ-90^\circ

right triangle, the hypotenuse length is

2r

, making the total radius of sector

AOB

equal to

2r + r = 3r = 12

, which yields

r = 4

. The area of circle

C

is

\pi (4^2) = 16\pi

. Subtracting the circle area from the sector area gives

24\pi - 16\pi = 8\pi

.

Step-by-Step Solution

1

Calculate the area of sector

AOB

Sector area

= \frac{60^\circ}{360^\circ} \times \pi (12^2) = \frac{1}{6} \times 144\pi = 24\pi

The area of a sector with central angle

\theta

and radius

R

is

\frac{\theta}{360^\circ} \pi R^2

.

2

Find the radius

r

of the inscribed circle

C

Distance from

O

to center of circle

C

is

OP = 2r

, so total radius

R = OP + r = 3r = 12 \implies r = 4

The line segment connecting center

O

to center

P

of circle

C

bisects the

60^\circ

angle into two

30^\circ

angles. A perpendicular dropped from

P

to radius

OA

forms a

30^\circ-60^\circ-90^\circ

right triangle where

\sin(30^\circ) = \frac{r}{OP} = \frac{1}{2}

, giving

OP = 2r

.

3

Calculate the area of circle

C

Area of circle

C = \pi r^2 = \pi (4^2) = 16\pi

The area of a circle with radius

r

is

\pi r^2

.

4

Subtract the area of circle

C

from the area of sector

AOB

24\pi - 16\pi = 8\pi

The desired region is the difference between the full sector area and the enclosed circle's area.

Key Concept

Inscribed circles in sectors and central angle sector area calculations

Estimated Time:2m 30s

Question 3Question →

An equilateral triangle $ABC$ with side length $6\sqrt{3}$ is inscribed in a circle with center $O$ . What is the area of sector $AOB$ , divided by $\pi$ ?

Show answer & explanation

Answer: 12

Answer

The area of sector

AOB

divided by

\pi

is 12.

Since triangle

ABC

is equilateral, its three vertices divide the circle into three congruent arcs of

120^\circ

each. Thus, central angle

\angle AOB = 120^\circ

. The relationship between the side length

s

of an inscribed equilateral triangle and the radius

R

of its circumscribed circle is

s = R\sqrt{3}

. Given

s = 6\sqrt{3}

, we solve for

R

to find

R = 6

. The area of sector

AOB

is

\frac{120^\circ}{360^\circ} \pi R^2 = \frac{1}{3} \pi (6^2) = 12\pi

. Dividing this area by

\pi

yields

12

.

Step-by-Step Solution

1

Find the central angle

\angle AOB

corresponding to side

AB

of the inscribed equilateral triangle.

\angle AOB = 120^\circ

An inscribed equilateral triangle divides the

360^\circ

circle into three equal central angles.

2

Calculate the radius

R

of the circumscribed circle from the given side length

s = 6\sqrt{3}

.

R = 6

In an inscribed equilateral triangle,

s = R\sqrt{3}

. Substituting

6\sqrt{3} = R\sqrt{3}

gives

R = 6

.

3

Compute the area of sector

AOB

and divide by

\pi

.

12

\text{Sector Area} = \frac{120^\circ}{360^\circ} \times \pi \times 6^2 = 12\pi

. Dividing by

\pi

leaves

12

.

Key Concept

Relationship between inscribed shapes, circle radii, and sector area

Question 4Question →

In circle $O$ , a sector with a central angle of $60^\circ$ has an area of $6\pi$ . In circle $P$ , a sector with a central angle of $120^\circ$ has an arc length equal to the arc length of the sector in circle $O$ . What is the area of the sector in circle $P$ ?

12\pi

6\pi

3\pi

9\pi

\frac{3\pi}{2}

Show answer & explanation

Answer: $3\pi$

Answer

3\pi

The sector area of

3\pi

is correctly calculated by first determining the radius of circle

O

(

r_1 = 6

), using it to find the arc length (

2\pi

), setting that as the arc length for the sector in circle

P

to find its radius (

r_2 = 3

), and finally computing its sector area

\frac{120^\circ}{360^\circ} \cdot \pi (3^2) = 3\pi

.

Step-by-Step Solution

1

Find the radius of circle

O

Radius

r_1 = 6

The sector area formula is

A = \frac{\theta}{360^\circ} \cdot \pi r^2

. Given

A = 6\pi

and

\theta = 60^\circ

, we have

\frac{60^\circ}{360^\circ} \cdot \pi r_1^2 = 6\pi \Rightarrow \frac{1}{6} r_1^2 = 6 \Rightarrow r_1^2 = 36 \Rightarrow r_1 = 6

.

2

Calculate the arc length of the sector in circle

O

Arc length

L = 2\pi

The arc length formula is

L = \frac{\theta}{360^\circ} \cdot 2\pi r

. Substituting

\theta = 60^\circ

and

r_1 = 6

, we get

L = \frac{1}{6} \cdot 2\pi (6) = 2\pi

.

3

Find the radius of circle

P

Radius

r_2 = 3

The sector in circle

P

has an arc length equal to

2\pi

and a central angle of

120^\circ

. Using

L = \frac{120^\circ}{360^\circ} \cdot 2\pi r_2

, we get

2\pi = \frac{1}{3} \cdot 2\pi r_2 \Rightarrow r_2 = 3

.

4

Calculate the area of the sector in circle

P

Sector area

= 3\pi

Using the sector area formula for circle

P

:

A_2 = \frac{120^\circ}{360^\circ} \cdot \pi r_2^2 = \frac{1}{3} \cdot \pi (3^2) = 3\pi

.

Key Concept

Relationship between central angles, radii, arc lengths, and sector areas in circles

Question 5Question →

In a circle centered at point $O$ , sector $AOB$ has a radius of length $r$ , a central angle of $\theta$ degrees, an arc length of $L$ , an area of $A_s$ , and a perimeter of $P$ . If the ratio of the sector area to the arc length is $\frac{A_s}{L} = 4$ , and the ratio of the sector area to the sector perimeter is $\frac{A_s}{P} = \frac{4}{3}$ , which of the following statements must be true? Select all that apply.

Select all that apply

The radius

r

of the circle is

8

.

The area

A_s

of sector

AOB

is

32

.

The central angle

\theta

is greater than

60^\circ

.

The perimeter

P

of sector

AOB

is

24

.

The arc length

L

of sector

AOB

is

8\pi

.

Show answer & explanation

Answer: The radius $r$ of the circle is $8$ .; The area $A_s$ of sector $AOB$ is $32$ .; The perimeter $P$ of sector $AOB$ is $24$ .

Answer

The true statements are that the radius of the circle is 8, the area of the sector is 32, and the perimeter of the sector is 24.

The ratio of sector area to arc length simplifies directly to half the radius,

\frac{r}{2} = 4

, giving a radius of

8

. Expressing the sector perimeter as the sum of the arc length and two radii (

L + 16

) and using

\frac{A_s}{P} = \frac{4L}{L + 16} = \frac{4}{3}

yields

L = 8

. From this, the sector area is

A_s = 32

and the sector perimeter is

P = 24

.

Step-by-Step Solution

1

Relate sector area

A_s

and arc length

L

to find the radius

r

.

A_s = \frac{\theta}{360}\pi r^2

and

L = \frac{\theta}{360}2\pi r

, so

\frac{A_s}{L} = \frac{r}{2} = 4 \implies r = 8

.

Dividing the sector area formula by the arc length formula cancels the central angle fraction and

\pi

.

2

Set up the ratio equation for

\frac{A_s}{P}

using

P = L + 2r

.

Since

r = 8

,

P = L + 16

. Given

A_s = 4L

, we have

\frac{4L}{L + 16} = \frac{4}{3} \implies 3L = L + 16 \implies 2L = 16 \implies L = 8

.

Expressing both sector area and perimeter in terms of arc length allows direct solution for

L

.

3

Calculate sector area

A_s

and perimeter

P

.

A_s = 4(8) = 32

and

P = 8 + 2(8) = 24

.

Substituting

L = 8

and

r = 8

gives the exact values for area and perimeter.

4

Determine the central angle

\theta

.

8 = \frac{\theta}{360} \cdot 2\pi(8) \implies 8 = \frac{16\pi \theta}{360} \implies \theta = \left(\frac{90}{\pi}\right)^\circ \approx 28.65^\circ

.

Plugging

L = 8

and

r = 8

into the arc length formula gives

\theta \approx 28.65^\circ

, which is less than

60^\circ

.

Key Concept

Relating sector area, arc length, and perimeter using proportional ratios and fundamental circle formulas.

Question 6Question →

A sector of a circle with a radius of $12$ units has an area of $24\pi$ square units. What is the measure, in degrees, of the central angle of the sector?

Show answer & explanation

Answer: 60

Answer

The measure of the central angle of the sector is

60^\circ

.

The area of the full circle is

\pi r^2 = \pi (12)^2 = 144\pi

. The sector area represents a fraction of the total area, specifically

\frac{24\pi}{144\pi} = \frac{1}{6}

. Multiplying this fraction by the full circle's central angle of

360^\circ

gives

\frac{1}{6} \times 360^\circ = 60^\circ

.

Step-by-Step Solution

1

Find the total area of the circle

The total area of the circle is

\pi \times 12^2 = 144\pi

.

The area of a full circle with radius

r

is given by

A = \pi r^2

.

2

Relate the sector area to the total circle area

The fraction of the circle represented by the sector is

\frac{24\pi}{144\pi} = \frac{1}{6}

.

The area of a sector is proportional to the fraction of the total central angle (

360^\circ

) it subtends.

3

Calculate the central angle in degrees

\theta = \frac{1}{6} \times 360^\circ = 60^\circ.

Multiply the fraction of the circle by

360^\circ

to get the central angle measure.

Key Concept

Relationship between central angle measure, total circle area, and sector area

Question 7Question →

In a circle, an arc of length $4\pi$ corresponds to a central angle of $40^\circ$ . If the area of the sector formed by this central angle is $k\pi$ , what is the value of $k$ ?

Show answer & explanation

Answer: 36

Answer

The value of

k

is 36.

Using the arc length equation

4\pi = \frac{40}{360} \cdot 2\pi r

, we solve for the radius

r = 18

. Substituting

r = 18

into the sector area formula

A = \frac{40}{360} \cdot \pi (18)^2

results in

A = 36\pi

. Thus,

k = 36

.

Step-by-Step Solution

1

Calculate the radius of the circle using the arc length formula

Radius

r = 18

Arc length is related to central angle and radius by

L = \frac{\theta}{360^\circ} \cdot 2\pi r

. Substituting

L = 4\pi

and

\theta = 40^\circ

gives

4\pi = \frac{1}{9} \cdot 2\pi r \implies r = 18

.

2

Calculate the area of the sector using the radius and central angle

Sector Area

A = 36\pi

Sector area is calculated using

A = \frac{\theta}{360^\circ} \cdot \pi r^2

. Substituting

\theta = 40^\circ

and

r = 18

gives

A = \frac{1}{9} \cdot \pi (18^2) = 36\pi

.

3

Extract the coefficient

k

from

k\pi

k = 36

Comparing

36\pi

to

k\pi

directly yields

k = 36

.

Key Concept

Relationship between central angle, arc length, radius, and sector area

Question 8Question →

In a circle centered at point $O$ , sector $OAB$ has a central angle of $60^\circ$ and a radius of $12$ . A smaller circle is inscribed inside sector $OAB$ such that it is tangent to radius $OA$ , radius $OB$ , and arc $AB$ . If the area of the region inside sector $OAB$ that lies outside the inscribed circle is expressed in the form $k\pi$ , what is the value of $k$ ?

Show answer & explanation

Answer: 8

Answer

The correct value of

k

is 8.

By using the geometry of the

30^\circ

-

60^\circ

-

90^\circ

right triangle formed by the angle bisector and the radius of tangency, the radius of the inscribed circle is found to be

r = 4

. Subtracting its area (

16\pi

) from the sector's area (

24\pi

) gives

8\pi

, so

k = 8

.

Step-by-Step Solution

1

Determine the relationship between the radius of the larger circle

R

and the radius of the inscribed circle

r

.

OP = 2r

and

R = 3r

.

The center

P

of the inscribed circle lies on the angle bisector of

\angle AOB = 60^\circ

, creating a

30^\circ

angle with radius

OA

. The perpendicular distance from

P

to radius

OA

is

r

, so

\sin(30^\circ) = \frac{r}{OP} = \frac{1}{2}

, giving

OP = 2r

. Since the inscribed circle touches arc

AB

,

OP + r = R

, so

3r = R

.

2

Calculate the radius

r

of the inscribed circle.

r = 4

.

Given

R = 12

, solving

3r = 12

yields

r = 4

.

3

Compute the area of sector

OAB

.

\text{Area}_{\text{sector}} = 24\pi

.

The formula for the area of a sector is

\frac{\theta}{360^\circ} \pi R^2

. Here,

\frac{60^\circ}{360^\circ} \pi (12^2) = \frac{1}{6} \times 144\pi = 24\pi

.

4

Compute the area of the inscribed circle.

\text{Area}_{\text{circle}} = 16\pi

.

The area of a circle with radius

r = 4

is

\pi r^2 = \pi (4^2) = 16\pi

.

5

Subtract the area of the inscribed circle from the area of sector

OAB

to find

k

.

k = 8

.

\text{Area}_{\text{region}} = 24\pi - 16\pi = 8\pi

, which means

k = 8

.

Key Concept

Inscribed shapes within sectors, arc length, and sector area relations

Question 9Question →

Two concentric circles centered at point $O$ have radii of $6$ units and $10$ units. A central angle $\theta$ defines a sector that intersects the region between the concentric circles, creating a region bounded by an outer arc, an inner arc, and two straight line segments. If the area of this bounded region is $16\pi$ square units, what is the perimeter of the region?

8\pi

4 + 8\pi

8 + 8\pi

8 + 16\pi

20 + 8\pi

Show answer & explanation

Answer: $8 + 8\pi$

Answer

The perimeter of the region is

8 + 8\pi

units.

The region between the concentric circles has an area equal to the fraction of the central angle times the area of the ring:

\frac{\theta}{360^\circ} \cdot \pi (10^2 - 6^2) = 64\pi \left(\frac{\theta}{360^\circ}\right)

. Setting this equal to

16\pi

gives

\frac{\theta}{360^\circ} = \frac{1}{4}

. The outer arc length is

\frac{1}{4}(2\pi \cdot 10) = 5\pi

, and the inner arc length is

\frac{1}{4}(2\pi \cdot 6) = 3\pi

. The region is bounded on the sides by two segments of length

10 - 6 = 4

units each. The total perimeter is

5\pi + 3\pi + 4 + 4 = 8 + 8\pi

.

Step-by-Step Solution

1

Express the area of the bounded region in terms of the central angle fraction

f = \frac{\theta}{360^\circ}

.

\text{Area} = f \cdot \pi(R^2 - r^2) = f \cdot \pi(10^2 - 6^2) = 64\pi f

.

The area of the region between two concentric sector arcs is the difference between the outer sector area and the inner sector area.

2

Solve for the fraction

f

using the given area of

16\pi

.

64\pi f = 16\pi \implies f = \frac{16\pi}{64\pi} = \frac{1}{4}

.

Equating the algebraic expression for the region's area to

16\pi

allows finding the proportion of the circle represented by the central angle.

3

Calculate the lengths of the outer arc, inner arc, and straight boundary segments.

\text{Outer Arc} = \frac{1}{4}(2\pi \cdot 10) = 5\pi

,

\text{Inner Arc} = \frac{1}{4}(2\pi \cdot 6) = 3\pi

,

\text{Segment Length} = 10 - 6 = 4

.

Arc lengths use the formula

f \cdot 2\pi r

, and each straight boundary segment is the radial distance between the two circles.

4

Sum all boundary components to find the total perimeter.

\text{Perimeter} = 5\pi + 3\pi + 4 + 4 = 8 + 8\pi

.

The complete perimeter of the bounded region consists of the outer arc, inner arc, and two radial segments.

Key Concept

Annular Sector Area and Perimeter

Estimated Time:1m 30s

Question 10Question →

A circle has a radius of $6$ . A central angle of $60^\circ$ intercepts an arc on the circle. What is the perimeter of the sector defined by this central angle?

2\pi

6\pi

2\pi + 6

2\pi + 12

12\pi

Show answer & explanation

Answer: $2\pi + 12$

Answer

2\pi + 12

The sector's perimeter includes the curved arc length and the two straight radii bounding it. The central angle of

60^\circ

represents

\frac{60}{360} = \frac{1}{6}

of the full circle. The arc length is

\frac{1}{6} \times 2\pi(6) = 2\pi

. Adding the two radii of length

6

gives

2\pi + 6 + 6 = 2\pi + 12

.

Step-by-Step Solution

1

Calculate the arc length of the sector

Arc length =

2\pi

The arc length formula is

\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r

. Substituting

\theta = 60^\circ

and

r = 6

yields

\frac{60}{360} \times 12\pi = \frac{1}{6} \times 12\pi = 2\pi

.

2

Calculate the perimeter of the sector by adding the arc length to the two radii

Perimeter =

2\pi + 12

The perimeter of a sector consists of the arc length plus two radii (

2r

). Thus,

\text{Perimeter} = 2\pi + 2(6) = 2\pi + 12

.

Key Concept

Perimeter of a sector equals arc length plus twice the radius (

L + 2r

).

Estimated Time:45s

Question 11Question →

A circle has a radius of $10$ units. A sector within this circle is defined by a central angle of $72^\circ$ . Which of the following statements about this sector are true? Select all such statements.

Select all that apply

The arc length of the sector is

4\pi

units.

The area of the sector is

20\pi

square units.

The area of the sector is

100\pi

square units.

The arc length of the sector is

20\pi

units.

The ratio of the sector's area to the circle's total area is

1

to

5

.

Show answer & explanation

Answer: The arc length of the sector is $4\pi$ units.; The area of the sector is $20\pi$ square units.; The ratio of the sector's area to the circle's total area is $1$ to $5$ .

Answer

The statements confirming an arc length of

4\pi

units, a sector area of

20\pi

square units, and a sector-to-total area ratio of

1

to

5

are correct.

The central angle fraction is

\frac{72^\circ}{360^\circ} = \frac{1}{5}

. Multiplying the full circumference

20\pi

by

\frac{1}{5}

gives an arc length of

4\pi

. Multiplying the total area

100\pi

by

\frac{1}{5}

gives a sector area of

20\pi

. The ratio of sector area to total area is also equal to

\frac{1}{5}

.

Step-by-Step Solution

1

Find the central angle fraction of the circle

The fraction is

\frac{72^\circ}{360^\circ} = \frac{1}{5}

Arc length and sector area are proportional to the central angle relative to a full

360^\circ

turn.

2

Calculate the arc length of the sector

Arc Length

= \frac{1}{5} \times 2\pi(10) = 4\pi

units

The arc length is the circle's circumference multiplied by the central angle fraction.

3

Calculate the area of the sector

Sector Area

= \frac{1}{5} \times \pi(10)^2 = 20\pi

square units

The sector area is the circle's total area multiplied by the central angle fraction.

4

Determine the area ratio

Ratio

= \frac{\text{Sector Area}}{\text{Total Area}} = \frac{20\pi}{100\pi} = \frac{1}{5}

The ratio of the sector area to the total area is identical to the central angle fraction.

Key Concept

Arc Length and Sector Area Formulas

Estimated Time:1m 0s

Question 12Question →

A circle has a circumference of $20\pi$ . A sector of this circle is defined by a central angle that intercepts an arc of length $5\pi$ . What is the area of this sector?

10\pi

25\pi

50\pi

100\pi

5\pi

Show answer & explanation

Answer: $25\pi$

Answer

The area of the sector is

25\pi

.

First, the radius is found using

2\pi r = 20\pi

, which gives

r = 10

. The total area of the circle is

\pi (10)^2 = 100\pi

. The fraction of the circle defined by the sector is

\frac{5\pi}{20\pi} = \frac{1}{4}

. Multiplying the total area by this fraction gives

\frac{1}{4} \times 100\pi = 25\pi

.

Step-by-Step Solution

1

Find the radius of the circle from the circumference.

Since

C = 2\pi r = 20\pi

, solving for

r

gives

r = 10

.

The radius is required to calculate the total circle area.

2

Determine the fraction of the circle that the sector represents.

\text{Fraction} = \frac{\text{Arc Length}}{\text{Circumference}} = \frac{5\pi}{20\pi} = \frac{1}{4}.

The ratio of arc length to circumference gives the proportion of the total circle occupied by the sector.

3

Calculate the total area of the circle.

A = \pi r^2 = \pi (10)^2 = 100\pi.

The area of a circle with radius 10 is

100\pi

.

4

Multiply the total area by the sector fraction.

\text{Sector Area} = \frac{1}{4} \times 100\pi = 25\pi.

Applying the proportional fraction yields the sector area.

Key Concept

The ratio of a sector's arc length to the full circumference is equal to the ratio of the sector's area to the full circle's area.

Question 13Question →

Circle $K$ has a radius of $10$ . Points $P$ and $Q$ lie on circle $K$ such that central angle $\angle POQ$ measures $72^\circ$ , where point $O$ is the center of circle $K$ . Which of the following statements are true? Select all that apply.

Select all that apply

The length of minor arc

PQ

is

4\pi

.

The area of sector

POQ

is

20\pi

.

The perimeter of sector

POQ

is

4\pi

.

The area of sector

POQ

is

40\pi

.

The ratio of the area of sector

POQ

to the total area of circle

K

is

1

to

4

.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ .; The area of sector $POQ$ is $20\pi$ .

Answer

The correct statements are that the length of minor arc

PQ

is

4\pi

, and the area of sector

POQ

is

20\pi

.

The central angle of

72^\circ

corresponds to

\frac{72^\circ}{360^\circ} = \frac{1}{5}

of the entire circle. The circle has a circumference of

2\pi(10) = 20\pi

and an area of

\pi(10)^2 = 100\pi

. Taking

\frac{1}{5}

of the circumference gives an arc length of

4\pi

, and taking

\frac{1}{5}

of the total area gives a sector area of

20\pi

. Both of these statements are mathematically accurate.

Step-by-Step Solution

1

Find the central angle fraction relative to the full circle

\text{Fraction} = \frac{72^\circ}{360^\circ} = \frac{1}{5}

A complete circle subtends

360^\circ

, so the arc and sector comprise one-fifth of the circle.

2

Calculate the circumference and length of minor arc

PQ

\text{Circumference} = 2\pi(10) = 20\pi, \quad \text{Arc Length } PQ = \frac{1}{5} \times 20\pi = 4\pi

Arc length is the central angle fraction multiplied by the total circumference.

3

Calculate the total circle area and area of sector

POQ

\text{Total Area} = \pi(10)^2 = 100\pi, \quad \text{Sector Area } POQ = \frac{1}{5} \times 100\pi = 20\pi

Sector area is the central angle fraction multiplied by the total area of the circle.

4

Evaluate sector perimeter and area ratio for remaining options

\text{Perimeter} = 4\pi + 2(10) = 4\pi + 20; \quad \text{Ratio} = \frac{20\pi}{100\pi} = \frac{1}{5}

Sector perimeter requires adding the two straight radii to the arc length, and the ratio of sector area to total area is

1:5

.

Key Concept

Arc Length and Sector Area of a Circle

Estimated Time:1m 30s

Question 14Question →

A sector of a circle has an area of $15\pi$ square units and a perimeter of $10 + 6\pi$ units. What is the measure of the central angle of the sector, in degrees?

108^\circ

135^\circ

144^\circ

216^\circ

270^\circ

Show answer & explanation

Answer: $216^\circ$

Answer

The correct answer is

216^\circ

, corresponding to the option stating

216^\circ

.

The perimeter of a sector is defined as

2r + s = 10 + 6\pi

, which yields a radius of

r = 5

and an arc length

s = 6\pi

. Checking with the sector area formula

\frac{1}{2}rs = \frac{1}{2}(5)(6\pi) = 15\pi

confirms these measurements. The total area of the circle is

\pi (5)^2 = 25\pi

. The sector thus constitutes

\frac{15\pi}{25\pi} = \frac{3}{5}

of the circle. Multiplying this fraction by

360^\circ

gives a central angle of

216^\circ

.

Step-by-Step Solution

1

Express the sector perimeter formula and solve for radius and arc length.

Radius

r = 5

and arc length

s = 6\pi

.

The perimeter of a sector equals two radii plus its arc length:

\text{Perimeter} = 2r + s = 10 + 6\pi

. Equating standard and

\pi

-termed components gives

2r = 10 \implies r = 5

and arc length

s = 6\pi

.

2

Verify consistency using the sector area formula.

\text{Area} = 15\pi

, matching the given information.

The area of a sector can also be calculated as

\frac{1}{2} r s = \frac{1}{2}(5)(6\pi) = 15\pi

.

3

Find the total area of the circle and the fraction of the circle occupied by the sector.

Total area

= 25\pi

, area fraction

= \frac{3}{5}

.

The full circle area is

\pi r^2 = \pi (5^2) = 25\pi

. The sector represents

\frac{15\pi}{25\pi} = \frac{3}{5}

of the entire circle.

4

Calculate the central angle

\theta

in degrees.

\theta = 216^\circ

.

Multiply the fraction by

360^\circ

:

\theta = \frac{3}{5} \times 360^\circ = 216^\circ

.

Key Concept

Perimeter, Arc Length, and Area of a Circle Sector

Question 15Question →

In a circle centered at point $O$ with radius $R$ , radii $OA$ and $OB$ bound a sector $AOB$ with a central angle of $120^\circ$ . Point $P$ lies on minor arc $AB$ such that the ratio of the length of arc $AP$ to the length of arc $PB$ is $1:3$ . Segment $PQ$ is drawn perpendicular to radius $OA$ , intersecting $OA$ at point $Q$ . What is the ratio of the area of the region bounded by line segment $PQ$ , line segment $AQ$ , and minor arc $AP$ to the area of sector $AOB$ ?

\frac{4\pi - 3\sqrt{3}}{16\pi}

\frac{4\pi - 3\sqrt{3}}{4\pi}

\frac{4\pi - 3\sqrt{3}}{8\pi}

\frac{\pi - 3}{4\pi}

\frac{1}{4}

Show answer & explanation

Answer: $\frac{4\pi - 3\sqrt{3}}{16\pi}$

Answer

\frac{4\pi - 3\sqrt{3}}{16\pi}

The correct answer

\frac{4\pi - 3\sqrt{3}}{16\pi}

is obtained by finding the central angle of sector

AOP

(

30^\circ

), subtracting the area of right triangle

OQP

(

\frac{\sqrt{3}}{16}R^2

) from the area of sector

AOP

(

\frac{\pi}{12}R^2

), and dividing this bounded area by the total sector

AOB

area (

\frac{\pi}{3}R^2

).

Step-by-Step Solution

1

Determine the central angle

\angle AOP

of sector

AOP

\angle AOP = \frac{1}{1+3} \times 120^\circ = 30^\circ

Arc length is directly proportional to central angle. A

1:3

ratio means sector

AOP

represents

\frac{1}{4}

of the

120^\circ

central angle.

2

Calculate the area of sector

AOP

and the total area of sector

AOB

\text{Area(Sector } AOP) = \frac{30^\circ}{360^\circ} \pi R^2 = \frac{\pi}{12} R^2

, and

\text{Area(Sector } AOB) = \frac{120^\circ}{360^\circ} \pi R^2 = \frac{\pi}{3} R^2

The area of a sector with central angle

\theta

is

\frac{\theta}{360^\circ} \pi R^2

.

3

Calculate the area of right triangle

OQP

\text{Area}(\triangle OQP) = \frac{\sqrt{3}}{16} R^2

In

\triangle OQP

,

\angle QOP = 30^\circ

and hypotenuse

OP = R

. Thus

OQ = R \cos(30^\circ) = \frac{R\sqrt{3}}{2}

and

PQ = R \sin(30^\circ) = \frac{R}{2}

.

\text{Area} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \left(\frac{R\sqrt{3}}{2}\right)\left(\frac{R}{2}\right) = \frac{\sqrt{3}}{16} R^2

.

4

Calculate the area of the bounded region and take the ratio to the area of sector

AOB

\text{Ratio} = \frac{\frac{\pi}{12}R^2 - \frac{\sqrt{3}}{16}R^2}{\frac{\pi}{3}R^2} = \frac{4\pi - 3\sqrt{3}}{16\pi}

Subtract the triangle area from sector

AOP

area to get the bounded region area, then divide by sector

AOB

area.

Key Concept

Sector Area, Arc Length Proportions, and Geometric Region Subdivision

Estimated Time:3m 0s

Question 16Question →

Two concentric circles centered at point $O$ have radii $r$ and $R$ , where $r < R$ . A sector bounded by radii $OA$ and $OB$ of the outer circle has central angle $\theta^\circ$ . Region $S$ is the region lying inside sector $AOB$ but outside the inner circle. The area of region $S$ is equal to $3$ times the area of the sector of the inner circle bounded by central angle $\theta^\circ$ . If the perimeter of region $S$ is equal to $\frac{11}{6}$ times the length of arc $AB$ , what is the value of $\theta$ ?

\frac{60}{\pi}

\frac{270}{\pi}

\frac{480}{\pi}

\frac{540}{\pi}

\frac{1080}{\pi}

Show answer & explanation

Answer: $\frac{540}{\pi}$

Answer

\frac{540}{\pi}

The area of region

S

is

\frac{\theta}{360}\pi(R^2 - r^2)

, which is equal to

3

times the area of the inner sector

\frac{\theta}{360}\pi r^2

. This simplifies to

R^2 - r^2 = 3r^2 \implies R = 2r

. The perimeter of region

S

consists of outer arc length

L_{\text{outer}}

, inner arc length

L_{\text{inner}} = \frac{1}{2} L_{\text{outer}}

, and two straight line segments each of length

R - r = r

. Setting the total perimeter

\frac{3}{2} L_{\text{outer}} + 2r

equal to

\frac{11}{6} L_{\text{outer}}

yields

2r = \frac{1}{3} L_{\text{outer}}

, or

L_{\text{outer}} = 6r

. Substituting

L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{4\pi r \theta}{360}

gives

\frac{4\pi r \theta}{360} = 6r

, which solves to

\theta = \frac{540}{\pi}

.

Step-by-Step Solution

1

Relate outer radius

R

to inner radius

r

using sector areas.

R = 2r

The area of region

S

is

\frac{\theta}{360}\pi(R^2 - r^2)

and the inner sector area is

\frac{\theta}{360}\pi r^2

. Setting

\frac{\theta}{360}\pi(R^2 - r^2) = 3 \cdot \frac{\theta}{360}\pi r^2

gives

R^2 - r^2 = 3r^2

, so

R^2 = 4r^2

and

R = 2r

.

2

Express the perimeter of region

S

in terms of outer arc length

L_{\text{outer}}

and radius

r

.

\text{Perimeter}(S) = \frac{3}{2} L_{\text{outer}} + 2r

Region

S

is bounded by outer arc

AB

(

L_{\text{outer}}

), inner arc

CD

(

L_{\text{inner}}

), and two straight segments

AC

and

BD

of length

R - r = 2r - r = r

. Since

R = 2r

,

L_{\text{inner}} = \frac{1}{2} L_{\text{outer}}

, making total perimeter

L_{\text{outer}} + \frac{1}{2} L_{\text{outer}} + 2(r) = \frac{3}{2} L_{\text{outer}} + 2r

.

3

Use the given perimeter relationship to express

L_{\text{outer}}

in terms of

r

.

L_{\text{outer}} = 6r

Setting

\frac{3}{2} L_{\text{outer}} + 2r = \frac{11}{6} L_{\text{outer}}

yields

2r = \left(\frac{11}{6} - \frac{9}{6}\right) L_{\text{outer}} = \frac{1}{3} L_{\text{outer}}

, so

L_{\text{outer}} = 6r

.

4

Solve for

\theta

using the definition of outer arc length.

\theta = \frac{540}{\pi}

L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{\theta}{360}(4\pi r)

. Setting

\frac{4\pi r \theta}{360} = 6r

simplifies to

\frac{\pi \theta}{90} = 6

, giving

\theta = \frac{540}{\pi}

.

Key Concept

Annular sector area and perimeter relations combining arc length formulas and concentric circle geometry.

Estimated Time:2m 30s

Question 17Question →

A circle has a radius of $9$ inches. What is the area, in square inches, of a sector of this circle formed by a central angle of $40^\circ$ ?

2\pi

9\pi

18\pi

27\pi

81\pi

Show answer & explanation

Answer: $9\pi$

Answer

The area of the sector is

9\pi

square inches.

The area of a sector with radius

r

and central angle

\theta

is calculated as

\text{Area} = \pi r^2 \left(\frac{\theta}{360^\circ}\right)

. Substituting

r = 9

and

\theta = 40^\circ

yields

\pi (9^2) \left(\frac{40^\circ}{360^\circ}\right) = 81\pi \left(\frac{1}{9}\right) = 9\pi

square inches.

Step-by-Step Solution

1

Calculate the area of the entire circle using the formula

A_{\text{circle}} = \pi r^2

.

A_{\text{circle}} = \pi (9^2) = 81\pi

square inches.

The full circle area provides the total measure from which the sector fraction is calculated.

2

Determine the fraction of the circle represented by the central angle of

40^\circ

.

\frac{40^\circ}{360^\circ} = \frac{1}{9}

.

A circle contains

360^\circ

in total, so the central angle over

360^\circ

gives the proportion of the circle covered by the sector.

3

Multiply the total area of the circle by the fraction of the sector.

\text{Sector Area} = 81\pi \times \frac{1}{9} = 9\pi

square inches.

Applying the fraction to the total area yields the specific sector area.

Key Concept

Sector Area of a Circle

Question 18Question →

Circle $C_1$ has radius $r_1$ and Circle $C_2$ has radius $r_2$ . An arc on Circle $C_1$ subtended by a central angle of $\theta_1^\circ$ has the exact same length as an arc on Circle $C_2$ subtended by a central angle of $\theta_2^\circ$ . The sector formed by this arc in Circle $C_1$ has an area of $54\pi$ , and the sector formed by this arc in Circle $C_2$ has an area of $36\pi$ . If $\theta_1 + \theta_2 = 150$ , what is the value of $r_1$ ?

Show answer & explanation

Answer: 18

Answer

The radius

r_1

of Circle

C_1

is 18.

Using the relationship

A = \frac{1}{2} L r

, the ratio of the two sector areas gives

\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = 1.5

, so

r_2 = \frac{2}{3} r_1

. Expressing arc length as

L = \frac{108\pi}{r_1}

, the central angles are

\theta_1 = \frac{19440}{r_1^2}

and

\theta_2 = \frac{29160}{r_1^2}

. Setting their sum equal to

150

leads to

\frac{48600}{r_1^2} = 150

, giving

r_1^2 = 324

and

r_1 = 18

.

Step-by-Step Solution

1

Relate sector area to arc length and radius

A_1 = \frac{1}{2} L r_1 = 54\pi

and

A_2 = \frac{1}{2} L r_2 = 36\pi

The area of a sector with arc length

L

and radius

r

is given by

A = \frac{\theta}{360}\pi r^2 = \frac{1}{2} L r

.

2

Find the ratio of

r_1

to

r_2

\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = \frac{3}{2} \implies r_2 = \frac{2}{3}r_1

Dividing the first area equation by the second cancels out

\frac{1}{2}L

.

3

Express central angles in terms of

r_1

\theta_1 = \frac{19440}{r_1^2}

and

\theta_2 = \frac{29160}{r_1^2}

Since

L = \frac{108\pi}{r_1}

, substituting into

\theta_1 = \frac{L}{2\pi r_1} \times 360

yields

\theta_1 = \frac{19440}{r_1^2}

, and substituting into

\theta_2 = \frac{L}{2\pi (\frac{2}{3}r_1)} \times 360

yields

\theta_2 = \frac{29160}{r_1^2}

.

4

Solve for

r_1

using the angle sum equation

r_1 = 18

Summing the angles gives

\frac{19440 + 29160}{r_1^2} = \frac{48600}{r_1^2} = 150 \implies r_1^2 = 324 \implies r_1 = 18

.

Key Concept

Relationship between arc length, radius, central angle, and sector area

Estimated Time:2m 30s

Question 19Question →

A square is inscribed in a circle with center $O$ . The perimeter of the square is $16\sqrt{2}$ units. A sector of this circle has an area equal to the total area of the region inside the circle that lies outside the square. What is the arc length of this sector?

4\pi - 8

8\pi - 16

16\pi - 32

16\pi - 8

8\pi - 32

Show answer & explanation

Answer: $8\pi - 16$

Answer

8\pi - 16

The correct answer is

8\pi - 16

. The inscribed square has side length

4\sqrt{2}

and diagonal

8

. Since the diagonal of an inscribed square is the circle's diameter, the circle has radius

r = 4

and total area

16\pi

. The square has area

32

, so the region outside the square has area

16\pi - 32

. For any sector of radius

r

, the arc length

L

and sector area

A

satisfy

L = \frac{2A}{r}

. Substituting

A = 16\pi - 32

and

r = 4

yields

L = \frac{2(16\pi - 32)}{4} = 8\pi - 16

.

Step-by-Step Solution

1

Find the side length and diagonal of the inscribed square.

Side length

s = \frac{16\sqrt{2}}{4} = 4\sqrt{2}

. Diagonal

d = s\sqrt{2} = (4\sqrt{2})\sqrt{2} = 8

.

The perimeter of a square is

4s

, and the diagonal of a square with side

s

is

s\sqrt{2}

.

2

Determine the radius and total area of the circle.

Diameter equals diagonal

d = 8

, so radius

r = 4

. Area of circle

A_{\text{circle}} = \pi r^2 = 16\pi

.

A square inscribed in a circle has its diagonal aligned with the diameter of the circle.

3

Calculate the area of the region inside the circle but outside the square.

Area of square

A_{\text{square}} = (4\sqrt{2})^2 = 32

. Area outside square

A_{\text{outside}} = 16\pi - 32

.

Subtract the area of the inscribed square from the total area of the circle.

4

Relate sector area to arc length using

L = \frac{2 A_{\text{sector}}}{r}

.

L = \frac{2(16\pi - 32)}{4} = \frac{32\pi - 64}{4} = 8\pi - 16

.

Since sector area

A_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2

and arc length

L = \frac{\theta}{360^\circ}(2\pi r)

, we have

L = \frac{2A_{\text{sector}}}{r}

.

Key Concept

Relationship between circle area, inscribed figures, sector area, and arc length

Estimated Time:1m 30s

Question 20Question →

In circle $O$ , minor arc $AB$ is subtended by a central angle of $120^\circ$ , and the radius of circle $O$ is $6$ . A smaller circle, circle $C$ , is constructed such that it is tangent to chord $AB$ at its midpoint and tangent to minor arc $AB$ at its midpoint. What is the area of a sector of circle $C$ subtended by a central angle of $90^\circ$ ?

\frac{9\pi}{16}

\frac{9\pi}{8}

\frac{9\pi}{4}

\frac{27\pi}{16}

9\pi

Show answer & explanation

Answer: $\frac{9\pi}{16}$

Answer

\frac{9\pi}{16}

The distance from the center of circle

O

to chord

AB

is

6 \cos(60^\circ) = 3

, leaving a height of

6 - 3 = 3

for the minor segment. Circle

C

is tangent to the chord and arc midpoints, making its diameter equal to

3

and its radius

\frac{3}{2}

. The area of a

90^\circ

sector of circle

C

is

\frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{9\pi}{16}

.

Step-by-Step Solution

1

Find the perpendicular distance from center

O

to chord

AB

.

In isosceles triangle

OAB

with

OA = OB = 6

and central angle

\angle AOB = 120^\circ

, the perpendicular bisector from

O

to

AB

meets

AB

at midpoint

M

.

\angle AOM = 60^\circ

, so

OM = OA \cos(60^\circ) = 6 \times \frac{1}{2} = 3

.

Determining

OM

allows us to find the sagitta (height) of the minor segment bounded by chord

AB

and minor arc

AB

.

2

Calculate the height of the minor segment (diameter of circle

C

).

The height of the minor segment along the radius passing through

M

to arc midpoint

N

is

MN = ON - OM = 6 - 3 = 3

. Since circle

C

is tangent to

AB

at

M

and to the arc at

N

, segment

MN

is a diameter of circle

C

, giving a diameter of

3

.

The space between the chord midpoint and arc midpoint bounds circle

C

, defining its diameter.

3

Find the radius of circle

C

.

Radius

r_C = \frac{\text{diameter}}{2} = \frac{3}{2}

.

The radius of circle

C

is needed to calculate its sector area.

4

Compute the area of the

90^\circ

sector of circle

C

.

\text{Area} = \frac{\theta}{360^\circ} \pi r_C^2 = \frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{1}{4} \pi \left(\frac{9}{4}\right) = \frac{9\pi}{16}.

Multiply the fraction of the full circle represented by the central angle by the total area of circle

C

.

Key Concept

Calculating sector area of an inscribed circle within a circle segment

Estimated Time:2m 0s

Circles, Arc Lengths, and Sector Areas

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