Circles, Arc Lengths, and Sector Areas Practice Questions — GRE General Test — Page 2

Question 21Question →

In a circle centered at point $O$ with a radius of $6$ units, points $P$ and $Q$ lie on the circle such that central angle $\angle POQ$ measures $120^\circ$ . Which of the following statements are true? Select all such statements.

Select all that apply

The length of minor arc

PQ

is

4\pi

.

The area of minor sector

POQ

is

12\pi

.

The perimeter of sector

POQ

is

4\pi

.

The area of sector

POQ

is

36\pi

.

The length of chord

PQ

is

6\sqrt{3}

.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ .; The area of minor sector $POQ$ is $12\pi$ .; The length of chord $PQ$ is $6\sqrt{3}$ .

Answer

The true statements are that the length of minor arc

PQ

is

4\pi

, the area of minor sector

POQ

is

12\pi

, and the length of chord

PQ

is

6\sqrt{3}

.

The arc length of a

120^\circ

central angle is

\frac{1}{3}

of the total circumference

12\pi

, which gives

4\pi

. The sector area is

\frac{1}{3}

of the total area

36\pi

, which gives

12\pi

. Furthermore, by geometry of an isosceles triangle with vertex angle

120^\circ

and congruent sides

6

, the base chord length is

2 \times 6 \sin(60^\circ) = 6\sqrt{3}

.

Step-by-Step Solution

1

Calculate the arc length of minor arc

PQ

Arc length =

4\pi

Arc length equals

\frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi(6) = 4\pi

.

2

Calculate the area of sector

POQ

Sector area =

12\pi

Sector area equals

\frac{\theta}{360^\circ} \times \pi r^2 = \frac{120^\circ}{360^\circ} \times \pi(6^2) = 12\pi

.

3

Determine the perimeter of sector

POQ

Perimeter =

4\pi + 12

The boundary of a sector consists of the curved arc length plus two straight radii:

4\pi + 6 + 6 = 4\pi + 12

.

4

Calculate the length of chord

PQ

Chord

PQ = 6\sqrt{3}

Using triangle

OPQ

with sides

OP = OQ = 6

and

\angle POQ = 120^\circ

, law of cosines or bisecting

\angle POQ

into two

30^\circ

-

60^\circ

-

90^\circ

triangles yields chord length

2 \times (6 \sin 60^\circ) = 2 \times 3\sqrt{3} = 6\sqrt{3}

.

Key Concept

Arc Length, Sector Area, and Chord Relationships in Circles

Estimated Time:1m 30s

Question 22Question →

In a circle with center $O$ and radius $12$ , radii $OA$ and $OB$ are perpendicular. Point $C$ lies on segment $OA$ such that $C$ is the midpoint of $OA$ . A line segment perpendicular to $OA$ is drawn from point $C$ to intersect minor arc $AB$ at point $D$ . What is the area of the region bounded by line segment $CD$ , line segment $CA$ , and minor arc $AD$ ?

12\pi

24\pi - 18

24\pi - 18\sqrt{3}

36\pi - 18\sqrt{3}

24\pi - 36\sqrt{3}

Show answer & explanation

Answer: $24\pi - 18\sqrt{3}$

Answer

24\pi - 18\sqrt{3}

The area of the region bounded by line segment

CD

, segment

CA

, and minor arc

AD

is obtained by subtracting the area of right triangle

OCD

from the area of sector

OAD

. With

OC = 6

and radius

OD = 12

, triangle

OCD

is a

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

right triangle, giving

CD = 6\sqrt{3}

and central angle

\angle AOD = 60^\circ

. The sector area is

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

, and the triangle area is

\frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}

. Subtracting the triangle area from the sector area yields

24\pi - 18\sqrt{3}

.

Step-by-Step Solution

1

Determine the length of

OC

and the height

CD

OC = 6

and

CD = 6\sqrt{3}

Since

C

is the midpoint of radius

OA = 12

,

OC = 6

. Triangle

OCD

is a right triangle at

C

with hypotenuse

OD = 12

(radius of circle). By the Pythagorean theorem,

CD = \sqrt{12^2 - 6^2} = \sqrt{108} = 6\sqrt{3}

.

2

Find the central angle

\angle AOD

\angle AOD = 60^\circ

In right triangle

OCD

,

\cos(\angle AOD) = \frac{OC}{OD} = \frac{6}{12} = \frac{1}{2}

, which implies

\angle AOD = 60^\circ

.

3

Calculate the area of sector

OAD

\text{Area(Sector } OAD) = 24\pi

$

The area of a sector with central angle

60^\circ

and radius

12

is

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

.

4

Calculate the area of right triangle

OCD

\text{Area(Triangle } OCD) = 18\sqrt{3}

The area of right triangle

OCD

is

\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}

.

5

Subtract the area of triangle

OCD

from the area of sector

OAD

\text{Area of shaded region} = 24\pi - 18\sqrt{3}

The bounded region is formed by removing triangle

OCD

from sector

OAD

.

Key Concept

Calculating the area of a region bounded by a circle arc and line segments by subtracting a right triangle area from a sector area.

Question 23Question →

A sector of a circle with a radius of $10$ units has a total perimeter of $20 + 5\pi$ units. What is the area of this sector?

25\pi

50\pi

5\pi

100\pi

100 + 25\pi

Show answer & explanation

Answer: $25\pi$

Answer

The area of the sector is

25\pi

.

The total perimeter of a sector with radius

r

and arc length

L

is given by

P = 2r + L

. Substituting

r = 10

gives

20 + 5\pi = 20 + L

, so

L = 5\pi

. Using the sector area formula

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

, the area is

\frac{1}{2} \times 10 \times 5\pi = 25\pi

.

Step-by-Step Solution

1

Set up the formula for the perimeter of a sector.

P = 2r + L

, where

r = 10

is the radius and

L

is the arc length.

A sector's perimeter is bounded by two straight radii and one curved arc.

2

Solve for the arc length

L

.

20 + 5\pi = 2(10) + L \implies L = 5\pi

.

Subtracting the combined length of the two radii (

20

) isolates the arc length.

3

Calculate the area of the sector.

Sector Area =

\frac{1}{2} r L = \frac{1}{2} (10)(5\pi) = 25\pi

.

The area of a sector can be directly evaluated using half the product of its radius and arc length.

Key Concept

Perimeter, Arc Length, and Area of a Circular Sector

Alternative Method

Find the central angle

\theta

first: Since

L = 5\pi

and circumference

C = 2\pi(10) = 20\pi

, the fraction of the circle is

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

, which corresponds to

\theta = 90^\circ

. The area is then

\frac{1}{4} \times \pi(10^2) = 25\pi

.

Estimated Time:1m 30s

Question 24Question →

Two concentric circles are centered at point $O$ . The inner circle has a radius of $6$ units, and the outer circle has a radius of $6\sqrt{3}$ units. Radii $OA$ and $OB$ of the outer circle form a central angle $\angle AOB = 60^\circ$ and intersect the inner circle at points $C$ and $D$ , respectively. Which of the following statements must be true regarding the region and boundary lengths defined by these figures? Select all that apply.

Select all that apply

The area of the region bounded by arc

AB

, arc

CD

, segment

AC

, and segment

BD

is equal to

12\pi

square units.

The straight-line distance between points

A

and

B

is

6\sqrt{3}

units.

The area of the circular segment bounded by chord

AB

and minor arc

AB

is

18\pi - 27\sqrt{3}

square units.

The ratio of the length of arc

CD

to the length of arc

AB

is

1 : 3

.

The perimeter of the region bounded by arc

AB

, arc

CD

, segment

AC

, and segment

BD

is

2\pi(1 + \sqrt{3}) + 6\sqrt{3}

units.

Show answer & explanation

Answer: The area of the region bounded by arc $AB$ , arc $CD$ , segment $AC$ , and segment $BD$ is equal to $12\pi$ square units.; The straight-line distance between points $A$ and $B$ is $6\sqrt{3}$ units.; The area of the circular segment bounded by chord $AB$ and minor arc $AB$ is $18\pi - 27\sqrt{3}$ square units.

Answer

The true statements are: the area of the region bounded by arc AB, arc CD, segment AC, and segment BD is 12π square units; the straight-line distance between points A and B is 6√3 units; and the area of the circular segment bounded by chord AB and minor arc AB is 18π - 27√3 square units.

The area of the region between the two concentric arcs is obtained by subtracting the inner sector area (6π) from the outer sector area (18π), giving 12π square units. The triangle OAB is equilateral because it has two sides of length 6√3 and an included angle of 60°, making chord AB equal to 6√3. Subtracting the area of this equilateral triangle (27√3) from the outer sector area (18π) yields the area of the circular segment bounded by chord AB and arc AB, which is 18π - 27√3.

Step-by-Step Solution

1

Calculate the area of sector OAB and sector OCD to evaluate the area of the annular sector region.

Sector OAB area = (60/360) * π * (6√3)^2 = (1/6) * 108π = 18π. Sector OCD area = (60/360) * π * (6^2) = (1/6) * 36π = 6π. Region area = 18π - 6π = 12π.

The area between two concentric sector arcs bounded by the same radii is the difference in sector areas.

2

Determine the length of chord AB using triangle properties.

In triangle OAB, OA = OB = 6√3 and angle AOB = 60°. An isosceles triangle with a 60° vertex angle is equilateral, so AB = 6√3.

All internal angles of an isosceles triangle with a 60° angle must equal 60°.

3

Calculate the area of circular segment AB.

Area of equilateral triangle OAB = (√3 / 4) * (6√3)^2 = 27√3. Segment area = Sector OAB area - Triangle OAB area = 18π - 27√3.

A circular segment's area is found by subtracting the area of the subtended triangle from the area of the corresponding sector.

4

Evaluate arc length ratio and total perimeter of the annular sector.

Arc CD / Arc AB = 6 / (6√3) = 1 / √3 ≠ 1 / 3. Total perimeter = Arc AB + Arc CD + 2*(R - r) = 2√3π + 2π + 2*(6√3 - 6) = 2π(1 + √3) + 12√3 - 12.

Verifies that statements regarding ratio 1:3 and incomplete perimeter calculations are mathematically false.

Key Concept

Concentric circle geometry, arc length proportions, sector area calculations, and circular segment area formulation.

Question 25Question →

In a circle centered at point $O$ , the ratio of the area of sector $AOB$ to the area of the entire circle is $3:8$ . If the total perimeter of sector $AOB$ is $24 + 9\pi$ , what is the radius of the circle?

Show answer & explanation

Answer: 12

Answer

The radius of the circle is 12.

Since the ratio of the sector area to the total circle area is

3:8

, the sector accounts for

\frac{3}{8}

of the entire circumference

2\pi r

, making the arc length

L = \frac{3}{8}(2\pi r) = \frac{3}{4}\pi r

. The perimeter of the sector is

2r + L = 2r + \frac{3}{4}\pi r

. Equating this to

24 + 9\pi

gives

2r = 24

, so

r = 12

.

Step-by-Step Solution

1

Relate sector area ratio to arc length

Arc length

L = \frac{3}{4}\pi r

The fraction of the circle occupied by the sector is

\frac{3}{8}

, so the arc length is

\frac{3}{8}

of the circle's circumference

2\pi r

.

2

Formulate the perimeter expression for sector

AOB

Perimeter =

2r + \frac{3}{4}\pi r

The boundary of a sector includes two straight radii of length

r

plus the curved arc length

L

.

3

Equate to given perimeter and solve for

r

r = 12

Comparing

2r + \frac{3}{4}\pi r

to

24 + 9\pi

, setting

2r = 24

yields

r = 12

, which also satisfies

\frac{3}{4}\pi (12) = 9\pi

.

Key Concept

Perimeter of a circle sector and fractional relationship between sector area, central angle, and arc length.

Question 26Question →

In circle $O$ , points $P$ and $Q$ lie on the circumference such that the ratio of the minor arc length $PQ$ to the radius $r$ of the circle is $\frac{5\pi}{6}$ . If the area of sector $POQ$ is $30\pi$ , what is the total perimeter of sector $POQ$ ?

12\sqrt{2} + 5\sqrt{2}\pi

12 + 5\pi

5\sqrt{2}\pi

12 + 10\pi

12\sqrt{2} + 12\sqrt{2}\pi

Show answer & explanation

Answer: $12\sqrt{2} + 5\sqrt{2}\pi$

Answer

12\sqrt{2} + 5\sqrt{2}\pi

The central angle

\theta

in radians is equal to the ratio of minor arc length to radius, so

\theta = \frac{5\pi}{6}

. Using the area of a sector formula

A = \frac{1}{2}r^2\theta

, we substitute

A = 30\pi

to get

\frac{1}{2}r^2\left(\frac{5\pi}{6}\right) = 30\pi

, which simplifies to

r^2 = 72

, or

r = 6\sqrt{2}

. The arc length is then

s = r\theta = (6\sqrt{2})\left(\frac{5\pi}{6}\right) = 5\sqrt{2}\pi

. The total perimeter of the sector includes both bounding radii and the arc length:

2r + s = 12\sqrt{2} + 5\sqrt{2}\pi

.

Step-by-Step Solution

1

Relate the central angle in radians to the given ratio of arc length to radius.

The central angle

\theta

in radians is given by

\theta = \frac{\text{arc length}}{r} = \frac{5\pi}{6}

.

By definition of radian measure, arc length

s = r\theta

, so

\frac{s}{r} = \theta

.

2

Use the sector area formula to solve for the radius

r

.

Setting

\frac{1}{2}r^2\left(\frac{5\pi}{6}\right) = 30\pi

yields

\frac{5\pi}{12}r^2 = 30\pi \implies r^2 = 72 \implies r = 6\sqrt{2}

.

The area of a sector with central angle

\theta

(in radians) is

A = \frac{1}{2}r^2\theta

.

3

Calculate the minor arc length

s

.

Arc length

s = r\theta = (6\sqrt{2})\left(\frac{5\pi}{6}\right) = 5\sqrt{2}\pi

.

Multiplying the radius by the central angle in radians gives the length of the subtended arc.

4

Calculate the total perimeter of sector

POQ

.

Perimeter

= 2r + s = 2(6\sqrt{2}) + 5\sqrt{2}\pi = 12\sqrt{2} + 5\sqrt{2}\pi

.

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

Key Concept

Relationship between radian measure, sector area, arc length, and sector perimeter

Estimated Time:2m 0s

Question 27Question →

In circle $O$ , sector $AOB$ has an area of $45\pi$ square units and arc $AB$ has a length of $6\pi$ units. What is the perimeter of sector $AOB$ ?

15 + 6\pi

30 + 6\pi

45 + 6\pi

60 + 6\pi

30\pi

Show answer & explanation

Answer: $30 + 6\pi$

Answer

30 + 6\pi

The area of a sector is given by

\frac{1}{2} r L

, where

r

is the radius and

L

is the arc length. Substituting

L = 6\pi

and Area

= 45\pi

yields

45\pi = \frac{1}{2} r (6\pi)

, which simplifies to

45\pi = 3\pi r

, so

r = 15

. The perimeter of the sector is

L + 2r = 6\pi + 2(15) = 30 + 6\pi

. Thus, the option equal to

30 + 6\pi

is correct.

Step-by-Step Solution

1

Express sector area and arc length in terms of radius

r

and central angle

\theta

in degrees.

Sector Area =

\frac{\theta}{360^\circ} \times \pi r^2 = 45\pi

and Arc Length =

\frac{\theta}{360^\circ} \times 2\pi r = 6\pi

.

Relating both formulas allows solving for the radius directly using the ratio of area to arc length.

2

Divide the sector area formula by the arc length formula to find radius

r

.

\frac{\text{Sector Area}}{\text{Arc Length}} = \frac{\frac{\theta}{360^\circ} \pi r^2}{\frac{\theta}{360^\circ} 2\pi r} = \frac{r}{2} = \frac{45\pi}{6\pi} = 7.5 \implies r = 15.

The central angle fraction

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

and

\pi

cancel out, leaving

\frac{r}{2} = 7.5

.

3

Calculate the total perimeter of sector

AOB

.

\text{Perimeter} = \text{Arc Length} + 2r = 6\pi + 2(15) = 30 + 6\pi.

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

OA

and

OB

).

Key Concept

The relationship between sector area, arc length, radius, and sector perimeter

Question 28Question →

In circle $O$ , line segments $AB$ and $CD$ are perpendicular diameters, each of length $12$ . An arc of a second circle, centered at point $A$ with radius $AC$ , is drawn from point $C$ to point $D$ through the interior of circle $O$ . What is the area of the crescent-shaped region bounded by the semicircle $CBD$ of circle $O$ and arc $CD$ of the second circle?

Show answer & explanation

Answer: 36

Answer

36

Circle

O

has radius

r = 6

, giving semicircle

CBD

an area of

\frac{1}{2}\pi(6^2) = 18\pi

. The distance

AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

is the radius of circle

A

. Because

\angle CAD = 90^\circ

, sector

ACD

has area

\frac{90^\circ}{360^\circ}\pi(6\sqrt{2})^2 = 18\pi

. Subtracting the area of triangle

ACD

(

\frac{1}{2} \times 12 \times 6 = 36

) yields a segment area of

18\pi - 36

. Subtracting this segment area from the semicircle area yields

18\pi - (18\pi - 36) = 36

.

Step-by-Step Solution

1

Find the radii of circle

O

and circle

A

.

Radius of circle

O

is

r = 6

. In right triangle

AOC

,

OA = OC = 6

, so radius

AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

.

Perpendicular diameters

AB

and

CD

intersect at center

O

, dividing each diameter into radii of length

6

.

2

Calculate the area of sector

ACD

of circle

A

and triangle

ACD

.

Sector area

= \frac{90^\circ}{360^\circ} \pi (6\sqrt{2})^2 = 18\pi

. Triangle area

= \frac{1}{2} \times 12 \times 6 = 36

.

Angle

\angle CAD = 90^\circ

because

\triangle ACD

is a right isosceles triangle with hypotenuse

CD = 12

.

3

Find the area of the circular segment bounded by chord

CD

and arc

CD

of circle

A

.

Segment Area

= 18\pi - 36

.

The area of a circular segment is equal to the sector area minus the triangle area.

4

Subtract the segment area from the area of semicircle

CBD

of circle

O

.

Region Area

= 18\pi - (18\pi - 36) = 36

.

Semicircle

CBD

has radius

6

and area

\frac{1}{2}\pi(6^2) = 18\pi

. Subtracting the segment area leaves the crescent region.

Key Concept

Area of circular sectors, segments, and compound regions (Lune of Hippocrates)

Question 29Question →

In a circle centered at point $O$ , the radius is $12$ units and central angle $\angle AOB$ measures $150^\circ$ . Which of the following statements regarding sector $AOB$ and minor arc $AB$ are correct? Select all such statements.

Select all that apply

The length of minor arc

AB

is

10\pi

units.

The area of sector

AOB

is

60\pi

square units.

The total perimeter of sector

AOB

is

24 + 10\pi

units.

The length of minor arc

AB

is

24\pi

units.

The perimeter of sector

AOB

is equal to

10\pi

units.

Show answer & explanation

Answer: The length of minor arc $AB$ is $10\pi$ units.; The area of sector $AOB$ is $60\pi$ square units.; The total perimeter of sector $AOB$ is $24 + 10\pi$ units.

Answer

The correct statements are those indicating that the minor arc length is

10\pi

units, the sector area is

60\pi

square units, and the perimeter of the sector is

24 + 10\pi

units.

The central angle fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

. Multiplying the full circumference (

24\pi

) by

\frac{5}{12}

gives an arc length of

10\pi

. Multiplying the full circle area (

144\pi

) by

\frac{5}{12}

gives a sector area of

60\pi

. Adding the two radii (

2 \times 12 = 24

) to the arc length (

10\pi

) gives the total sector perimeter of

24 + 10\pi

.

Step-by-Step Solution

1

Determine the central angle fraction of the circle.

The fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

.

Arc length and sector area are proportional to the ratio of the central angle to

360^\circ

.

2

Calculate the circumference of the circle and the arc length of minor arc

AB

.

Circumference =

2\pi(12) = 24\pi

. Arc length =

\frac{5}{12} \times 24\pi = 10\pi

units.

Arc length equals full circumference multiplied by the central angle fraction.

3

Calculate the total area of the circle and the area of sector

AOB

.

Circle Area =

\pi(12^2) = 144\pi

. Sector Area =

\frac{5}{12} \times 144\pi = 60\pi

square units.

Sector area equals total circle area multiplied by the central angle fraction.

4

Calculate the total perimeter of sector

AOB

.

Perimeter =

\text{arc length} + 2r = 10\pi + 2(12) = 24 + 10\pi

units.

The perimeter of a sector consists of the bounding arc plus the two radii.

Key Concept

Circles, Arc Lengths, and Sector Areas

Question 30Question →

In a circle centered at point $O$ , the radius is $18$ units. Points $P$ and $Q$ lie on the circle such that the area of sector $POQ$ is $54\pi$ square units. What is the ratio of the length of minor arc $PQ$ to the total perimeter of sector $POQ$ ?

\frac{\pi}{\pi + 6}

\frac{1}{6}

\frac{\pi}{\pi + 3}

\frac{6\pi}{\pi + 6}

\frac{9\pi}{\pi + 6}

Show answer & explanation

Answer: $\frac{\pi}{\pi + 6}$

Answer

The ratio of the length of minor arc

PQ

to the total perimeter of sector

POQ

is

\frac{\pi}{\pi + 6}

.

The full circle area is

324\pi

, making the sector

54\pi / 324\pi = 1/6

of the circle. The arc length is

1/6 \times 36\pi = 6\pi

. The sector perimeter is the arc length plus two radii (

6\pi + 36

). Taking the ratio of arc length to sector perimeter yields

6\pi / (6\pi + 36) = \pi / (\pi + 6)

.

Step-by-Step Solution

1

Calculate the total area of the circle and determine the fractional size of sector

POQ

.

Total area =

\pi(18)^2 = 324\pi

. Sector fraction =

\frac{54\pi}{324\pi} = \frac{1}{6}

.

Determining the fraction of the circle represented by the sector is required to find the arc length.

2

Calculate the length of minor arc

PQ

and the total perimeter of sector

POQ

.

Minor arc

PQ = \frac{1}{6} \times 2\pi(18) = 6\pi

. Sector perimeter =

6\pi + 2(18) = 6\pi + 36

.

The sector perimeter consists of the curved arc length plus the two straight radii

OP

and

OQ

.

3

Form and simplify the ratio of arc length to sector perimeter.

\frac{6\pi}{6\pi + 36} = \frac{6\pi}{6(\pi + 6)} = \frac{\pi}{\pi + 6}

.

Factoring out 6 from the numerator and denominator simplifies the expression to its lowest form.

Key Concept

Arc Length and Sector Perimeter Calculations

Question 31Question →

In a circle centered at point $O$ , the radius is $6$ units and central angle $\angle POQ$ measures $120^\circ$ . Which of the following statements regarding sector $POQ$ and minor arc $PQ$ must be true? Select all such statements.

Select all that apply

The length of minor arc

PQ

is

4\pi

units.

The area of sector

POQ

is

12\pi

square units.

The area of sector

POQ

is

36\pi

square units.

The length of minor arc

PQ

is

12\pi

units.

The perimeter of sector

POQ

is

4\pi + 6

units.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ units.; The area of sector $POQ$ is $12\pi$ square units.

Answer

The correct statements are that the length of minor arc

PQ

is

4\pi

units and the area of sector

POQ

is

12\pi

square units.

The correct statements accurately apply the arc length and sector area formulas by multiplying the full circumference (

12\pi

) and full area (

36\pi

) by the central angle fraction

\frac{120^\circ}{360^\circ} = \frac{1}{3}

, yielding an arc length of

4\pi

units and a sector area of

12\pi

square units.

Step-by-Step Solution

1

Calculate the central angle fraction of the circle.

The fraction of the circle represented by sector

POQ

is

\frac{120^\circ}{360^\circ} = \frac{1}{3}

.

Arc length and sector area are proportional to the ratio of the central angle to

360^\circ

.

2

Calculate the arc length of minor arc

PQ

.

\text{Arc length} = \frac{1}{3} \times 2\pi(6) = 4\pi \text{ units}.

Arc length equals the central angle fraction times the total circumference

2\pi r

.

3

Calculate the area of sector

POQ

.

\text{Sector area} = \frac{1}{3} \times \pi(6^2) = 12\pi \text{ square units}.

Sector area equals the central angle fraction times the total circle area

\pi r^2

.

4

Evaluate the given statements against the calculated values.

Statements asserting an arc length of

4\pi

units and a sector area of

12\pi

square units are true. Other statements miscalculate by omitting the fraction or adding incorrect boundary components.

Comparing calculated values confirms the valid choices.

Key Concept

Arc Length and Sector Area Formulas

Circles, Arc Lengths, and Sector Areas

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Alternative Method

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