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HiSET: Math : Arc length and area of a sector

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Example Questions

Example Question #21 : Measurement And Geometry

Circle area

The depicted circle has a radius of 3 cm. The arc length between the two points shown on the circle is $\pi$ cm. Find the area of the enclosed sector (highlighted in green).

Possible Answers:

$\frac{1}{6}\; cm^2$

$\frac{\pi}{6}\; cm^2$

$27\; cm^2$

$\frac{3\pi}{2}\; cm^2$

$\pi\; cm^2$

Correct answer:

$\frac{3\pi}{2}\; cm^2$

Explanation:

First, find the area and circumference of the circle using the radius and the following formulae for circles:

$Circumference = 2\pi r$

$Area = \pi r^2$

Substituting in 3 for yields:

$Circumference = 6\pi \ cm$

$Area = 9\pi \ cm^2$

Next, find what fraction of the total circumference is between the two points on the circle (the arc length).

$\frac{Arc \ Length }{Circumference } = \frac{\pi}{6\pi} = \frac{1}{6}$

Finally, use this fraction to calculate the area of the enclosed sector. Note that this area is proportional to the above fraction. In other words:

$\frac{Arc \ Length }{Circumference } = \frac{Sector \ Area}{Total \ Area}$

So the Sector Area is one sixth of the total area.

$\frac{1 }{6} = \frac{Sector \ Area}{9\pi}$

Cross multiply:

$Sector \ Area\times6 = 9\pi$

Divide both sides by 6, then simplify to get the final answer:

$Sector \ Area = \frac{9\pi}{6} = \frac{3\pi}{2}\; cm^2$

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Example Question #1 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 6 . Evaluate the length of $\overarc{AB}$ .

Possible Answers:

$\frac{15}{2} \pi$

$\frac{9}{4} \pi$

$6 \pi$

$\frac{15}{4} \pi$

$\frac{9}{2} \pi$

Correct answer:

$\frac{9}{2} \pi$

Explanation:

The radius of the circle is given to be r= AO = 6 . The total circumference of the circle is $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (6)= 12\pi$ .

The length of $\overarc{AB}$ is equal to

$l = \frac{m \overarc {AB}}{360} \cdot C$

Thus, it is first necessary to find the degree measure of $\overarc{AB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

360- 2 x = 90

Subtract 360 from both sides:

360- 2 x - 360 = 90 - 360

-2x = -270

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

x = 135 ,

the degree measure of $\overarc{AB}$ .

Thus, the length of $\overarc{AB}$ is

$l = \frac{m \overarc {AB}}{360} \cdot C$

$= \frac{225}{360} \cdot (12 \pi)$

$= \frac{3}{8} (12\pi)$

$= \frac{36}{8} \pi$

$= \frac{9}{2} \pi$

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Example Question #1 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 8 . Evaluate the area of the shaded sector.

Possible Answers:

$12 \pi$

$40 \pi$

The information given is insufficient to answer the question.

$24 \pi$

$20 \pi$

Correct answer:

$24 \pi$

Explanation:

The radius of the circle is given to be r= AO = 8 . The total area of the circle can be found using the area formula:

$A = \pi r^{2}$

$A = \pi (8^{2})$

$A =64 \pi$

The area of the sector is equal to

$a = \frac{m \overarc {AB}}{360} \cdot A$

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

360- 2 x = 90

Subtract 360 from both sides:

360- 2 x - 360 = 90 - 360

-2x = -270

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

x = 135 ,

the degree measure of $\overarc{AB}$ .

Thus, the area of the shaded sector is

$a = \frac{m \overarc {AB}}{360} \cdot A$

$= \frac{135}{360} \cdot (64 \pi)$

$= \frac{3}{8} \cdot 64 \pi$

$= \frac{192}{8} \pi$

$=24 \pi$

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Example Question #1 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 6 . Evaluate the area of the shaded sector.

Possible Answers:

$\frac{45}{2} \pi$

$\frac{45}{4} \pi$

$\frac{27}{4} \pi$

$\frac{27}{2} \pi$

The information given is insufficient to answer the question.

Correct answer:

$\frac{45}{2} \pi$

Explanation:

The radius of the circle is given to be r= AO = 6 . The total area of the circle can be found using the area formula:

$A = \pi r^{2}$

$A = \pi (6^{2})$

$A =36\pi$

The area of the sector is equal to

$a = \frac{m \overarc {ACB}}{360} \cdot A$

Thus, it is first necessary to find the degree measure of $\overarc{ACB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{x- (360- x ) }{2} =45$

$\frac{ 2 x -360 }{2} =45$

Multiply both sides by 2:

$\frac{2 x -360}{2} \cdot 2 =45 \cdot 2$

2 x -360 = 90

Add 360 to both sides:

2 x -360+ 360 = 90 + 360

2 x= 450

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

x = 225 ,

the degree measure of $\overarc{ACB}$ .

Therefore, the area of the shaded sector is

$a = \frac{m \overarc {ACB}}{360} \cdot A$

$= \frac{225}{360} \cdot 36 \pi$

$= \frac{5}{8 } \cdot 36 \pi$

$= \frac{180}{8 } \pi$

$= \frac{45}{2} \pi$

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Example Question #1 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 8 . Evaluate the length of $\overarc{ACB}$ .

Possible Answers:

$10 \pi$

$6 \pi$

$5 \pi$

$12 \pi$

$3 \pi$

Correct answer:

$10 \pi$

Explanation:

The radius of the circle is given to be r= AO = 8 .

The total circumference of the circle is $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (8)= 16\pi$ .

The length of $\overarc{ACB}$ is equal to

$l= \frac{m \overarc {ACB}}{360} \cdot C$

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{x- (360- x ) }{2} =45$

$\frac{ 2 x -360 }{2} =45$

Multiply both sides by 2:

$\frac{2 x -360}{2} \cdot 2 =45 \cdot 2$

2 x -360 = 90

Add 360 to both sides:

2 x -360+ 360 = 90 + 360

2 x= 450

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

x = 225 ,

the degree measure of $\overarc{ACB}$ .

Thus, the length of $\overarc{ACB}$ is

$l = \frac{m \overarc {ACB}}{360} \cdot C$

$= \frac{225}{360} \cdot 16 \pi$

$= \frac{5}{8} \cdot 16 \pi$

$= \frac{80}{8} \pi$

$= 10 \pi$

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Example Question #1 : Arc Length And Area Of A Sector

Intercepted

Refer to the above figure. Give the ratio of the area of Sector 2 to that of Sector 1.

Possible Answers:

5:1

25:1

6:1

16:1

4:1

Correct answer:

5:1

Explanation:

The ratio of the area of the larger Sector 2 to that of smaller Sector 1 is equal to the ratio of their respective arc measures - that is,

$m \overarc{ACB} : m \overarc{AD}$ .

Therefore, it is sufficient to find these arc measures.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{x- (360- x ) }{2} =45$

$\frac{ 2 x -360 }{2} =45$

Multiply both sides by 2:

$\frac{2 x -360}{2} \cdot 2 =45 \cdot 2$

2 x -360 = 90

Add 360 to both sides:

2 x -360+ 360 = 90 + 360

2 x= 450

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

x = 225 ,

the degree measure of $\overarc{ACB}$ .

It follows that

$m \overarc{AB} = 360 - 225 = 135$

By the Arc Addition Principle,

$m \overarc{AB} = m \overarc{AD} + m \overarc{DB}$

Since $\angle DOB$ , the central angle which intercepts $\overarc{DB}$ , is a right angle, $m \overarc{DB} = 90$ . By substitution,

$135 = m \overarc{AD} + 90$ ,

and

$135 - 90 = m \overarc{AD} + 90 - 90$

$45= m \overarc{AD}$

The ratio $m \overarc{ACB} : m \overarc{AD}$ is equal to

$\frac{225}{45} = \frac{225 \div 45}{45 \div 45 } = \frac{5}{1}$ ,

a 5 to 1 ratio.

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HiSET: Math : Arc length and area of a sector

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #21 : Measurement And Geometry

Example Question #1 : Arc Length And Area Of A Sector

Example Question #1 : Arc Length And Area Of A Sector

Example Question #1 : Arc Length And Area Of A Sector

Example Question #1 : Arc Length And Area Of A Sector

Example Question #1 : Arc Length And Area Of A Sector

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