Arc length and area of a sector - HiSET

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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).

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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$$

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

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The radius of the circle is given to be . 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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$

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

Tap to see back →

The radius of the circle is given to be . 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$

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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$

$=24 \pi$

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

Tap to see back →

The radius of the circle is given to be . 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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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$

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

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The radius of the circle is given to be .

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$

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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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$

$= 10 \pi$

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

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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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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.

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).

Tap to see back →

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$$

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

Tap to see back →

The radius of the circle is given to be . 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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$

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

Tap to see back →

The radius of the circle is given to be . 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$

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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$

$=24 \pi$

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

Tap to see back →

The radius of the circle is given to be . 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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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$

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

Tap to see back →

The radius of the circle is given to be .

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$

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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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$

$= 10 \pi$

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

Tap to see back →

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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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.

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).

Tap to see back →

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$$

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

Tap to see back →

The radius of the circle is given to be . 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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$

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

Tap to see back →

The radius of the circle is given to be . 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$

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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$

$=24 \pi$

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

Tap to see back →

The radius of the circle is given to be . 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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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$

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

Tap to see back →

The radius of the circle is given to be .

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$

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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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$

$= 10 \pi$

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

Tap to see back →

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$

Add 360 to both sides:

Divide both sides by 2:

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

,

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.

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).

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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$$

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

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The radius of the circle is given to be . 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$

Subtract 360 from both sides:

Divide both sides by :

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

,

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$