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All HiSET: Math Resources

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Flashcards: Arc length and area of a sector

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).

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

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

$27\; cm^2$

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

$\pi\; cm^2$

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Flashcards: Arc length and area of a sector

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).

Your answer:

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

Your answer:

$27\; cm^2$

Your answer:

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

Your answer:

$\pi\; cm^2$

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:

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

Flashcards: Arc length and area of a sector

Intercepted 2

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

$6 \pi$

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

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

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

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

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Flashcards: Arc length and area of a sector

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

Your answer:

$6 \pi$

Your answer:

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

Your answer:

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

Your answer:

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

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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Flashcards: Arc length and area of a sector

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Flashcards: Arc length and area of a sector

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Flashcards: Arc length and area of a sector

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