How to find the area of a sector

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ISEE Upper Level Quantitative Reasoning › How to find the area of a sector

Questions 1 - 3

1

Sector

Give the area of the white region of the above circle if $\overarc{AB}$ has length $12 \pi$ .

$567 \pi$

$81 \pi$

$729 \pi$

$243 \pi$

Explanation

If we let be the circumference of the circle, then the length of $\overarc{AB}$ is $\frac{80}{360} = \frac{2}{9 }$ of the circumference, so

$\frac{2}{9 } C = 12 \pi$

$\frac{9 }{2}\cdot \frac{2}{9 } C =\frac{9 }{2}\cdot 12 \pi$

$C =54\pi$

The radius is the circumference divided by $2 \pi$ :

$r= 54 \pi \div 2 \pi = 27$

Use the formula to find the area of the entire circle:

$A = \pi r ^{2} = \pi \left (27 \right )^{2} = 729 \pi$

The area of the white region is $1 - \frac{2}{9} = \frac{7}{9}$ of that of the circle, or

$\frac{7}{9} \cdot 729 \pi = 567 \pi$

2

Sector

The circumference of the above circle is $30 \pi$ . Give the area of the shaded region.

$50 \pi$

$25 \pi$

$100 \pi$

$400 \pi$

Explanation

The radius of a circle is found by dividing the circumference $C = 30 \pi$ by $2 \pi$ :

$r= \frac{C}{2 \pi}= \frac{30 \pi}{2 \pi} = 15$

The area of the entire circle can be found by substituting for in the formula:

$A = \pi r ^{2} = \pi \cdot 15 ^{2} = 225 \pi$ .

The area of the shaded $80 ^{\circ }$ sector is $\frac{80 }{360 }$ of the total area:

$\frac{80 }{360 } \times 225 \pi = \frac{2}{9} \times 225 \pi = 50 \pi$

3

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . If the screen has a radius of 4 inches, what is the area of the highlighted wedge?

$2.89\pi in^2$

$14.23\pi in^2$

$9.08\pi in^2$

Explanation

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . If the screen has a radius of 4 inches, what is the area of the highlighted wedge?

To begin, let's recall our formula for area of a sector.

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

Now, we have theta and r, so we just need to plug them in and simplify!

$A_{sector}=\frac{65^{\circ}}{360^{\circ}}\pi (4in)^2=2.89\pi in^2$

So our answer is $2.89\pi in^2$

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