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

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

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

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:

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

Circle 1

The above circle, which is divided into sectors of equal size, has diameter 20. Give the area of the shaded region.

$\frac{125 \pi }{4 }$

$125 \pi$

$\frac{125 \pi }{16 }$

$\frac{125 \pi }{8}$

Explanation

The radius of a circle is half its diameter; the radius of the circle in the diagram is half of 20, or 10.

To find the area of the circle, set in the area formula:

$A = \pi r ^{2} = \pi \cdot 10 ^{2} = 100 \pi$

The circle is divided into sixteen sectors of equal size, five of which are shaded; the shaded portion is

$\frac{5}{16} \cdot 100 \pi = \frac{500}{16} \pi = \frac{125 \pi }{4 }$ .

Icecreamcone 2

Refer to the above figure, Which is the greater quantity?

(a) The area of $\bigtriangleup ABC$

(b) The area of the orange semicircle

(a) is the greater quantity

(b) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

Explanation

$\bigtriangleup ABC$ has two angles of degree measure 45; the third angle must measure 90 degrees, making $\bigtriangleup ABC$ a right triangle.

For the sake of simplicity, let ; the reasoning is independent of the actual length. The legs of a 45-45-90 triangle are congruent, so . The area of a right triangle is half the product of its legs, so

$A = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} = 0.5$

Also, if , then the orange semicircle has diameter 1 and radius $\frac{1}{2}$ . Its area can be found by substituting $r = \frac{1}{2}$ in the formula:

$A = \frac{1}{2} \cdot \pi r^{2}$

$= \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}$

$= \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )$

$= \frac{1}{8} \cdot \pi$

$\approx 0. 125 \cdot 3.14$

$\approx 0. 3925$

$\bigtriangleup ABC$ has a greater area than the orange semicircle.

Generalsector-12

What is the area, rounded to the nearest hundredth, of the sector shaded in circle O in the diagram above?

Explanation

To find the area of a sector, you need to find a percentage of the total area of the circle. You do this by dividing the sector angle by the total number of degrees in a full circle (i.e. ˚). Thus, for our circle, which has a sector with an angle of ˚, we have a percentage of:

$\frac{81}{360}$

Now, we will multiply this by the total area of the circle. Recall that we find such an area according to the equation:

$A = \pi * r^2$

For our problem,

Therefore, our equation is:

$\frac{81}{360} * \pi * 12.3^2 = \frac{12254.49\pi}{360}$

Using your calculator, you can determine that this is approximately .

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

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 $2.89\pi in^2$

Generalsector-12

What is the area, rounded to the nearest hundredth, of the sector shaded in circle O in the diagram above?

Explanation

$\frac{122}{360}$

Now, we will multiply this by the total area of the circle. Recall that we find such an area according to the equation:

$A = \pi * r^2$

For our problem,

Therefore, our equation is:

$\frac{122}{360} * \pi * 9.12^2 = \frac{10147.2768\pi}{360}$

Using your calculator, you can determine that this is approximately .

Refer to the above figure, Which is the greater quantity?

(a) The area of the orange semicircle

(b) The area of $\bigtriangleup ABC$

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

Explanation

$\bigtriangleup ABC$ has two angles of degree measure 60; its third angle must also have measure 60, making $\bigtriangleup ABC$ an equilateral triangle

For the sake of simplicity, let ; the reasoning is independent of the actual length. The area of equilateral $\bigtriangleup ABC$ can be found by substituting in the formula

$A = \frac{s^{2}\sqrt{3}}{4}$

$= \frac{1^{2}\sqrt{3}}{4}$

$= \frac{1 \cdot \sqrt{3}}{4}$

$\approx \frac{1.7}{4}$

$\approx 0.425$

Also, if , then the orange semicircle has diameter 1 and radius $\frac{1}{2}$ . Its area can be found by substituting $r = \frac{1}{2}$ in the formula:

$A = \frac{1}{2} \cdot \pi r^{2}$

$= \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}$

$= \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )$

$= \frac{1}{8} \cdot \pi$

$\approx 0. 125 \cdot 3.14$

$\approx 0. 3925$

$\bigtriangleup ABC$ has a greater area than the orange semicircle.

Circle A has twice the radius of Circle B. Which is the greater quantity?

(a) The area of a $90^{\circ }$ sector of Circle A

(b) The area of Circle B

(a) and (b) are equal.

It is impossible to tell from the information given.

(a) is greater.

(b) is greater.

Explanation

Let be the radius of Circle B. The radius of Circle A is therefore .

A $90^{\circ }$ sector of a circle comprises $\frac{90}{360} = \frac{1}{4}$ of the circle. The $90^{\circ }$ sector of circle A has area $\frac{1}{4} \pi (2r)^{2} = \frac{1}{4} \cdot 4 \pi r^{2} = \pi r^{2}$ , the area of Circle B. The two quantities are equal.

How to find the area of a sector

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