Sectors

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ISEE Upper Level Quantitative Reasoning › Sectors

Questions 1 - 10

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 TYP occupies 43% of a circle. Find the degree measure of angle TYP.

$154.8^{\circ}$

$15.5^{\circ}$

$430^{\circ}$

$366^{\circ}$

Explanation

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

Use the following formula and solve for x:

$\frac{x}{360^{\circ}}100=43%$

Begin by dividing over the 100

$\frac{x}{360^{\circ}}=.43$

Then multiply by 360

$x=.43*360^{\circ}=154.8^{\circ}$

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

$154.8^{\circ}$

$15.5^{\circ}$

$430^{\circ}$

$366^{\circ}$

Explanation

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

Use the following formula and solve for x:

$\frac{x}{360^{\circ}}100=43%$

Begin by dividing over the 100

$\frac{x}{360^{\circ}}=.43$

Then multiply by 360

$x=.43*360^{\circ}=154.8^{\circ}$

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Explanation

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Recall that there is a total of 360 degrees in a circle. SOW occupies 45 of them. To find the percentage, simply do the following:

$\frac{45^{\circ}}{360^{\circ}}*100=.125*100=12.5%$

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Explanation

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Recall that there is a total of 360 degrees in a circle. SOW occupies 45 of them. To find the percentage, simply do the following:

$\frac{45^{\circ}}{360^{\circ}}*100=.125*100=12.5%$

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:

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

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:

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

A giant clock has a minute hand six feet long. How far, in inches, did the tip move between noon and 1:20 PM?

$192 \pi \textrm{ in}$

$48 \pi \textrm{ in}$

$384 \pi \textrm{ in}$

$768 \pi \textrm{ in}$

$96\pi \textrm{ in}$

Explanation

The distance that the tip of the minute hand moves during one hour is the circumference of a circle with radius 6 feet. This circumference is $2\pi r = 2 \pi \times 6 = 12\pi$ feet. One hour and twenty minutes is $1 \frac{1}{3}$ hours, so the tip of the hand moved $1 \frac{1}{3} \times 12 \pi = 16 \pi$ feet, or $16 \pi \times 12 = 192 \pi$ inches.