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High School Math : Sectors

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Example Question #11 : Sectors

Find the area of the shaded region:

Screen_shot_2014-02-27_at_6.35.30_pm

Possible Answers:

$108\pi-81\sqrt{2}m^2$

$108\pi-108\sqrt{3}m^2$

$108\pi-81\sqrt{3}m^2$

$81\pi-108\sqrt{3}m^2$

$81\pi-81\sqrt{3}m^2$

Correct answer:

$108\pi-81\sqrt{3}m^2$

Explanation:

To find the area of the shaded region, you must subtract the area of the triangle from the area of the sector.

The formula for the shaded area is:

$A=A_{sector}-A_{triangle}$

$A = \pi(r^2)(part)-\frac{1}{2}(b)(h)$ ,

where is the radius of the circle, part is the fraction of the sector, is the base of the triangle, and is the height of the triangle.

In order to the find the base and height of the triangle, use the formula for a 30-60-90 triangle:

$a-a\sqrt{3}-2a$ , where is the side opposite the $\measuredangle30$ .

Plugging in our final values, we get:

$A = \pi(18m^2)(\frac{120}{360})-\frac{1}{2}(18\sqrt{3}m)(9m)$

$A = 108\pi-81\sqrt{3}m^2$

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Example Question #12 : Sectors

Find the area of the following sector:

Possible Answers:

$3 \pi\ m^2$

$18 \pi\ m^2$

$6 \pi\ m^2$

$9 \pi\ m^2$

$12 \pi\ m^2$

Correct answer:

$9 \pi\ m^2$

Explanation:

The formula for the area of a sector is

$A = \pi r^2 (part)$ ,

where is the radius of the circle and part is the fraction of the sector.

Plugging in our values, we get:

$A = \pi (6\ m)^2 (\frac{90}{360})$

$A=9 \pi\ m^2$

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Example Question #1 : Sectors

Circle

The radius of the circle above is and $\angle A=45^{\circ}$ . What is the area of the shaded section of the circle?

Possible Answers:

$8\pi$

$\pi$

$16\pi$

$2\pi$

$4\pi$

Correct answer:

$2\pi$

Explanation:

Area of Circle = πr² = π4²= 16π

Total degrees in a circle = 360

Therefore 45 degree slice = 45/360 fraction of circle = 1/8

Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π

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Example Question #13 : Sectors

Find the area of the shaded segment of the circle. The right angle rests at the center of the circle.

Possible Answers:

$9\pi-18$

$9\pi-36$

$36\pi-18$

$9\pi+36$

$9\pi$

Correct answer:

$9\pi-18$

Explanation:

We know that the right angle rests at the center of the circle; thus, the sides of the triangle represent the radius of the circle.

r=6

Because the sector of the circle is defined by a right triangle, the region corresponds to one-fourth of the circle.

$\frac{degrees\ in\ angle}{degrees\ in\ circle}=\frac{90^o}{360^o}=\frac{1}{4}$

First, find the total area of the circle and divide it by four to find the area of the depicted sector.

$\small A(circle)=\pi(r)^2$

$\small A(sector)=\frac{\pi(r)^2}{4}$

$\small A(sector)=\frac{\pi(6)^2}{4}=\frac{\pi(36)}{4}=9\pi$

Next, calculate the area of the triangle.

$\small A(tri)=\frac{1}{2}bh=\frac{1}{2}(6)(6)=18$

Finally, subtract the area of the triangle from the area of the sector.

$\small A(shaded)=9\pi-18$

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Example Question #5 : Sectors

Square-missing

SRAD is a square.

$\overline{AB}=\overline{BC}=\overline{CD}$

The arc from to is a semicircle with a center at the midpoint of $\overline{BC}$ .

All units are in feet.

The diagram shows a plot of land.

The cost of summer upkeep is $2.50 per square foot.

In dollars, what is the total upkeep cost for the summer?

Possible Answers:

$\$(2250-31.25\pi )$

$\$900$

$\$(2250-250\pi )$

$\$(900-12.5\pi )$

$\$(900-25\pi )$

Correct answer:

$\$(2250-31.25\pi )$

Explanation:

To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.

The area of the square is straightforward:

30 * 30 = 900 square feet

Because each side is 30 feet long, AB + BC + CD = 30.

We can substitute BC for AB and CD since all three lengths are the same:

BC + BC + BC = 30

3BC = 30

BC = 10

Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.

The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 5²π = 25π, so the area of the semi-circle is half of that, or 12.5π.

The total area of the plot is the square less the semicircle: 900 - 12.5π square feet

The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).

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Example Question #261 : Plane Geometry

In the figure, PQ is the arc of a circle with center O. If the area of the sector is $3\pi$ what is the perimeter of sector?

Possible Answers:

$3 + 2\pi$

$12 + \pi$

$12 + 2\pi$

$1 + \pi$

$6 + \pi$

Correct answer:

$12 + \pi$

Explanation:

First, we figure out what fraction of the circle is contained in sector OPQ: $\frac{30^{\circ}}{360^{\circ}}= \frac{1}{12}$ , so the total area of the circle is $\dpi{100} \small 12\times 3\pi=36$ .

Using the formula for the area of a circle, ${\pi}r^{2}$ , we can see that $\dpi{100} \small r=6$ .

We can use this to solve for the circumference of the circle, $2{\pi}r$ , or $12{\pi}$ .

Now, OP and OQ are both equal to r, and PQ is equal to $\dpi{100} \small \frac{1}{12}$ of the circumference of the circle, or ${\pi}$ .

To get the perimeter, we add OP + OQ + PQ, which give us $12+{\pi}$ .

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Example Question #14 : Sectors

If a quarter of the area of a circle is $9\pi$ , then what is a quarter of the circumference of the circle?

Possible Answers:

$12\pi$

$2.25\pi$

$3\pi$

$9\pi$

Correct answer:

$3\pi$

Explanation:

If a quarter of the area of a circle is $9\pi$ , then the area of the whole circle is $36\pi$ . This means that the radius of the circle is 6. The diameter is 12. Thus, the circumference of the circle is $12\pi$ . One fourth of the circumference is $3\pi$ .

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Example Question #8 : Sectors

Figure not drawn to scale.

In the figure above, circle C has a radius of 18, and the measure of angle ACB is equal to 100°. What is the perimeter of the red shaded region?

Possible Answers:

36 + 20π

18 + 10π

18 + 36π

36 + 36π

36 + 10π

Correct answer:

36 + 10π

Explanation:

The perimeter of any region is the total distance around its boundaries. The perimeter of the shaded region consists of the two straight line segments, AC and BC, as well as the arc AB. In order to find the perimeter of the whole region, we must add the lengths of AC, BC, and the arc AB.

The lengths of AC and BC are both going to be equal to the length of the radius, which is 18. Thus, the perimeter of AC and BC together is 36.

Lastly, we must find the length of arc AB and add it to 36 to get the whole perimeter of the region.

Angle ACB is a central angle, and it intercepts arc AB. The length of AB is going to equal a certain portion of the circumference. This portion will be equal to the ratio of the measure of angle ACB to the measure of the total degrees in the circle. There are 360 degrees in any circle. The ratio of the angle ACB to 360 degrees will be 100/360 = 5/18. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2πr, according to the formula for circumference.

length of arc AB = (5/18)(2πr) = (5/18)(2π(18)) = 10π.

Thus, the length of arc AB is 10π.

The total length of the perimeter is thus 36 + 10π.

The answer is 36 + 10π.

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Example Question #15 : Sectors

Find the arc length of a sector that has an angle of 120 degrees and radius of 3.

Possible Answers:

$36\Pi$

$16\Pi$

$2\Pi$

$4\Pi$

$6\Pi$

Correct answer:

$2\Pi$

Explanation:

The equation for the arc length of a sector is $(\frac{x}{360})2\Pi r$ .

Substitute the given radius for and the given angle for to get the following equation:

$(\frac{120}{360})2\Pi(3)$

Simplify:

$=(\frac{1}{3})2\Pi (3)$

$=2\Pi$

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Example Question #16 : Sectors

$\textup{What is the length, in inches, of an arc with a central angle of }$

$72\textup{ degrees on a circle with a radius of 20 inches?}$

Possible Answers:

$20\pi$

$10\pi$

$4\pi$

$8\pi$

$12\pi$

Correct answer:

$8\pi$

Explanation:

$\frac{\textup{length of arc}}{\textup{circumference}}=\frac{\textup{central angle}}{360}$

$\textup{Find circumference: }C=2\pi r = 2\pi 20 = 40\pi$

$\frac{x}{40\pi }= \frac{72}{360}\:\:\textup{reduce: }\frac{x}{40\pi }=\frac{1}{5}\:\textup{ cross-multiply: }\:5x=40\pi$

$x=8\pi$

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High School Math : Sectors

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : Sectors

Example Question #12 : Sectors

Example Question #1 : Sectors

Example Question #13 : Sectors

Example Question #5 : Sectors

Example Question #261 : Plane Geometry

Example Question #14 : Sectors

Example Question #8 : Sectors

Example Question #15 : Sectors

Example Question #16 : Sectors

All High School Math Resources

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