Math - How to find the area of a sector

To the nearest tenth, give the area of a $60^{\circ }$ sector of a circle with diameter 18 centimeters.

$42.4 \textrm{ cm}^{2}$

$84.8 \textrm{ cm}^{2}$

$169.6 \textrm{ cm}^{2}$

$21.2 \textrm{ cm}^{2}$

$254.5 \textrm{ cm}^{2}$

Explanation

The radius of a circle with diameter 18 centimeters is half that, or 9 centimeters. The area of a $60^{\circ }$ sector of the circle is

$A = \frac{60}{360 }\cdot \pi \cdot 9^{2} =\frac{1}{6} \cdot \pi \cdot 81 \approx 42.4 \textrm{ cm}^{2}$

Find the area of the shaded region:

Screen_shot_2014-02-27_at_6.35.30_pm

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

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

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

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

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

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, 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 triangle:

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

Plugging in our final values, we get:

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

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

Find the area of the following sector:

$9 \pi\ m^2$

$18 \pi\ m^2$

$6 \pi\ m^2$

$3 \pi\ m^2$

$12 \pi\ m^2$

Explanation

The formula for the area of a sector is

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

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

Plugging in our values, we get:

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

$A=9 \pi\ m^2$

What is the area of the following sector of a full circle?

Arch

Note: Figure is not drawn to scale.

$13.5\pi m^2$

$27\pi m^2$

$21.5\pi m^2$

$23.5\pi m^2$

$10\pi m^2$

Explanation

In order to find the fraction of a sector from an angle, you need to know that a full circle is $360^{\circ}$ .

Therefore, we can find the fraction by dividing the angle of the sector by $360^{\circ}$ :

$Percentage = \frac{\measuredangle}{360^{\circ}}$

$Percentage = \frac{60^{\circ}}{360^{\circ}}=\frac{1}{6}$

The formula to find the area of a sector is:

$A=\pi (r^2)(fraction\ of\ whole\ circle)$

where is the radius of the circle.

Plugging in our values, we get:

$A=\pi (r^2)(fraction)$

$A=\pi (9m^2)\left(\frac{1}{6}\right)=13.5\pi m^2$

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

$3\Pi$

$9\Pi$

$\frac{1}{3}\Pi$

$27\Pi$

$12\Pi$

Explanation

The equation to find the area of a sector is $(\frac{x}{360})\Pi r^2$ .

Substitute the given radius in for and the given angle in for to get:

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

Simplify the equation to get the area:

$(\frac{1}{3})\Pi9=3\Pi$

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?

$12 + \pi$

$1 + \pi$

$3 + 2\pi$

$6 + \pi$

$12 + 2\pi$

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 $12+{\pi}$ .

If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?

$100\pi$

$5\pi$

$200$

$256\pi$

$144$

Explanation

Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:

$AB^2=AC^2+BC^2\rightarrow AB=\sqrt{AC^2+BC^2}=\sqrt{16^2+12^2}=\sqrt{400}=20$

Since AB is a radius of B, we can find the area of circle B via:

$Area=\pi R^2=\pi(20^2)=400\pi$

Angle DBE is a right angle, and therefore $\frac{1}{4}$ of the circle so it follows:

$Area(DBE)=\frac{400}{4}\pi=100\pi$

Circle

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

$2\pi$

$4\pi$

$8\pi$

$16\pi$

$\pi$

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

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

$9\pi-18$

$9\pi$

$9\pi-36$

$36\pi-18$

$9\pi+36$

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.

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.

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

Square-missing

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?

$$(2250-250\pi )$

$$(900-25\pi )$

$$(900-12.5\pi )$

$$(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).

How to find the area of a sector

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