What if you were given a circle with two radii in which the region between those two radii was shaded? How could you find the area of that shaded region of the circle? After completing this Concept, you'll be able to use the formula for the area of a circle's sector to solve problems like this one.

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Area of Sectors and Segments CK-12

### Guidance

A **sector of a circle** is the area bounded by two radii and the arc between the endpoints of the radii. If \begin{align*}r\end{align*} is the radius and @$\begin{align*}\widehat{AB}\end{align*}@$ is the arc bounding a sector, then the **area** of the sector is @$\begin{align*}A=\frac{m \widehat{AB}}{360^\circ} \cdot \pi r^2\end{align*}@$.

A **segment of a circle** is the area of a circle that is bounded by a chord and the arc with the same endpoints as the chord. The **area** of a segment is @$\begin{align*}A_{segment}=A_{sector}-A_{\triangle ABC}\end{align*}@$

#### Example A

Find the area of the blue sector. Leave your answer in terms of @$\begin{align*}\pi\end{align*}@$.

In the picture, the central angle that corresponds with the sector is @$\begin{align*}60^\circ\end{align*}@$. @$\begin{align*}60^\circ\end{align*}@$ is @$\begin{align*}\frac{1}{6}\end{align*}@$ of @$\begin{align*}360^\circ\end{align*}@$, so this sector is @$\begin{align*}\frac{1}{6}\end{align*}@$ of the total area. @$\begin{align*}area \ of \ blue \ sector=\frac{1}{6} \cdot \pi 8^2=\frac{32}{3} \pi\end{align*}@$

#### Example B

The area of a sector is @$\begin{align*}8\pi\end{align*}@$ and the radius of the circle is 12. What is the central angle?

Plug in what you know to the sector area formula and then solve for the central angle, which we will call @$\begin{align*}x\end{align*}@$.

@$$\begin{align*}8 \pi &= \frac{x}{360^\circ} \cdot \pi 12^2\\ 8 \pi & =\frac{x}{360^\circ} \cdot 144 \pi\\ 8 &= \frac{2x}{5^\circ}\\ x &= 8 \cdot \frac{5^\circ}{2}=20^\circ\end{align*}@$$

#### Example C

Find the area of the blue segment below.

The area of the segment is the area of the sector minus the area of the isosceles triangle made by the radii. If we split the isosceles triangle in half, each half is a 30-60-90 triangle, where the radius is the hypotenuse. The height of @$\begin{align*}\triangle ABC\end{align*}@$ is 12 and the base is @$\begin{align*}2 \left(12 \sqrt{3}\right)=24 \sqrt{3}\end{align*}@$.

@$$\begin{align*}A_{sector} &= \frac{120}{360} \pi \cdot 24^2 && A_\triangle =\frac{1}{2} \left(24 \sqrt{3} \right)(12)\\ &= 192 \pi && \quad \ =144 \sqrt{3}\end{align*}@$$

The area of the segment is @$\begin{align*}A=192 \pi - 144 \sqrt{3} \approx 353.8\end{align*}@$ units.

Area of Sectors and Segments CK-12

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### Guided Practice

1. The area of a sector is @$\begin{align*}135 \pi\end{align*}@$ and the arc measure is @$\begin{align*}216^\circ\end{align*}@$. What is the radius of the circle?

2. Find the area of the shaded region. The quadrilateral is a square.

3. Find the area of the blue sector of @$\begin{align*}\bigodot A\end{align*}@$.

**Answers:**

1. Plug in what you know to the sector area formula and solve for @$\begin{align*}r\end{align*}@$.

@$$\begin{align*}135 \pi &= \frac{216^\circ}{360^\circ} \cdot \pi r^2\\ 135 &= \frac{3}{5} \cdot r^2\\ \frac{5}{3} \cdot 135 &= r^2\\ 225 &= r^2 \rightarrow r=\sqrt{225}=15\end{align*}@$$

2. The radius of the circle is 16, which is also half of the diagonal of the square. So, the diagonal is 32 and the sides would be @$\begin{align*}\frac{32}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=16 \sqrt{2}\end{align*}@$ because each half of a square is a 45-45-90 triangle.

@$$\begin{align*}A_{circle} &= 16^2 \pi =256 \pi\\ A_{square} &= \left(16 \sqrt{2} \right)^2 = 256 \cdot 2=512\end{align*}@$$

The area of the shaded region is @$\begin{align*}256 \pi -512 \approx 292.25\end{align*}@$

3. The right angle tells us that this sector represents @$\begin{align*}\frac{1}{4}\end{align*}@$ of the circle. The area of the whole circle is @$\begin{align*}A=\pi 8^2=64\pi\end{align*}@$. So, the area of the sector is @$\begin{align*}\frac{1}{4}64\pi = 16\pi\end{align*}@$.

### Explore More

Find the area of the blue sector or segment in @$\begin{align*}\bigodot A\end{align*}@$. Leave your answers in terms of @$\begin{align*}\pi\end{align*}@$. Round any decimal answers to the nearest hundredth.

Find the radius of the circle. Leave your answer in terms of @$\begin{align*}\pi\end{align*}@$.

Find the central angle of each blue sector. Round any decimal answers to the nearest tenth.

- Find the area of the sector in @$\begin{align*}\bigodot A\end{align*}@$. Leave your answer in terms of @$\begin{align*}\pi\end{align*}@$.
- Find the area of the equilateral triangle.
- Find the area of the segment. Round your answer to the nearest hundredth.
- Find the area of the sector in @$\begin{align*}\bigodot A\end{align*}@$. Leave your answer in terms of @$\begin{align*}\pi\end{align*}@$.
- Find the area of the right triangle.
- Find the area of the segment. Round your answer to the nearest hundredth.