### Area of Sectors and Segments

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*}**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*}

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?

### Examples

#### Example 1

The area of a sector is \begin{align*}135 \pi\end{align*}

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*}

#### Example 2

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

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*}

\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*}

#### Example 3

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*}

#### Example 4

The area of a sector is \begin{align*}8\pi\end{align*}

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 5

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*}

\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*}

### Review

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.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 10.11.