What if you wanted to find the area of a pizza, this time taking into consideration the area of the crust? In another Concept, we found the length of the crust for a 14 in pizza. However, crust typically takes up some area on a pizza. Leave your answers in terms of \begin{align*}\pi\end{align*}

a) Find the area of the crust of a deep-dish 16 in pizza. A typical deep-dish pizza has 1 in of crust around the toppings.

b) A thin crust pizza has \begin{align*}\frac{1}{2}\end{align*}

c) Which piece of pizza has more crust? A twelfth of the deep dish pizza or a fourth of the thin crust pizza?

After completing this Concept, you will be able to answer these questions.

### Watch This

CK-12 Foundation: Chapter10AreaofSectorsandSegmentsA

Learn more about the area of a sector by watching the video at this link.

Watch this video to learn even more about the area of a sector.

### Guidance

A **sector of a circle** is the area bounded by two radii and the arc between the endpoints of the radii.

The area of a sector is a fractional part of the area of the circle, just like arc length is a fractional portion of the circumference. The **Area of a sector** is \begin{align*}A= \frac{m\widehat{AB}}{360^\circ} \cdot \pi r^2\end{align*}

The last part of a circle that we can find the area of is called a segment, not to be confused with a line segment. 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*}area \ of \ blue \ sector=\frac{1}{6} \cdot \pi 8^2=\frac{32}{3} \pi\end{align*}

#### Example B

Find the area of the blue segment below.

As you can see from the picture, 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, we see that each half is a 30-60-90 triangle, where the radius is the hypotenuse. Therefore, 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*}.

#### Example C

The area of a sector of circle is \begin{align*}50 \pi\end{align*} and its arc length is \begin{align*}5 \pi\end{align*}. Find the radius of the circle.

First substitute what you know to both the sector formula and the arc length formula. In both equations we will call the central angle, “\begin{align*}CA\end{align*}.”

\begin{align*}50 \pi &= \frac{CA}{360} \pi r^2 && \quad \ 5 \pi =\frac{CA}{360} 2 \pi r\\ 50 \cdot 360 &= CA \cdot r^2 && 5 \cdot 180=CA \cdot r\\ 18000 &= CA \cdot r^2 && \quad 900=CA \cdot r\end{align*}

Now, we can use substitution to solve for either the central angle or the radius. Because the problem is asking for the radius we should solve the second equation for the central angle and substitute that into the first equation for the central angle. Then, we can solve for the radius. Solving the second equation for \begin{align*}CA\end{align*}, we have: \begin{align*}CA=\frac{900}{r}\end{align*}. Plug this into the first equation.

\begin{align*}18000 &= \frac{900}{r} \cdot r^2\\ 18000 &= 900r\\ r &= 20\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter10AreaofSectorsandSegmentsB

#### Concept Problem Revisited

The area of the crust for a deep-dish pizza is \begin{align*}8^2 \pi - 7^2 \pi =15 \pi\end{align*}. The area of the crust of the thin crust pizza is \begin{align*}8^2 \pi - 7.5^2 \pi =\frac{31}{4} \pi\end{align*}. One-twelfth of the deep dish pizza has \begin{align*}\frac{15}{12} \pi\end{align*} or \begin{align*}\frac{5}{4} \pi \ in^2\end{align*} of crust. One-fourth of the thin crust pizza has \begin{align*}\frac{31}{16} \pi \ in^2\end{align*}. To compare the two measurements, it might be easier to put them both into decimals. \begin{align*}\frac{5}{4} \pi \approx 3.93 \ in^2\end{align*} and \begin{align*}\frac{31}{16} \pi \approx 6.09 \ in^2\end{align*}. From this, we see that one-fourth of the thin-crust pizza has more crust than one-twelfth of the deep dish pizza.

### Vocabulary

A ** circle** is the set of all points that are the same distance away from a specific point, called the

**. A**

*center***is the distance from the center to the outer rim of the circle. A**

*radius***is a line segment whose endpoints are on a circle. A**

*chord***is a chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.**

*diameter***is the amount of space inside a figure and is measured in square units. \begin{align*}\pi\end{align*}, or**

*Area***is the ratio of the circumference of a circle to its diameter. A**

*“pi”***is the area bounded by two radii and the arc between the endpoints of the radii. A**

*sector 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.**

*segment of a circle*### 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*}.

### Practice

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*}. You may use decimals or fractions in your answers, but do not round.

Find the radius of the circle. Leave your answer in simplest radical form.

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

- The area of a sector of a circle is \begin{align*}54 \pi\end{align*} and its arc length is \begin{align*}6\pi\end{align*}. Find the radius of the circle.
- Find the central angle of the sector from #13.
- The area of a sector of a circle is \begin{align*}2304 \pi\end{align*} and its arc length is \begin{align*}32 \pi\end{align*}. Find the central angle of the sector.