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10.5: Areas of Circles and Sectors

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Find the area of circles, sectors, and segments.

Review Queue

1. Find the area of both squares.
2. Find the area of the shaded region.
3. The triangle to the right is an equilateral triangle.
1. Find the height of the triangle.
2. Find the area of the triangle.

Know What? Back to the pizza. In the previous section, we found the length of the crust for a 14 in pizza. However, crust typically takes up some area on a pizza. Round your answers to the nearest hundredth.

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*}-in of crust around the edge of the pizza. Find the area of a thin crust 16 in pizza.

Area of a Circle

Take a circle and divide it into several wedges. Then, unfold the wedges so they are in a line, with the points at the top.

The height of the wedges is the radius and the length is the circumference of the circle. Now, take half of these wedges and flip them upside-down and place them so they all fit together.

Now our circle looks like a parallelogram. The area of this parallelogram is \begin{align*}A=bh=\pi r \cdot r=\pi r^2\end{align*}.

To see an animation of this derivation, see http://www.rkm.com.au/ANIMATIONS/animation-Circle-Area-Derivation.html, by Russell Knightley.

Area of a Circle: If \begin{align*}r\end{align*} is the radius of a circle, then \begin{align*}A=\pi r^2\end{align*}.

Example 1: Find the area of a circle with a diameter of 12 cm.

Solution: If \begin{align*}d = 12 \ cm\end{align*}, then \begin{align*}r = 6 \ cm\end{align*}. The area is \begin{align*}A=\pi \left(6^2 \right)=36 \pi \ cm^2\end{align*}.

Example 2: If the area of a circle is \begin{align*}20 \pi\end{align*}, what is the radius?

Solution: Plug in the area and solve for the radius.

\begin{align*}20 \pi &= \pi r^2\\ 20 &= r^2\\ r &= \sqrt{20}=2 \sqrt{5}\end{align*}

Just like the circumference, we will leave our answers in terms of \begin{align*}\pi\end{align*}, unless otherwise specified.

Example 3: A circle is inscribed in a square. Each side of the square is 10 cm long. What is the area of the circle?

Solution: The diameter of the circle is the same as the length of a side of the square. Therefore, the radius is 5 cm.

\begin{align*}A=\pi 5^2=25 \pi \ cm^2\end{align*}

Example 4: Find the area of the shaded region.

Solution: The area of the shaded region would be the area of the square minus the area of the circle.

\begin{align*}A=10^2-25 \pi =100-25 \pi \approx 21.46 \ cm^2\end{align*}

Area of a Sector

Sector of a Circle: The area bounded by two radii and the arc between the endpoints of the radii.

Area of a Sector: If \begin{align*}r\end{align*} is the radius and \begin{align*}\widehat{AB}\end{align*} is the arc bounding a sector, then \begin{align*}A=\frac{m \widehat{AB}}{360^\circ} \cdot \pi r^2\end{align*}.

Example 5: Find the area of the blue sector. Leave your answer in terms of \begin{align*}\pi\end{align*}.

Solution: In the picture, the central angle that corresponds with the sector is \begin{align*}60^\circ\end{align*}. \begin{align*}60^\circ\end{align*} would be \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*}

Another way to write the sector formula is \begin{align*}A=\frac{central \ angle}{360^\circ} \cdot \pi r^2\end{align*}.

Example 6: 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?

Solution: Plug in what you know to the sector area formula and then solve for the central angle, we will call it \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 7: 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?

Solution: 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 8: Find the area of the shaded region. The quadrilateral is a square.

Solution: 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*}

Segments of a Circle

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.

Segment of a Circle: The area of a circle that is bounded by a chord and the arc with the same endpoints as the chord.

\begin{align*}A_{segment}=A_{sector}-A_{\triangle ABC}\end{align*}

Example 9: Find the area of the blue segment below.

Solution: 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 would be \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*}.

Know What? 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*}.

Review Questions

• Questions 1-10 are similar to Examples 1 and 2.
• Questions 11-16 are similar to Example 5.
• Questions 17-19 are similar to Example 7.
• Questions 20-22 are similar to Example 6.
• Questions 23-25 are similar to Examples 3, 4, and 8.
• Questions 26-31 are similar to Example 9.

Fill in the following table. Leave all answers in terms of \begin{align*}\pi\end{align*}.

1. 2
2. \begin{align*}16 \pi\end{align*}
3. \begin{align*}10\pi\end{align*}
4. \begin{align*}24\pi\end{align*}
5. 9
6. \begin{align*}90\pi\end{align*}
7. \begin{align*}35\pi\end{align*}
8. \begin{align*}\frac{7}{\pi}\end{align*}
9. 60
10. 36

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.

1. Find the area of the sector in \begin{align*}\bigodot A\end{align*}. Leave your answer in terms of \begin{align*}\pi\end{align*}.
2. Find the area of the equilateral triangle.
3. Find the area of the segment. Round your answer to the nearest hundredth.
4. Find the area of the sector in \begin{align*}\bigodot A\end{align*}. Leave your answer in terms of \begin{align*}\pi\end{align*}.
5. Find the area of the right triangle.
6. Find the area of the segment. Round your answer to the nearest hundredth.

1. \begin{align*}8^2 - 4^2 = 64 - 16 = 48\end{align*}
2. \begin{align*}6(10) - \frac{1}{2} (7)(3) = 60 - 10.5 = 49.5\end{align*}
3. \begin{align*}\frac{1}{2} (6) \left (3 \sqrt{3} \right) = 9 \sqrt{3}\end{align*}
4. \begin{align*}\frac{1}{2} (s) \left( \frac{1}{2} s \sqrt{3} \right ) = \frac{1}{4} s^2 \sqrt{3}\end{align*}

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