### Area and Circumference of a Circle

A **circle** is a set of points equidistant from a given point. The **radius** of a circle, \begin{align*}r\end{align*}, is the distance from the center of the circle to the circle. All circles are geometrically **similar**.

A **regular** **polygon** is a closed figure that is both equilateral and equiangular. As the number of sides of a regular polygon increases, the polygon looks more and more like a circle.

Previously you have learned that the **area** of a circle with radius \begin{align*}r\end{align*} is given by \begin{align*}\pi r^2\end{align*} and the **circumference** of a circle with radius \begin{align*}r\end{align*} is given by \begin{align*}2 \pi r\end{align*}. In the examples and guided practice, you will derive these formulas by looking at the area and perimeter of regular polygons.

#### Finding the Area

1. Find the area of a regular octagon inscribed in a circle with radius 1 unit.

Break the octagon into 8 congruent triangles. You will find the area of one triangle and multiply that by 8 to find the area of the whole octagon. Draw in the height for one of the triangles. The angle formed by the height and the radius of the circle is \begin{align*}\frac{360^\circ}{16}=22.5^\circ\end{align*}. Remember that \begin{align*}360\end{align*} is the number of degrees in a full circle.

Now, you can use trigonometry to find the height and base of the triangle.

- \begin{align*}\sin 22.5=\frac{0.5b}{1} \rightarrow b=2 \sin 22.5 \rightarrow b \approx 0.7654\end{align*}
- \begin{align*}\cos 22.5=\frac{h}{1} \rightarrow h=\cos 22.5 \rightarrow h \approx 0.9239\end{align*}

The area of the triangle is:

\begin{align*}A=\frac{bh}{2}=\frac{(0.7654)(0.9239)}{2} \approx 0.3536 \ un^2\end{align*}

Therefore, the area of the octagon is:

\begin{align*}A=8(0.3536) \approx 2.8288 \ un^2\end{align*}

2. Find the area of a regular 60-gon inscribed in a circle with radius 1 unit and the area of a regular 120-gon inscribed in a circle with radius 1 unit.

While you can't accurately draw a regular 60-gon or a regular 180-gon, you can use the method from #1 to find the area of each.

**Regular 60-gon:** Divide the polygon into 60 congruent triangles. Consider one of those triangles and draw in its height. You will focus on finding the area of this triangle. The angle formed by the height and the radius of the circle is \begin{align*}\frac{360^\circ}{120}=3^\circ\end{align*}.

Use trigonometry to find the height and base of the triangle.

- \begin{align*}\sin 3=\frac{0.5b}{1} \rightarrow b=2 \sin 3 \rightarrow b \approx 0.1047\end{align*}
- \begin{align*}\cos 3=\frac{h}{1} \rightarrow h=\cos 3 \rightarrow h \approx 0.9986\end{align*}

The area of the triangle is:

\begin{align*}A=\frac{bh}{2}=\frac{(0.1047)(0.9986)}{2} \approx 0.0523 \ un^2\end{align*}

Therefore, the area of the 60-gon is:

\begin{align*}A=60(0.0523) \approx 3.138 \ un^2\end{align*}

**Regular 120-gon:** Divide the polygon into 180 congruent triangles. Consider one of those triangles and draw in its height. You will focus on finding the area of this triangle. The angle formed by the height and the radius of the circle is \begin{align*}\frac{360^\circ}{360}=1^\circ\end{align*}.

Use trigonometry to find the height and base of the triangle.

- \begin{align*}\sin 1=\frac{0.5b}{1} \rightarrow b=2 \sin 1 \rightarrow b \approx 0.0349048\end{align*}
- \begin{align*}\cos 1=\frac{h}{1} \rightarrow h=\cos 1 \rightarrow h \approx 0.9998477\end{align*}

The area of the triangle is:

\begin{align*}A=\frac{bh}{2}=\frac{(0.0349048)(0.9998477)}{2} \approx 0.0174482 \ un^2\end{align*}

Therefore, the area of the 180-gon is:

\begin{align*}A=180(0.0174482) \approx 3.141 \ un^2\end{align*}

Now, let's explore how area can change based on the number of sides.

What happens to the area of the regular polygon as the number of sides increases? Make a conjecture about the area of a circle with radius 1 unit.

You have the following information:

Number of Sides |
8 |
60 |
180 |

Area |
2.8288 |
3.138 |
3.141 |

As the number of sides increases, the regular polygon will get closer and closer to the circle that inscribes it. Therefore, the area of the regular polygon will get closer and closer to the area of the circle. Notice that as the number of sides went from 60 to 180, the area barely changed, staying around 3.14. If you increase the number of sides to 1000, you will find that the area of the regular polygon is still approximately 3.14157. You should recognize these numbers as approximately the value of \begin{align*}\pi\end{align*}.

As the number of sides increases, the polygon becomes closer and closer to a circle, and the area gets closer and closer to \begin{align*}\pi \ un^2\end{align*}. A conjecture would be that the area of a circle with radius 1 unit is \begin{align*}\pi \ un^2\end{align*}.

**Examples**

**Example 1**

Earlier, you were asked where the \begin{align*}r^2\end{align*} came from in the formula for the area of a circle.

You can use regular polygons with an increasing number of sides to help explain why a circle of radius 1 unit has an area of \begin{align*}\pi \ un^2\end{align*}. Remember that all circles are similar. To create another circle with radius \begin{align*}r\end{align*} from a circle with radius 1 unit, apply a similarity transformation with scale factor \begin{align*}k=radius\end{align*}.

The area of the transformed circle is \begin{align*}k^2\end{align*} times the area of the original circle. Because the area of the original circle is \begin{align*}\pi\end{align*} and the scale factor is equal to \begin{align*}r\end{align*}, the area of the transformed circle is \begin{align*}\pi r^2\end{align*}. This is one way to explain why the formula for the area of a circle is \begin{align*}\pi r^2\end{align*}.

#### Example 2

Find the perimeter of a regular octagon inscribed in a circle with radius 1 unit.

The base of one triangle was \begin{align*}b=2 \sin 22.5 \rightarrow b \approx 0.7654\end{align*}. Therefore, the perimeter of the octagon is \begin{align*}P \approx 8(0.7654)=6.1232 \ un\end{align*}.

#### Example 3

Find the perimeter of a regular 60-gon inscribed in a circle with radius 1 unit.

The base of one triangle was \begin{align*}b=2 \sin 3 \rightarrow b \approx 0.1047\end{align*}. Therefore, the perimeter of the 60-gon is \begin{align*}P \approx 60(0.1047)=6.282 \ un\end{align*}.

#### Example 4

Find the perimeter of a regular 180-gon inscribed in a circle with radius 1 unit.

The base of one triangle was \begin{align*}b=2 \sin 1 \rightarrow b \approx 0.0349048\end{align*}. Therefore, the perimeter of the 180-gon is \begin{align*}P \approx 180(0.0349048)=6.283 \ un\end{align*}.

#### Example 5

Make a conjecture about the circumference of a circle with radius 1 unit.

The perimeters from #2, #3, and #4 are approaching \begin{align*}6.283 \approx 2 \pi\end{align*}. A conjecture would be that the circumference of a circle with radius 1 unit is \begin{align*}2 \pi \ un\end{align*}.

### Review

Consider a regular \begin{align*}n\end{align*}-gon inscribed in a circle with radius 1 unit for questions 1-8.

1. What's the measure of the angle between the radius and the height of one triangle in terms of \begin{align*}n\end{align*}?

2. What's the length of the base of one triangle in terms of sine and \begin{align*}n\end{align*}?

3. What's the height of one triangle in terms of cosine and \begin{align*}n\end{align*}?

4. What's the area of one triangle in terms of sine, cosine, and \begin{align*}n\end{align*}?

5. What's the area of the polygon in terms of sine, cosine, and \begin{align*}n\end{align*}?

6. What's the perimeter of the polygon in terms of sine, cosine, and \begin{align*}n\end{align*}?

7. Let \begin{align*}n=10,000\end{align*}. What is the area of the polygon? What is the perimeter of the polygon? Use your calculator and your answers to #6 and #7.

8. Let \begin{align*}n=1,000,000\end{align*}. What is the area of the polygon? What is the perimeter of the polygon? Are you convinced that the area of a circle with radius 1 unit is \begin{align*}\pi\end{align*} and the circumference of a circle with radius 1 unit is \begin{align*}2 \pi\end{align*}?

9. Explain why the area of a regular polygon inscribed in a circle with radius 1 unit gets closer to \begin{align*}\pi\end{align*} as the number of sides increases.

10. Explain why the perimeter of a regular polygon inscribed in a circle with radius 1 unit gets closer to \begin{align*}2 \pi\end{align*} as the number of sides increases.

11. Use similarity and the fact that the circumference of a circle with radius 1 unit is \begin{align*}2 \pi\end{align*} to explain why the formula for the circumference of a circle with radius \begin{align*}r\end{align*} is \begin{align*}2 \pi r\end{align*}.

12. A circle with radius 3 units is transformed into a circle with radius 5 units. What is the ratio of their areas? What is the ratio of their circumferences?

13. The ratio of the areas of two circles is \begin{align*}25:4\end{align*}. The radius of the smaller circle is 2 units. What's the radius of the larger circle?

14. The ratio of the areas of two circles is \begin{align*}25:9\end{align*}. The radius of the larger circle is 10 units. What's the radius of the smaller circle?

15. The ratio of the areas of two circles is \begin{align*}5:4\end{align*}. The radius of the larger circle is 8 units. What's the circumference of the smaller circle?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 8.2.