Area of a sector of a circle

Question

Could someone explain to me why the following holds?

$$ \frac{\text{Area of sector}}{\text{Area of circle}} = \frac{\theta}{2\pi} $$

When deriving the area of a sector my book just quotes the above but doesn't explain why it holds. Could someone explain?

Also, a similar argument is used when deriving arc length, which I don't understand.

By the way, I know that $\theta$ is the angle subtended by the arc and $2\pi$ is the angle in a full circle.

Image from book

How does the ratio of the areas make it equivalent to the ratio of their angles?

Answer 1

This is more intuitive than axiomatic. If you cut a pie into $n$ equal pieces then each piece will have the following properties.

1. The central angle $\theta$ will be $\theta = \frac{2\pi}{n}$
2. The arc length $L$ of the each sector will be the same, i.e. $L=\frac{2\pi r}{n}$
3. The area $A$ of each piece will be the same, i.e. $A=\frac{\pi r^2}{n}$

Thus, the quantities angle, area and length will be $\frac{1}{n}$ of $2\pi$, the area of the circle and the circumference of the circle, respectively.

Using this we can derive the usual formulas:

$A = \frac{\pi r^2}{n}=\pi r^2\frac{\theta}{2\pi}=\frac{r^2\theta}{2}$

$L = \frac{2\pi r}{n}=2\pi r \frac{\theta}{2\pi}=r\theta$

Answer 2

We have the area of a sector given by $A_S = \frac{\theta r^2}{2}$, and the area of a circle by $A_T = \pi r^2$. Hence

$\frac{A_S}{A_C} = \frac{\frac{\theta r^2}{2}}{\pi r^2}= \frac{\theta}{2 \pi}$

Answer 3

How about some calculus? This is the area of the sector of radius $r$ and angle $\theta$:

$$A_{sector} = \int_{r'=0}^{r}\int_{\theta '=0}^{\theta} r' dr' d\theta ' = r^2/2 \cdot \theta.$$

For the full circle, integrate $\theta$ from $0$ to $2\pi$; the ratio of the two answers will get you the relation in your question.

Answer 4

The area of the circle is $\pi R^2$ and it is proportional to the angle, indeed if we divide the circle in $n$ equals slices, by symmetry, the area of each part is $\frac{\pi R^2}{n}$ and the angles of each slice is $\frac{2\pi}{n}$, then

$$\frac{\text{Area circle}}{2\pi}=\frac{\text{Area sector}}{\theta}\implies \frac{\text{Area sector}}{\text{Area circle}}=\frac{\theta }{2\pi}$$