# What is the relationship between the area of a triangle and an area of a segment of a circle?

I had a very smart physics teacher in the past remind us of the area of a segment of circle through this 'derivation':

"well, if you put two of those together doesn't it kind of look like a rectangle? what's the area of a rectangle? now divide that in half."

It is greatly bothering me that this explanation works and assume there's some relationship I don't see please help!

• Your picture shows a sector and not a segment. A sector is a "piece of pizza". A segment is any shape you get by making a straight cut across a pizza. In particular, this cut does not need to pass through the centre. – Fly by Night Jun 7 '14 at 23:56

The area of a circle is

$A = \pi r^2$

Now, ask yourself the following: What's the area of a half-circle?

$\text{Area (half circle)} = \frac{1}{2} A = \frac{1}{2} \pi r^2$

How about one fourth of a circle?

$\text{Area (fourth of a circle)} = \frac{1}{4} A = \frac{1}{4} \pi r^2$

In general, we need to multiply the area of the circle by the fraction of the circle we are talking about. We can find that fraction of the circle by dividing $\theta$, the area of the sector, by the number of radians in the whole circle, $2\pi$.

So in general, the area of a sector with angle $\theta$ is:

$A(\theta) = (\frac{\theta}{2\pi}) A = (\frac{\theta}{2\pi}) \pi r^2 = \frac{1}{2} r^2 \theta$

• This doesn't seem to address the question though. This is certainly a good derivation, but doesn't address why the teacher's derivation seems to work. – Hayden Jun 7 '14 at 23:57
• Yes, this is the derivation I generally go by – 2c2c Jun 8 '14 at 4:48
• @Hayden I agree. In fact, if you don't use calculus then the formula $A=\pi r^2$ depends on the teacher's derivation. You can cut a circle up into little sectors and then re-arrange them to make a rectangle. – Fly by Night Jun 8 '14 at 13:04

The formula for the area of a sector comes from the formula $$\frac{1}{2}ab\sin C$$ for the area of a triangle where $a$ and $b$ are the lengths of two sides and $C$ is the included angle. In the picture below, the sides $OA$ and $OB$ are both radii and so have length $r$. Imagine that the angle $\angle AOB$ is a tiny angle, say $\delta\theta$. The area of the triangle is then $$\frac{1}{2}r^2\,\delta\theta$$ If the angle is tiny then the area of the triangle $\triangle OAB$ is pretty much the same as the area of the sector $OAB$. To find the area of a larger sector, we need to "add-up" lots if tiny sectors.

This is exactly what an integral is used for. If you have lots of little triangles, all with angles $\delta\theta_i$ which cover $\theta_1 < \theta < \theta_2$ in polar coordinates then

$$\sum \frac{1}{2}r^2\,\delta\theta_i = \frac{1}{2}\int_{\theta_1}^{\theta_2} r^2~\mathrm{d}\theta$$