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!
 A: 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:
$A(\theta) = (\frac{\theta}{2\pi}) A = (\frac{\theta}{2\pi}) \pi r^2 = \frac{1}{2} r^2 \theta$
A: 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
$$\sum \frac{1}{2}r^2\,\delta\theta_i = \frac{1}{2}\int_{\theta_1}^{\theta_2} r^2~\mathrm{d}\theta$$

