Area of Pentagon question Suppose that a regular pentagon circumscribes a circle of radius $r$. Show that the area of the pentagon is $5r^2\tan(36°)$.
I know that the area of a triangle is $\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. But how do I solve this problem and prove this theory?
This was my attempt and I'm stuck. 
$5r^2\tan(36°) = \frac{1}{2}bh \cdot \sin(36°)$

The central angle of each of the five triangles that make up the circumscribed pentagon is $72^{\circ}$.  Half of this is $36^{\circ}.$
Cut each of these triangles in half to make ten scalene right triangles, with angles $36, 54, \text{ and } 90$ degrees.
The distance from the center of the circle (and of the pentagon) to the edge of the circle is $r$.  This is also the distance to the centers of each of the sides of the pentagon.
Half the length of the pentagon's side by trigonometry, then, is $x = r \tan 36^{\circ},$ making the area of each of the right triangles
$$A = \frac{rx}{2} = \frac{r \cdot r \tan 36^{\circ}}{2}.$$
There are $10$ such triangles in the pentagon, so $10A = 5r^2 \tan 36^{\circ}$ is your answer.
