# What is the basic formula for finding the area of an isosceles triangle?

With the base and height: $\frac{1}{2} b h$.
The leg and $\frac{1}{2}$ of the base form $2$ sides of a right triangle. The height, the third side, is equivalent to $\frac{\sqrt{4 {l}^{2} - {b}^{2}}}{2}$ though the Pythagorean theorem. Thus, the area of an isosceles triangle given a base and a leg is $\frac{b \sqrt{4 {l}^{2} - {b}^{2}}}{4}$.
I could come up with more if you are given angles. Just ask—they can all be figured out through manipulation, but the most important thing to remember is $A = \frac{1}{2} b h$ for all triangles.