Area of an isosceles triangle

Text

An isosceles triangle has base $12$ and the two equal sides of length $10$. Compute its area.

Solution

The height to the base meets the base at its midpoint, forming two right triangles with leg $\frac{12}{2}=6$ and hypotenuse $10$. By the Pythagorean theorem the height is $h=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8.$ Hence the area is $A=\frac{12\cdot 8}{2}=\boxed{48}.$