[Helper area. The area helps solve the task] Pythagorean Theorem (direct and inverse) ]

Find the area of an isosceles triangle if the height dropped to the base is 10, and the height dropped to the side is 12.
Compose an equation with respect to half the base of the triangle.

## Solution

Let $B K$ and $C M$ be the altitudes of the isosceles triangle $A B C (A B=B C, B K=10, C M=12)$. Denote $A K=$ КС $=x$. Then

$$
\begin{gathered}
A B^{2}=B K^{2}+A K^{2}=100+x^{2} \\
S_{\Delta \mathrm{ABC}}=\frac{1}{2} A C \cdot B K=\frac{1}{2} A B \cdot C M .
\end{gathered}
$$

From this equation, we find that $x=\frac{15}{2}$. Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A C \cdot B K=75
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-15.jpg?height=406&width=575&top_left_y=394&top_left_x=747)

Answer

75.