10.162. Calculate the area of an isosceles triangle if the length of its height, drawn to the lateral side, is $12 \mathrm{~cm}$, and the length of the base is $15 \mathrm{cm}$.
## Solution.

Let $A K$ be the perpendicular to the side $B C$ (Fig. 10.136). Then $A K=12, A C=15$ (cm); $K C=\sqrt{A C^{2}-A K^{2}}=\sqrt{225-144}=9$ (cm). Let $A B=B C=x$. Then $A B^{2}=A K^{2}+13 x^{2}$ or $x^{2}=144+(x-9)^{2}$, from which $x=\frac{15}{2}$ (cm). Then $S_{A B C}=\frac{1}{2} A K \cdot B C=\frac{1}{2} 12 \cdot \frac{15}{2}=75\left(\mathrm{~cm}^{2}\right)$.

Answer: $75 \mathrm{~cm}^{2}$.