10.166. One of the legs of a right triangle is 15 cm, and the radius of the circle inscribed in the triangle is 3 cm. Find the area of the triangle.
## Solution.

Let $K, L, M$ be the points of tangency of the inscribed circle (Fig. 10.138). Then $K O=L O=M O=3$ (cm), $C A=15$ (cm). From this, $C M=K C=3$ (cm), $M A=C A-C M=15-3=12$ (cm). Therefore, $L A=$ $=12$ (cm). Let $B K=B L=x$. Then the area of the triangle $S=2 S_{B K O}+2 S_{O L A}+S_{K O M C}=$ $=2 \cdot \frac{1}{2} x \cdot 3+2 \cdot \frac{1}{2} \cdot 12 \cdot 3+3 \cdot 3=3 x+45$. On the other hand, $S=(x+3) \cdot 15 \cdot \frac{1}{2}=\frac{15}{2} x+\frac{45}{2}$. Then $\frac{15}{2} x+\frac{45}{2}=3 x+45$, from which $x=5$. Thus, the area of the triangle $S=15 \cdot 8 \cdot \frac{1}{2}=60\left(\mathrm{~cm}^{2}\right)$.

Answer: $60 \mathrm{~cm}^{2}$.