10.193. Inside a right angle, there is a point $M$, the distances from which to the sides of the angle are 4 and $8 \mathrm{~cm}$. A line passing through point $M$ cuts off a triangle from the right angle with an area of $100 \mathrm{~cm}^{2}$. Find the legs of the triangle.
Solution.

Given $\angle C=90^{\circ}, M P=4 \mathrm{~cm}, M Q=8 \mathrm{~cm}, S_{\triangle A B C}=100 \mathrm{~cm}^{2}$ (Fig. 10.3); we need to find $B C$ and $A C$. Let $B C=x, A C=y$; then $0.5 x y=100$, i.e., $x y=200$. Since $\triangle B P M \sim \triangle M Q A$, we have $\frac{M P}{A Q}=\frac{B P}{M Q}$ or $\frac{4}{y-4}=\frac{x-8}{8}$.

We have the system of equations

$$
\left\{\begin{array}{l}
\frac{4}{y-4}=\frac{x-8}{8}, \\
x y=200
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+2 y=50, \\
x y=200
\end{array} \Rightarrow y^{2}-25 y+100=0, y_{1}=5 \right.\right. \text { cm }
$$

$y_{2}=20 \mathrm{~cm}$. We obtain two solutions: $x_{1}=40 \mathrm{~cm}, y_{1}=5 \mathrm{~cm} ; x_{2}=10 \mathrm{~cm}$, $y_{2}=20 \mathrm{~cm}$.

Answer: 40 and 5 cm or 10 and 20 cm.