1.36. 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.
1.36. According to the condition, $\angle C=90^{\circ}, M P=$ $=4 \text{ cm}, M Q=8 \text{ cm}, S_{\triangle A B C}=100 \text{ cm}^{2}$ (Fig. $1.30$); it is required 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

$$
\frac{M P}{A Q}=\frac{B P}{M Q}, \text { or } \frac{4}{y-4}=\frac{x-8}{8} . \text { Therefore, }
$$

we have the system of equations

$$
\left\{\begin{array}{l}
\frac{4}{y-4}=\frac{x-8}{8}, \\
x y=200
\end{array} \text { or } \left\{\begin{array}{l}
x+2 y=50 \\
x y=200
\end{array}\right.\right.
$$

which reduces to the quadratic equation $y^{2}-25 y+100=0$, from which we find $y_{1}=5 \text{ cm} ; y_{2}=20 \text{ cm}$. As a result, we get two solutions: $x_{1}=40 \text{ cm}, y_{1}=5 \text{ cm} ; x_{2}=10 \text{ cm}, y_{2}=20 \text{ cm}$.

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