10.316. In a right-angled triangle, the distance from the midpoint of the hypotenuse to one of the legs is 5 cm, and the distance from the midpoint of this leg to the hypotenuse is 4 cm. Calculate the area of the triangle.
Solution.

Let in $\triangle A B C$ (Fig. 10.105) $\angle A C B=90^{\circ}, M$ - the midpoint of $A B$, $M N \perp B C, M N=5 \text{ cm}$. Then $N$ - the midpoint of $B C$ and $A C=2 M N=10 \text{ cm}$. Drop perpendiculars $N K$ and $C P$ to $A B$. Then $N K=4 \text{ cm}, K$ - the midpoint of $B P$ and $C P=2 N K=8 \text{ cm}$. From $\triangle A P C\left(\angle A P C=90^{\circ}\right): A P=\sqrt{A C^{2}-C P^{2}}=6 \text{ cm}$. From $\triangle A C B\left(\angle A C B=90^{\circ}\right): A C^{2}=A P \cdot A B ; A B=\frac{A C^{2}}{A P}=\frac{50}{3} \text{ cm}$. Then $S_{\triangle A B C}=\frac{1}{2} A B \cdot C P=\frac{200}{3} \text{ cm}^{2}$.

Answer: $\frac{200}{3} \text{ cm}^{2}$.