10.064. The lateral side of an isosceles triangle is $10 \mathrm{~cm}$, and the base is $12 \mathrm{~cm}$. Tangents are drawn to the inscribed circle of the triangle, parallel to the height of the triangle, and cutting off two right triangles from the given triangle. Find the lengths of the sides of these triangles.
## Solution.

The area of the triangle is found using the formula (Fig. 10.63):

$$
S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{16(16-10)(16-10)(16-12)}=48
$$

The radius of the inscribed circle $r=\frac{S}{p}=\frac{48}{16}=3$ (cm); $r=D H . H C=D C-$ $-D H=6-3=3 \text{ cm} ; \triangle B C D$ and $\triangle F C H$ are similar, so $\frac{D C}{H C}=\frac{B C}{F C}$. Therefore, $F C=\frac{B C \cdot H C}{D C}=\frac{10 \cdot 3}{6}=5$ cm. From $\triangle F C H$ it follows that $F C^{2}=$ $=F H^{2}+H C^{2} ; F H=\sqrt{F C^{2}-H C^{2}}=\sqrt{5^{2}-3^{2}}=4$ (cm).

## Answer: 3 cm, 4 cm, 5 cm.