10.295. The legs of a right triangle are 6 and 8 cm. A circle is drawn through the midpoint of the smaller leg and the midpoint of the hypotenuse, touching the hypotenuse. Find the area of the circle bounded by this circle.
## Solution.

Given in $\triangle ABC$ (Fig. 10.88) $\angle ABC=90^{\circ}, AB=6$ cm, $BC=8$ cm, points $E$ and $D$ are the midpoints of $AB$ and $AC$, respectively, and $O$ is the center of the circle mentioned in the problem. $AC=\sqrt{AB^{2}+BC^{2}}=10$ cm. Then $ED$ is the midline of $\triangle ABC$, $ED=\frac{1}{2} BC=4$ cm. $K$ is the midpoint of $ED$. Then $KD=2$ cm. $OK \perp ED$, $BC \| ED$. Then $OK \perp BC$. $OD \perp AC$. Therefore, $\angle KOD=\angle BCA$ - as angles with mutually perpendicular sides, and right triangles $OKD$ and $CBA$ are similar. Then $\frac{OD}{AC}=\frac{KD}{AB}$, $OD=\frac{AC \cdot KD}{AB}=\frac{10}{3}$ cm. The area of the circle $S=\pi \cdot OD^{2}=\frac{100 \pi}{9} \text{ cm}^{2}$. Answer: $\frac{100 \pi}{9} \text{ cm}^{2}$.