10.094. Find the area of a circle inscribed in a right triangle (Fig. 10.90), if the projections of the legs on the hypotenuse are 9 and 16 m.
Solution.

$$
A C=A D+D C=25\left(\text { m); } \triangle A D B \text { and } \triangle A B C \text { - similar: } \frac{A D}{A B}=\frac{A B}{A C} .\right) \text { Hence }
$$

from $\frac{9}{x}=\frac{x}{25}, x^{2}=9 \cdot 25=225, x=\sqrt{225}=15($ m), $A B=15$ (m). $\triangle B D C$ and $\triangle A B C$ are similar: $\frac{D C}{B C}=\frac{B C}{A C}$. Hence $\frac{16}{y}=\frac{y}{25}, y^{2}=16 \cdot 25=400$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-601.jpg?height=388&width=416&top_left_y=90&top_left_x=160)

Fig. 10.91

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-601.jpg?height=356&width=530&top_left_y=108&top_left_x=687)

Fig. 10.92

$y=\sqrt{400}=20$ (m), $B C=20$ (m). Radius of the inscribed circle: $r=\frac{a+b-c}{2}=\frac{15+20-25}{2}=5$ (m). The required area of the circle $S=\pi r^{2}=25 \pi\left(\mathbf{M}^{2}\right)$.

Answer: $25 \pi \mathrm{M}^{2}$.