10.104. The perimeter of a right-angled triangle is 24 cm, and its area is $24 \mathrm{~cm}^{2}$. Find the area of the circumscribed circle.
## Solution.

The hypotenuse of the triangle $c=2 R$ (Fig. 10.100).

Solve the system $\left\{\begin{array}{l}p=x+y+2 R, \\ S=\frac{1}{2} x y, \\ x^{2}+y^{2}=4 R^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}24=x+y+2 R, \\ 24=\frac{1}{2} x y, \\ x^{2}+y^{2}=4 R^{2},\end{array}\right.\right.$ we get $x=6$ (cm), $y=8$ (cm). The radius of the circle $R=\frac{2 \cdot 24-x-y}{2}=5$ (cm). The area of the circumscribed circle $S=\pi R^{2}=25 \pi\left(\mathrm{cm}^{2}\right)$.

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

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-606.jpg?height=292&width=648&top_left_y=138&top_left_x=156)

Fig. 10.101

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-606.jpg?height=334&width=310&top_left_y=93&top_left_x=873)

Fig. 10.102