10.089. Find the area of the circle circumscribed around an isosceles triangle if the base of this triangle is 24 cm and the lateral side is $13 \mathrm{~cm}$.
Solution.

The area of a circle $S=\pi R^{2}$ (Fig.10.86). The radius of the circumscribed circle $R=\frac{a b c}{4 S_{\triangle A B C}}$, where $S_{\triangle A B C}=\frac{1}{2} A C \cdot B D$.

From $\triangle B D C$ we get $B D=\sqrt{B C^{2}-D C^{2}}=\sqrt{13^{2}-12^{2}}=5$ (cm). Then $S_{\triangle A B C}=\frac{1}{2} 24 \cdot 5=60$ (cm ${ }^{2}$ ). Therefore, $R=\frac{13 \cdot 13 \cdot 24}{4 \cdot 60}=\frac{169}{10}$ (cm). Thus, $S=\frac{169^{2}}{100} \pi=285.61 \pi\left(\mathrm{cm}^{2}\right)$.

Answer: $285.61 \pi \mathrm{cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-599.jpg?height=458&width=1140&top_left_y=91&top_left_x=96)

Fig. 10.87