11.021. Each of the lateral edges of the pyramid is $269 / 32$ cm. The base of the pyramid is a triangle with sides 13, 14, and 15 cm. Find the volume of the pyramid.
Solution.

Let $S A=S B=S C=\frac{269}{32}$ (cm) (Fig. 11.4), and $A B=c=15$ (cm), $B C=a=13$ (cm), $A C=b=14$ (cm), $S H$ - height. This means that $\angle S H A=\angle S H B=\angle S H C$. Then $A H=B H=C H$ (from the equality of $\triangle A S H$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-662.jpg?height=414&width=436&top_left_y=145&top_left_x=156)

Fig. 11.16

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-662.jpg?height=512&width=520&top_left_y=94&top_left_x=730)

Fig. 11.17

$\Delta B S H, \triangle C S H)$. This means that $H$ is the center of the circumscribed circle around $\triangle A B C$ and $R=\frac{a b c}{4 S}$. The area of $\triangle A B C$ can be found using Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}=21$ (cm). Then $S=\sqrt{21(21-13)(21-14)(21-15)}=84\left(\mathrm{~cm}^{2}\right)$. We calculate the radius of the circumscribed circle $R=\frac{13 \cdot 14 \cdot 15}{4 \cdot 84}=\frac{65}{8}$ (cm) and the height of the pyramid

$$
S H=\sqrt{A S^{2}-A H^{2}}=\sqrt{\left(\frac{269}{32}\right)^{2}-\left(\frac{65}{8}\right)^{2}}=\frac{69}{32}(\mathrm{~cm})
$$

The volume $V=\frac{1}{3} S_{\text {base }} \cdot S H=\frac{1}{3} \cdot 84 \cdot \frac{69}{32}=60.375 \mathrm{~cm}^{3}$.

Answer: $60.375 \mathrm{~cm}^{3}$.