11.096. Determine the surface area of the sphere circumscribed around a cone with a base radius of $R$ and height $h$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-708.jpg?height=402&width=372&top_left_y=411&top_left_x=830)

Fig. 11.83

The surface area of the sphere is: \( S_{\mathrm{u}} = 4 \pi r^{2} \) (Fig. 11.83). The cross-section of the cone is an isosceles triangle. For this triangle, the radius of the circumscribed circle is \( r = \frac{a \cdot b \cdot c}{4 S} \). For \( \triangle ABC \), \( S = \frac{1}{2} AC \cdot BD = \frac{1}{2} 2R \cdot h = Rh \). The sides \( AB = BC = \sqrt{R^{2} + h^{2}} \). Therefore, \( r = \frac{2R(R^{2} + h^{2})}{4Rh} = \frac{(R^{2} + h^{2})}{2h} \). Finally, we get \( S_{\mathrm{m}} = 4 \pi \frac{(R^{2} + h^{2})^{2}}{4h^{2}} = \frac{\pi (R^{2} + h^{2})^{2}}{h^{2}} \).

Answer: \( \frac{\pi (R^{2} + h^{2})^{2}}{h^{2}} \)