11.076. The height of the cone is equal to the diameter of its base. Find the ratio of the area of its base to the lateral surface area.
## Solution.

The area of the base of the cone: $S_{\text {osn }}=\pi R^{2}$ (Fig. 11.68). The lateral surface area of the cone $S_{\text {bok }}=\pi R l$, where

$l=\sqrt{H^{2}+R^{2}}=\sqrt{4 R^{2}+R^{2}}=R \sqrt{5}$. Therefore, $S_{\text {bok }}=\pi R^{2} \sqrt{5}$. Finally, we get $\frac{S_{\text {osn }}}{S_{\text {bok }}}=\frac{\pi R^{2}}{\pi R^{2} \sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$.

Answer: $\frac{\sqrt{5}}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-697.jpg?height=542&width=392&top_left_y=91&top_left_x=188)

Fig. 11.69

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-697.jpg?height=430&width=368&top_left_y=197&top_left_x=768)

Fig. 11.70