10,11

The base of the pyramid $S A B C D$ is a trapezoid $A B C D$ with bases $B C$ and $A D$, and $B C: A D=2: 5$.

The diagonals of the trapezoid intersect at point $E$, and the center $O$ of the sphere inscribed in the pyramid lies on the segment $S E$ and divides it in the ratio $S O: O E=7: 2$. Find the total surface area of the pyramid if the area of the lateral face $S B C$ is 8.
Let

$$
\begin{gathered}
S_{\triangle B S C}=S 1, S_{\triangle A S B}=S 2, S_{\triangle A S D}=S 3, S_{\triangle C S D}=S 4, S_{\text {base }}=S_{A B C D}=S, \\
V_{S B E C}=V 1, V_{S A E B}=V 2, V_{S A E D}=V 3, V_{S C E D}=V 4, V_{S A B C D}=V .
\end{gathered}
$$

By the given condition, $S 1=8$. Note that

$$
\frac{S_{1}}{S_{2}}=\frac{C E^{\prime}}{A E}=\frac{B C}{A D}=\frac{2}{5}, \frac{S_{1}}{S_{3}}=\left(\frac{B C}{A D}\right) 2=\frac{4}{25}, \frac{S_{1}}{S_{1}}=\frac{B E}{D E}=\frac{B C}{A D}=\frac{2}{5}
$$

therefore

$$
S 2=\frac{5}{2} S 1=\frac{5}{2} \cdot 8=20, S 3=\frac{25}{4} S 1=\frac{25}{4} \cdot 8=50, S 4=\frac{5}{2} S 1=\frac{5}{2} \cdot 8=20 .
$$

Thus,

$$
S_{\text {lateral }}=S 1+S 2+S 3+S 4=8+20+50+20=98 .
$$

Let $r$ be the radius of the inscribed sphere in the pyramid, and $h$ be the height of the pyramid. Then

$$
V=\frac{1}{3} S_{\text {total }} r=\frac{1}{3} S_{\text {base }} h,
$$

and since the center $O$ of the inscribed sphere lies on the segment $S E$, we have $\frac{r}{h}=\frac{O E}{S E}=\frac{2}{7}$, therefore

$$
\frac{S}{S}=\frac{h}{r}=\frac{9}{2}
$$

On the other hand, the heights of the pyramids $SBEC$, $SAEB$, $SAED$, and $SCED$, drawn from the common vertex $E$, are $\frac{7}{7} r$, so

$$
\begin{gathered}
V=V_{S B E C}+V_{S A E B}+V_{S A E D}+V_{S C E D}=V 1+V 2+V 3+V 4= \\
=\frac{1}{3} S 1 \cdot \frac{9}{7} r+\frac{1}{3} S 2 \cdot \frac{9}{7} r+\frac{1}{3} S 3 \cdot \frac{9}{7} r+\frac{1}{3} S 4 \cdot \frac{9}{7} r=\frac{1}{3} \cdot \frac{9}{7} r(S 1+S 2+S 3+S 4) r= \\
=\frac{3}{7} S_{\text {lateral }} r=\frac{1}{3} S_{\text {total }} r,
\end{gathered}
$$

from which $\frac{S}{S} =\frac{9}{7}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-22.jpg?height=106&width=532&top_left_y=2594&top_left_x=763)

## Answer

126.00

## Problem