11.157. In a regular truncated quadrilateral pyramid, the sides of the bases are equal to $a$ and $b$, and the lateral surface area is half of the total surface area. Find the volume of the pyramid.
Solution.

Let $O$ and $O_{1}$ be the centers of the lower $ABCD$ and upper $A_{1}B_{1}C_{1}D_{1}$ bases of a regular truncated pyramid (Fig. 11.52),

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0862.jpg?height=422&width=539&top_left_y=1339&top_left_x=703)

Fig. 11.52
$M$ is the midpoint of $CD$, and $M_{1}$ is the midpoint of $C_{1}D_{1}$. Then $M M_{1}$ is the slant height of the pyramid. Let $M M_{1}=x$. The lateral surface area of the pyramid $S_{6}=4 \cdot \frac{CD+C_{1}D_{1}}{2} \cdot M M_{1}=2(a+b) x$. The area of the upper base $S_{1}=b^{2}$, and the area of the lower base $S_{2}=a^{2}$.

The total surface area $S=S_{1}+S_{2}+S_{6}$. According to the problem, $S=2 S_{6}$. Therefore, $S_{6}=S_{1}+S_{2} ; 2(a+b) x=a^{2}+b^{2} ; x=\frac{a^{2}+b^{2}}{2(a+b)}$. Drop a perpendicular $M_{1}P$ to the plane of the lower base. Then $P$ lies on $OM$,

$$
PM=\frac{a-b}{2} \cdot \text{ and in } \Delta M_{1}PM\left(\angle M_{1}PM=90^{\circ}\right): M_{1}P=\sqrt{M M_{1}^{2}-P M^{2}}=
$$

$$
=\sqrt{\frac{\left(a^{2}+b^{2}\right)^{2}}{4(a+b)}-\frac{(a-b)^{2}}{4}}=\frac{\sqrt{\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}-b^{2}\right)^{2}}}{2(a+b)}=\frac{ab}{a+b}.
$$

The volume of the pyramid $V=\frac{1}{3} M_{1}P\left(S_{1}+S_{2}+\sqrt{S_{1}S_{2}}\right)=\frac{ab\left(a^{2}+ab+b^{2}\right)}{3(a+b)}$.

Answer: $\quad \frac{ab\left(a^{2}+ab+b^{2}\right)}{3(a+b)}$.