11.147. The base of the pyramid is a parallelogram with sides equal to 10 and 8 m, and one of the diagonals is 6 m. The height of the pyramid passes through the point of intersection of the diagonals of the base and is 4 m. Determine the total surface area of the pyramid.
## Solution.

Given $A B=8 \mathrm{M}, A D=10 \mathrm{M}, B D=6 \mathrm{M}$ (Fig. 11.44). Since $6^{2}+8^{2}=10^{2}$, $\triangle A B D$ is a right triangle and $S_{\text {base }}=8 \cdot 6=48\left(\mathrm{M}^{2}\right)$.

Since $B D \perp A B$, then $S B \perp A B$, and $S_{\triangle A S B}=\frac{1}{2} A B \sqrt{S O^{2}+O B^{2}}=$ $=\frac{1}{2} \cdot 8 \sqrt{4^{2}+3^{2}}=20\left(\mathrm{M}^{2}\right)$.

Draw $S K \perp A D$; then $S_{\triangle A S D}=\frac{1}{2} A D \cdot S K=\frac{1}{2} A D \sqrt{S O^{2}+O K^{2}}$. To find $O K$, we use the fact that $\triangle O K D \sim \triangle A B D$;

then $O K: A B=O D: A D \Rightarrow O K=\frac{8 \cdot 3}{10}=\frac{12}{5}$ (m). Therefore, $S K=\sqrt{16+\frac{144}{25}}=\frac{4 \sqrt{34}}{5}$ (m) and $S_{\triangle A S D}=\frac{1}{2} \cdot 10 \cdot \frac{4 \sqrt{34}}{5}=4 \sqrt{34}$ (m²). Thus, $S_{\text {lateral }}=48+2 \cdot 20+2 \cdot 4 \sqrt{34}=8(11+\sqrt{34})\left(\mathrm{M}^{2}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0856.jpg?height=460&width=785&top_left_y=61&top_left_x=284)

Fig. 11.44

Answer: $\quad 8(11+\sqrt{34})\left(\mathrm{M}^{2}\right)$.