2.35. The base of the pyramid is a parallelogram with sides equal to 10 and $8 \mathrm{M}$, and one of the diagonals is $6 \mathrm{~m}$. 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.
2.35. Given that $A B = 8 \mathrm{M}, A D = 10 \mathrm{M}, B D = 6 \mathrm{M}$ (Fig. 2.30). From the equality $6^{2} + 8^{2} = 10^{2}$, it follows that $\triangle A B D$ is a right triangle and $S_{\text{base}} = 8 \cdot 6 = 48 \left(\mathrm{M}^{2}\right)$. Since $B O \perp A B$, then $S B \perp A B$, and

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-474.jpg?height=356&width=590&top_left_y=426&top_left_x=561)

Fig. 2.30

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

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 $\triangle O K D \sim \triangle A B D$; we have $O K : A B = O D : A D$, from which $O K = \frac{8 \cdot 3}{10} = \frac{12}{5}$ (m). Therefore, $S K = \sqrt{16 + \frac{144}{25}} = \frac{4 \sqrt{34}}{5}$ (m) and $S_{\triangle A S D} = \frac{1}{2} \cdot 10 \cdot \frac{4 \sqrt{34}}{5} = 4 \sqrt{34} \left(\mathrm{M}^{2}\right)$. Thus,

$$
S_{\text{poly}} = 48 + 2 \cdot 20 + 2 \cdot 4 \sqrt{34} = 8(11 + \sqrt{34}) \left(\mathrm{M}^{2}\right).
$$

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