### 12.458 The sides of the lower and upper bases of a regular triangular truncated pyramid are equal to $a$ and $\boldsymbol{b}(a>b)$, respectively. The lateral face forms an angle $\alpha$ with the plane of the base. Find the area of the section of the pyramid by a plane passing through the midline of the lateral face and the center of the lower base.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-324.jpg?height=426&width=380&top_left_y=676&top_left_x=87)

Let $MN$ be the midline of the lateral face of the given regular truncated pyramid $ABC A_1 B_1 C_1$, $O$ and $O_1$ be the centers of the bases of the pyramid, and $MN L K$ be the section of the pyramid by a plane passing through $O$ and $MN$. Then $MN \| AB \| KL$ (Theorem 14). If now $P$ and $P_1$ are the midpoints of $AB$ and $A_1 B_1$, then the plane $OPP_1 O_1$ is perpendicular to the bases (the pyramid is regular). Let $OQ$ be the line of intersection of $OPP_1 O_1$ and $MN L K$. Since $AB \perp PP_1, AB \perp OP$, then $\angle OPQ = \alpha, MN \perp OQ$, i.e., $OQ$ is the height of the trapezoid $MN L K$,

$$
S_{MN L K} = OQ \cdot \frac{1}{2}(MN + KL)
$$

But $PP_1 = \frac{PD}{\cos \alpha} = \frac{PO - P_1 O_1}{\cos \alpha} = \frac{a - b}{2 \sqrt{3} \cos \alpha} \Rightarrow PQ = \frac{PP_1}{2} = \frac{a - b}{4 \sqrt{3} \cos \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-324.jpg?height=272&width=303&top_left_y=1250&top_left_x=98)

Since $PO = \frac{a}{2 \sqrt{3}}$, by the cosine theorem for

$\triangle PQO, OQ = \sqrt{PO^2 + PQ^2 - 2 \cdot PO \cdot PQ \cdot \cos \alpha} = $

$$
= \frac{\sqrt{3(a^2 + b^2 + 2ab \cos 2\alpha)}}{12 \cos \alpha}
$$

But $MN = \frac{a + b}{2}, KL = \frac{2}{3}a$. Therefore, we have

Answer: $S_{MN L K} = \frac{7a + 3b}{144 \cos \alpha} \cdot \sqrt{3(a^2 + b^2 + 2ab \cos 2\alpha)}$.