12.227. The base of the pyramid is an isosceles trapezoid, where the lateral side is equal to $a$, and the acute angle is equal to $\alpha$. All lateral faces form the same angle $\beta$ with the base of the pyramid. Find the total surface area of the pyramid.
## Solution.

Let trapezoid $ABCD$ be the base of the pyramid $FABCD$ (Fig. 12.94), $FO$ - the height, $AB=CD=a, \angle ABC=\alpha, 0^{\circ}<\alpha<90^{\circ}$.

Since all lateral faces of the pyramid form the same angle $\beta$ with the base, point $O$ is equidistant from all sides of trapezoid $ABCD$ and is the center of the circle inscribed in this trapezoid. Hence, $AD + BC = AB + CD = 2a$.

If $AE$ is the height of trapezoid $ABCD$, then $AE = a \sin \alpha$.

The area of trapezoid $ABCD$ is $S = \frac{AD + BC}{2} \cdot AE = a^2 \sin \alpha$, and the area of the lateral surface of the pyramid is $S_{\text{lat.}} = \frac{S}{\cos \beta} \Rightarrow$

$$
\Rightarrow S_{n} = S + S_{6} = S + \frac{S}{\cos \beta} = \frac{S(1 + \cos \beta)}{\cos \beta} = \frac{2a^2 \sin \alpha \cos^2 \frac{\beta}{2}}{\cos \beta}
$$

$$
\text{Answer: } \frac{2a^2 \sin \alpha \cos^2 \frac{\beta}{2}}{\cos \beta}
$$