11.124. The base of a right prism is an isosceles triangle, the base of which is equal to $a$, and the angle at it is $45^{\circ}$. Determine the volume of the prism if its lateral surface area is equal to the sum of the areas of the bases.
Solution.

The base of the prism is an isosceles right triangle, its area \( S = \frac{a^2}{4} \), perimeter \( P = a(\sqrt{2} + 1) \), and the lateral surface area \( S_6 = P H = a H (\sqrt{2} + 1) \), where \( H \) is the height of the prism.

Since \( S_6 = 2 S \), then \( a H (\sqrt{2} + 1) = \frac{a^2}{2} ; H = \frac{a}{2(\sqrt{2} + 1)} = \frac{a(\sqrt{2} - 1)}{2} \), the volume of the prism \( V = S H = \frac{a^3 (\sqrt{2} - 1)}{8} \).

Answer: \( \frac{a^3 (\sqrt{2} - 1)}{8} \).