2.50. The base of a right prism is a right-angled triangle with a hypotenuse equal to $c$ and an acute angle of $30^{\circ}$. A plane is drawn through the hypotenuse of the lower base and the vertex of the right angle of the upper base, forming an angle of $45^{\circ}$ with the base plane. Determine the volume of the triangular pyramid cut off from the prism by the plane.
2.50. The desired volume $V=\frac{1}{3} S_{\text {base }} \cdot C C_{1}$ (Fig. 2.42). Considering that $\angle A B C=30^{\circ}$, we find $A C=\frac{c}{2}, B C=\frac{c \sqrt{3}}{2}$, and thus, $S_{\text {base }}=\frac{c^{2} \sqrt{3}}{8}$. On the other hand, $S_{\text {base }}=\frac{1}{2} A B \cdot C D$, where $C D \perp A B$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-482.jpg?height=465&width=389&top_left_y=104&top_left_x=765)

Fig. 2.42

$C D=\frac{1}{2} B C=\frac{c \sqrt{3}}{4}$. Since $C D \perp A B$, then $C_{1} D \perp A B$, i.e., $\angle C_{1} D C=45^{\circ}$ (by the condition); therefore, in $\triangle C_{1} D C$ we have $C C_{1}=C D=\frac{c \sqrt{3}}{4}$. Thus,

$$
V=\frac{1}{3} \cdot \frac{c^{2} \sqrt{3}}{8} \cdot \frac{c \sqrt{3}}{4}=\frac{c^{3}}{32}
$$

Answer: $\frac{c^{3}}{32}$.