11.144. 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.
## Solution.

The desired volume $V=\frac{1}{3} S_{\text {base }} \cdot C C_{1}$ (Fig. 11.41). Since $\angle A B C=30^{\circ}$, then $A C=\frac{c}{2}, B C=\frac{c \sqrt{3}}{2}$, hence $S_{\text {base }}=\frac{c^{2} \sqrt{3}}{8}$. On the other hand, $\quad S_{\text {base }}=\frac{1}{2} A B \cdot C D$, where $C D \perp A B$ and $C D=\frac{1}{2} B C=\frac{c \sqrt{3}}{4}$. Since

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0853.jpg?height=559&width=535&top_left_y=68&top_left_x=96)

Fig. 11.41

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0853.jpg?height=467&width=620&top_left_y=158&top_left_x=643)

Fig. 11.42

$C D \perp A B$, 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 $C C_{1}=C D=\frac{c \sqrt{3}}{4}$. We obtained $V=\frac{1}{3} \cdot \frac{c^{2} \sqrt{3}}{8} \cdot \frac{c \sqrt{3}}{4}=\frac{c^{3}}{32}$.

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