## Task B-3.5.

Calculate the volume of a regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ if $|A B|=a$ and $\varangle C_{1} A B + \varangle C_{1} A C = 135^{\circ}$.
## First Solution.

Let $\varangle C_{1} A B=\alpha, \varangle C_{1} A C=\beta$, and the height of the prism be $v$. Therefore, $\alpha+\beta=135^{\circ}$.

From the right triangle $C_{1} A C$, we have $\cos \beta=\frac{a \sqrt{2}}{\left|A C_{1}\right|}$,

and from the right triangle $C_{1} A B$, we have $\cos \alpha=\frac{a}{\left|A C_{1}\right|}$.

Thus, $\frac{a \sqrt{2}}{\cos \beta}=\frac{a}{\cos \alpha}$, which simplifies to $\cos \beta=\sqrt{2} \sin \alpha$ and we get

$$
\cos \beta=\sqrt{2} \cos \left(135^{\circ}-\beta\right)=\sqrt{2}\left(\cos 135^{\circ} \cos \beta+\sin 135^{\circ} \sin \beta\right)=-\cos \beta+\sin \beta
$$

From $\cos \beta=-\cos \beta+\sin \beta$, it follows that $\operatorname{tg} \beta=2$.

From the right triangle $C_{1} A C$, we have $\operatorname{tg} \beta=\frac{v}{a \sqrt{2}}$ so $v=\operatorname{tg} \beta \cdot a \sqrt{2}=2 a \sqrt{2}$.

Therefore, the volume of the prism is $V=B \cdot v=a^{2} \cdot 2 a \sqrt{2}=2 a^{3} \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_1c7eac828d9f4d45279fg-15.jpg?height=617&width=445&top_left_y=1031&top_left_x=811)