11.045. The largest diagonal of a regular hexagonal prism is $d$ and forms an angle of $30^{\circ}$ with the lateral edge of the prism. Find the volume of the prism.
Solution.

$$
V=S H . \mathrm{B} \triangle A D D_{1} \text { (Fig. 11.40) } \angle A D D_{1}=90^{\circ}, \angle A D_{1} D=30^{\circ}, A D_{1}=d
$$

then $H=D_{1} D=d \cdot \cos 30^{\circ}=d \sqrt{3} / 2 ; A D=d \sin 30^{\circ}=d / 2$. Let $B C=x$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-678.jpg?height=494&width=534&top_left_y=91&top_left_x=121)

Fig. 11.41

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-678.jpg?height=488&width=534&top_left_y=92&top_left_x=685)

Fig. 11.42

$$
\begin{aligned}
& A E=\left(\frac{d}{2}-x\right): 2=\frac{d}{4}-\frac{x}{2}, \angle E B A=30^{\circ} ; x=\frac{A E}{\sin 30^{\circ}}=2 A E, x=\frac{d}{2}-x \\
& 2 x=\frac{d}{2} ; x=\frac{d}{4} ; S=\frac{3}{2} d^{2} \sqrt{3}=\frac{3 d^{2} \sqrt{3}}{32} ; V=S H=\frac{3 d^{2} \sqrt{3}}{32} \cdot \frac{d \sqrt{3}}{2}=\frac{9 d^{3}}{64}
\end{aligned}
$$

Answer: $\frac{9 d^{3}}{64}$.