11.030. The base of a right parallelepiped is a parallelogram with sides 1 and 4 cm and an acute angle of $60^{\circ}$. The larger diagonal of the parallelepiped is $5 \mathrm{~cm}$. Determine its volume.
Solution.

Let $AB=1$ (cm), $BC=4$ (cm) (Fig. 11.25); $\angle BAD=60^{\circ}$, then $\angle ADC=120^{\circ}$. Therefore, $AC$ can be found using the cosine theorem: $AC^{2}=$ $=AD^{2}+DC^{2}-2AD \cdot DC \cos 120^{\circ}=1+16+2 \cdot 4 \cdot \frac{1}{2}=21, AC_{1}=5$ (cm). Therefore, $CC_{1}^{2}=AC_{1}^{2}-AC^{2}=25-21=4\left(\mathrm{~cm}^{2}\right)$. Then $CC_{1}=2$ (cm). Volume $V=S_{\text {base }} \cdot CC_{1} ; \quad S_{\text {base }}=AB \cdot AD \sin 60^{\circ}=1 \cdot 4 \cdot \frac{\sqrt{3}}{2}=2 \sqrt{3}\left(\mathrm{~cm}^{2}\right) . \quad$ Then $V=2 \sqrt{3} \cdot 2=4 \sqrt{3}(\mathrm{cm})^{3}$.

Answer: $4 \sqrt{3} \mathrm{~cm}^{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-669.jpg?height=512&width=562&top_left_y=94&top_left_x=127)

Fig. 11.25

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-669.jpg?height=430&width=452&top_left_y=173&top_left_x=750)

Fig. 11.26