Quadrilateral prism

The height of a regular quadrilateral prism is v = 10 cm, the deviation of the body diagonal from the base is 60°. Determine the length of the base edges, the surface, and the volume of the prism.

Correct result:

a =  4.0825 cm
S =  196.6326 cm²
V =  166.6667 cm³

Solution:

$$\begin{gathered} v=10\text{ cm } \\ \theta =60^{\circ } \\ \tan \theta =v:u \\ u=v\mathrm{/}\tan {\theta }^{\circ }=v\mathrm{/}\tan 60^{\circ }=10\mathrm{/}\tan 60^{\circ }=10\mathrm{/}1.732051=5.7735\text{ cm } \\ u=\sqrt{2}a \\ a=u\mathrm{/}\sqrt{2}=5.7735\mathrm{/}\sqrt{2}=4.0825\text{ cm} \end{gathered}$$

$$\begin{gathered} S_1=a^2=4.0825^2={\displaystyle \frac{50}{3}}\doteq 16.6667{\text{cm}}^2 \\ {S}_2=4\cdot a\cdot v=4\cdot 4.0825\cdot 10\doteq 163.2993{\text{cm}}^2 \\ S=2\cdot S_1+S_2=2\cdot 16.6667+163.2993=196.6326{\text{cm}}^2 \end{gathered}$$

$V=S_1\cdot v=16.6667\cdot 10={\displaystyle \frac{500}{3}}=166.6667{\text{cm}}^3$

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