Base diagonal

In a regular 4-sided pyramid, the side edge forms an angle of 55° with the base's diagonal. The length of the side edge is eight meters. Calculate the surface area and volume of the pyramid.

Correct answer:

S =  137.0161 m²
V =  91.9869 m³

Step-by-step explanation:

$$\begin{gathered} s=8\text{ m } \\ \phi =55\mathrm{°}\to \text{ rad}=55\mathrm{°}\cdot {\displaystyle \frac{\pi }{180}}=55\mathrm{°}\cdot {\displaystyle \frac{3.1415926}{180}}=0.95993=11\pi \mathrm{/}36 \\ \sin \phi =h:s \\ h=s\cdot \sin (\phi )=8\cdot \sin (0.9599)\doteq 6.5532\text{ m } \\ \cos \phi =u\mathrm{/}2:s \\ u=2\cdot s\cdot \cos (\phi )=2\cdot 8\cdot \cos (0.9599)\doteq 9.1772\text{ m } \\ {u}^2=a^2+a^2 \\ a=u\mathrm{/}\sqrt{2}=9.1772\mathrm{/}\sqrt{2}\doteq 6.4893\text{ m } \\ {h}_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{6.5532^2+(6.4893\mathrm{/}2{)}^2}\doteq 7.3125\text{ m } \\ {S}_1=a^2=6.4893^2\doteq 42.1107{\text{dm}}^2 \\ {S}_2={\displaystyle \frac{a\cdot h_2}{2}}={\displaystyle \frac{6.4893\cdot 7.3125}{2}}\doteq 23.7263{\text{m}}^2 \\ S=S_1+4\cdot S_2=42.1107+4\cdot 23.7263=137.0161{\text{m}}^2 \end{gathered}$$

$V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 42.1107\cdot 6.5532=91.9869{\text{m}}^3$

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