Base diagonal

In a regular four-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 pyramid's surface area and volume.

Final Answer:

S =  137.0161 m²
V =  91.9869 m³

Step-by-step explanation:

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

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

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