Regular quadrangular pyramid

How many square meters are needed to cover the tower the shape of a regular quadrangular pyramid base edge 10 meters, if the deviation lateral edges from the base plane are 68 °? Calculate coverage of waste by 10%.

Correct result:

S =  400.4379 m²

Solution:

$$\begin{gathered} h=(a\mathrm{/}2)\cdot \sqrt{2}\cdot \tan 68^{\circ }=17.502 \\ {h}_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=18.202 \\ {S}_1=a{h}_2\mathrm{/}2=91.01 \\ S=4\cdot S_1\cdot 1.1=400.4379{\text{m}}^2 \end{gathered}$$

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