Regular quadrangular pyramid

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

Result

S =  400.438 m²

Solution:

$h = (a/2)\cdot \sqrt2 \cdot \tan 68 ^\circ = 17.502 \ \\ h_2 = \sqrt{ h^2 + (a/2)^2 } = 18.202 \ \\ S_1 = a h_2 / 2 = 91.01 \ \\ S = 4 \cdot S_1 \cdot 1.1 = 400.438 \ \text{m}^2$

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