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%.

Correct 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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