House roof

The roof of the house has the shape of a regular quadrangular pyramid with a base edge 17 m. How many m2 is needed to cover roof if roof pitch is 57° and we calculate 11% of waste, connections and overlapping of area roof?

Result

S =  589 m2

Solution:

a=17 m cos57=a/2h=17/2h h=172cos57=15.61 m  S1=12ah=121715.61=132.685 m2  S=4S1(1+11100)=a2cos57=4132.6851.11=589 m2a = 17 \ m \ \\ \cos 57 ^\circ = \dfrac{a/2}{h} = \dfrac{ 17/2}{h} \ \\ h = \dfrac{ 17 }{ 2 \cos 57 ^\circ } = 15.61 \ m \ \\ \ \\ S_1 = \dfrac12 ah = \dfrac12 \cdot 17 \cdot 15.61 = 132.685 \ m^2 \ \\ \ \\ S = 4S_1(1+\dfrac{ 11 }{100}) = \dfrac{a^2}{\cos 57 ^\circ }= 4 \cdot 132.685 \cdot 1.11 = 589 \ \text{m}^2



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