House roof

The house's roof is a regular quadrangular pyramid with a base edge 17 m. If the roof pitch is 57° and we calculate 11% of waste, connections, and overlapping of the area roof, how much m² is needed to cover the roof?

Correct answer:

S =  589 m²

Step-by-step explanation:

$$\begin{gathered} a=17m \\ \cos 57^{\circ }=\frac{a/2}{h}=\frac{17/2}{h} \\ h=\frac{17}{2\cos 57^{\circ }}=15.61m \\ {S}_1=\frac{1}{2}ah=\frac{1}{2}\cdot 17\cdot 15.61=132.685m^2 \\ S=4S_1(1+\frac{11}{100})=\frac{a^2}{\cos 57^{\circ }}=4\cdot 132.685\cdot 1.11=589{\text{m}}^2 \end{gathered}$$

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