Regular quadrangular pyramid

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

Correct answer:

S =  400.4379 m²

Step-by-step explanation:

$$\begin{gathered} a=10\text{m} \\ \alpha =68{}^{\circ } \\ q=100\%+10\%=1+\frac{10}{100}=1.1 \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 10=10\sqrt{2}\text{m}\doteq 14.1421\text{m} \\ \tan \alpha =h:(u/2) \\ h=u/2\cdot \tan \alpha =u/2\cdot \tan 68°=14.1421/2\cdot \tan 68°=14.1421/2\cdot 2.475087=17.50151\text{m} \\ {w}^2=h^2+(a/2{)}^2 \\ w=\sqrt{h^2+(a/2{)}^2}=\sqrt{17.5015^2+(10/2{)}^2}\doteq 18.2017\text{m} \\ {S}_1=\frac{a\cdot w}{2}=\frac{10\cdot 18.2017}{2}\doteq 91.0086{\text{m}}^2 \\ S=4\cdot S_1\cdot q=4\cdot 91.0086\cdot 1.1=400.4379{\text{m}}^2 \end{gathered}$$

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