Church roof

The roof of the church tower has the shape of a regular tetrahedral pyramid with base edge length 5.4 meters and a height 5 m. It was found that needs to be corrected 27% covering of the roof area. What amount of material will be required?

Correct result:

S =  16.57 m²

Solution:

$$\begin{gathered} a=5.4 \\ h=5 \\ {h}_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{5^2+(5.4\mathrm{/}2{)}^2}\doteq 5.6824 \\ {S}_1=a\cdot h_2\mathrm{/}2=5.4\cdot 5.6824\mathrm{/}2\doteq 15.3426 \\ {S}_2=4\cdot S_1=4\cdot 15.3426\doteq 61.3702 \\ S=S_2\cdot 27\mathrm{/}100=61.3702\cdot 27\mathrm{/}100=16.57{\text{m}}^2 \end{gathered}$$

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