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?

Result

S =  16.57 m2

Solution:

a=5.4 h=5 h2=h2+(a/2)2=52+(5.4/2)25.6824 S1=a h2/2=5.4 5.6824/215.3426 S2=4 S1=4 15.342661.3702 S=S2 27/100=61.3702 27/10016.5716.57 m2a=5.4 \ \\ h=5 \ \\ h_{2}=\sqrt{ h^2+(a/2)^2 }=\sqrt{ 5^2+(5.4/2)^2 } \doteq 5.6824 \ \\ S_{1}=a \cdot \ h_{2}/2=5.4 \cdot \ 5.6824/2 \doteq 15.3426 \ \\ S_{2}=4 \cdot \ S_{1}=4 \cdot \ 15.3426 \doteq 61.3702 \ \\ S=S_{2} \cdot \ 27/100=61.3702 \cdot \ 27/100 \doteq 16.57 \doteq 16.57 \ \text{m}^2



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