Top of the tower

The top of the tower has the shape of a regular hexagonal pyramid. The base edge has a length of 1.2 m, the pyramid height is 1.6 m. How many square meters of sheet metal is needed to cover the top of the tower if 15% extra sheet metal is needed for joints, overlap and waste?

Result

S =  7.074 m2

Solution:

a=1.2 m h=1.6 m n=6  h2=h2+(a/2)2=1.62+(1.2/2)21.7088 m S1=n a h22=6 1.2 1.708826.1517 m2 q=15%=1+15100=1.15  S=q S1=1.15 6.15177.07447.074 m2a=1.2 \ \text{m} \ \\ h=1.6 \ \text{m} \ \\ n=6 \ \\ \ \\ h_{2}=\sqrt{ h^2+(a/2)^2 }=\sqrt{ 1.6^2+(1.2/2)^2 } \doteq 1.7088 \ \text{m} \ \\ S_{1}=n \cdot \ \dfrac{ a \cdot \ h_{2} }{ 2 }=6 \cdot \ \dfrac{ 1.2 \cdot \ 1.7088 }{ 2 } \doteq 6.1517 \ \text{m}^2 \ \\ q=15 \%=1 + \dfrac{ 15 }{ 100 }=1.15 \ \\ \ \\ S=q \cdot \ S_{1}=1.15 \cdot \ 6.1517 \doteq 7.0744 \doteq 7.074 \ \text{m}^2



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