Roof cover

Above the pavilion with a square ground plan with a side length of a = 12 m is a pyramid-shaped roof with a height v = 4.5 m. Calculate how much m2 of sheet metal is needed to cover this roof if 5.5% of the sheet we must add for joints and waste.

Result

S =  189.9 m2

Solution:

a=12 m v=4.5 m q=1+5.5100=211200=1.055  h2=v2+(a/2)2=4.52+(12/2)2=152=7.5 m  S1=a h2/2=12 7.5/2=45 m2  S2=4 S1=4 45=180 m2 S=S2 q=180 1.055=189910=189.9=189.9 m2a = 12 \ m \ \\ v = 4.5 \ m \ \\ q = 1 + \dfrac{ 5.5 }{ 100 } = \dfrac{ 211 }{ 200 } = 1.055 \ \\ \ \\ h_{ 2 } = \sqrt{ v^2 + (a/2)^2 } = \sqrt{ 4.5^2 + (12/2)^2 } = \dfrac{ 15 }{ 2 } = 7.5 \ m \ \\ \ \\ S_{ 1 } = a \cdot \ h_{ 2 } /2 = 12 \cdot \ 7.5 /2 = 45 \ m^2 \ \\ \ \\ S_{ 2 } = 4 \cdot \ S_{ 1 } = 4 \cdot \ 45 = 180 \ m^2 \ \\ S = S_{ 2 } \cdot \ q = 180 \cdot \ 1.055 = \dfrac{ 1899 }{ 10 } = 189.9 = 189.9 \ m^2



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