Heptagonal pyramid

A hardwood for a column is in the form of a frustum of a regular heptagonal pyramid. The lower base edge is 18 cm and the upper base of 14 cm. The altitude is 30 cm. Determine the weight in kg if the density of the wood is 10 grams/cm3.

Result

m =  17.397 kg

Solution:

s1=18 cm s2=14 cm h=30 cm   n=7  S1=14 n s1 cotg(π/n)=14 7 18 cotg(3.1416/7)65.4104 cm2 S2=14 n s2 cotg(π/n)=14 7 14 cotg(3.1416/7)50.8748 cm2  V=13 h (S1+S2+S1 S2)=13 30 (65.4104+50.8748+65.4104 50.8748)1739.7177 cm3   ρ=10 g/cm3  m1=ρ V=10 1739.717717397.177 g  m=m1kg=m1/1000 kg=17.39718 kg=17.397  kg s_{ 1 } = 18 \ cm \ \\ s_{ 2 } = 14 \ cm \ \\ h = 30 \ cm \ \\ \ \\ \ \\ n = 7 \ \\ \ \\ S_{ 1 } = \dfrac{ 1 }{ 4 } \cdot \ n \cdot \ s_{ 1 } \cdot \ \cotg(\pi/n) = \dfrac{ 1 }{ 4 } \cdot \ 7 \cdot \ 18 \cdot \ \cotg(3.1416/7) \doteq 65.4104 \ cm^2 \ \\ S_{ 2 } = \dfrac{ 1 }{ 4 } \cdot \ n \cdot \ s_{ 2 } \cdot \ \cotg(\pi/n) = \dfrac{ 1 }{ 4 } \cdot \ 7 \cdot \ 14 \cdot \ \cotg(3.1416/7) \doteq 50.8748 \ cm^2 \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ h \cdot \ ( S_{ 1 }+S_{ 2 } + \sqrt{ S_{ 1 } \cdot \ S_{ 2 } }) = \dfrac{ 1 }{ 3 } \cdot \ 30 \cdot \ ( 65.4104+50.8748 + \sqrt{ 65.4104 \cdot \ 50.8748 }) \doteq 1739.7177 \ cm^3 \ \\ \ \\ \ \\ ρ = 10 \ g/cm^3 \ \\ \ \\ m_{ 1 } = ρ \cdot \ V = 10 \cdot \ 1739.7177 \doteq 17397.177 \ g \ \\ \ \\ m = m_{ 1 } \rightarrow kg = m_{ 1 } / 1000 \ kg = 17.39718 \ kg = 17.397 \ \text{ kg }



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