# 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:

$s_{ 1 } = 18 \ cm \ \\ s_{ 2 } = 14 \ cm \ \\ h = 30 \ cm \ \\ \ \\ \ \\ n = 7 \ \\ \ \\ S_{ 1 } = \dfrac{ 1 }{ 4 } \cdot \ n \cdot \ s_{ 1 } \cdot \ \cotg(\pi/n) = \dfrac{ 1 }{ 4 } \cdot \ 7 \cdot \ 18 \cdot \ \cotg(3.1416/7) \doteq 65.4104 \ cm^2 \ \\ S_{ 2 } = \dfrac{ 1 }{ 4 } \cdot \ n \cdot \ s_{ 2 } \cdot \ \cotg(\pi/n) = \dfrac{ 1 }{ 4 } \cdot \ 7 \cdot \ 14 \cdot \ \cotg(3.1416/7) \doteq 50.8748 \ cm^2 \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ h \cdot \ ( S_{ 1 }+S_{ 2 } + \sqrt{ S_{ 1 } \cdot \ S_{ 2 } }) = \dfrac{ 1 }{ 3 } \cdot \ 30 \cdot \ ( 65.4104+50.8748 + \sqrt{ 65.4104 \cdot \ 50.8748 }) \doteq 1739.7177 \ cm^3 \ \\ \ \\ \ \\ ρ = 10 \ g/cm^3 \ \\ \ \\ m_{ 1 } = ρ \cdot \ V = 10 \cdot \ 1739.7177 \doteq 17397.177 \ g \ \\ \ \\ m = m_{ 1 } \rightarrow kg = m_{ 1 } / 1000 \ kg = 17.3971769886 \ kg = 17.397 \ \text { kg }$

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