Tetrahedral pyramid

Calculate the surface S and the volume V of a regular tetrahedral pyramid with the base side a = 5 m and a body height of 14 m.

Correct result:

V =  116.667 m3
S =  167.215 m2

Solution:

a=5 m h=14 m  S1=a2=52=25 m2  V=13 S1 h=13 25 14=3503=116.667 m3a=5 \ \text{m} \ \\ h=14 \ \text{m} \ \\ \ \\ S_{1}=a^2=5^2=25 \ \text{m}^2 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 25 \cdot \ 14=\dfrac{ 350 }{ 3 }=116.667 \ \text{m}^3
h2=h2+(a/2)2=142+(5/2)214.2215 m  S2=a h22=5 14.2215235.5537 m2  S=S1+4 S2=25+4 35.5537=167.215 m2h_{2}=\sqrt{ h^2 + (a/2)^2 }=\sqrt{ 14^2 + (5/2)^2 } \doteq 14.2215 \ \text{m} \ \\ \ \\ S_{2}=\dfrac{ a \cdot \ h_{2} }{ 2 }=\dfrac{ 5 \cdot \ 14.2215 }{ 2 } \doteq 35.5537 \ \text{m}^2 \ \\ \ \\ S=S_{1} + 4 \cdot \ S_{2}=25 + 4 \cdot \ 35.5537=167.215 \ \text{m}^2



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