# 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.

Result

V =  116.667 m3
S =  167.215 m2

#### Solution:

$a=5 \ \text{m} \ \\ h=14 \ \text{m} \ \\ \ \\ S_{ 1 }=a^2=5^2=25 \ \text{m}^2 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{ 1 } \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 25 \cdot \ 14 \doteq \dfrac{ 350 }{ 3 } \doteq 116.6667 \doteq 116.667 \ \text{m}^3$
$h_{ 2 }=\sqrt{ h^2 + (a/2)^2 }=\sqrt{ 14^2 + (5/2)^2 } \doteq 14.2215 \ \text{m} \ \\ \ \\ S_{ 2 }=\dfrac{ a \cdot \ h_{ 2 } }{ 2 }=\dfrac{ 5 \cdot \ 14.2215 }{ 2 } \doteq 35.5537 \ \text{m}^2 \ \\ \ \\ S=S_{ 1 } + 4 \cdot \ S_{ 2 }=25 + 4 \cdot \ 35.5537 \doteq 167.2146 \doteq 167.215 \ \text{m}^2$

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