# Tetrahedral pyramid

What is the surface of a regular tetrahedral (four-sided) pyramid if the base edge a=16 and height v=16?

Result

S =  828.4

#### Solution:

$a=16 \ \\ v=16 \ \\ \ \\ w_{1}=\sqrt{ v^2 + (\dfrac{ a }{ 2 } )^2 }=\sqrt{ 16^2 + (\dfrac{ 16 }{ 2 } )^2 } \doteq 8 \ \sqrt{ 5 } \doteq 17.8885 \ \\ \ \\ S_{1}=a^2=16^2=256 \ \\ \ \\ S_{2}=\dfrac{ a \cdot \ w_{1} }{ 2 }=\dfrac{ 16 \cdot \ 17.8885 }{ 2 } \doteq 64 \ \sqrt{ 5 } \doteq 143.1084 \ \\ \ \\ S=S_{1} + 4 \cdot \ S_{2}=256 + 4 \cdot \ 143.1084 \doteq 828.4334 \doteq 828.4$

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