Tetrahedral pyramid

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

Correct result:

S =  828.4

Solution:

a=16 v=16  w1=v2+(a2)2=162+(162)2=8 5≐17.8885  S1=a2=162=256  S2=a⋅ w12=16⋅ 17.88852=64 5≐143.1084  S=S1+4⋅ S2=256+4⋅ 143.1084=828.4

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