Tetrahedral pyramid

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

Correct answer:

S =  828.4

Step-by-step explanation:

$$\begin{gathered} a=16 \\ v=16 \\ {w}_1=\sqrt{v^2+({\displaystyle \frac{a}{2}}{)}^2}=\sqrt{16^2+({\displaystyle \frac{16}{2}}{)}^2}=8\sqrt{5}\doteq 17.8885 \\ {S}_1=a^2=16^2=256 \\ {S}_2={\displaystyle \frac{a\cdot w_1}{2}}={\displaystyle \frac{16\cdot 17.8885}{2}}=64\sqrt{5}\doteq 143.1084 \\ S=S_1+4\cdot S_2=256+4\cdot 143.1084=828.4 \end{gathered}$$

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