Triangular pyramid

It is given perpendicular regular triangular pyramid: base side a = 5 cm, height v = 8 cm, volume V = 28.8 cm³. What is it content (surface area)?

Correct answer:

S =  71.79 cm²

Step-by-step explanation:

$$\begin{gathered} a=5\text{ cm } \\ h=8\text{ cm } \\ V=28.8{\text{cm}}^3 \\ {h}_1=\sqrt{a^2-(a\mathrm{/}2{)}^2}=\sqrt{5^2-(5\mathrm{/}2{)}^2}\doteq 4.3301\text{ cm } \\ {S}_1={\displaystyle \frac{a\cdot h_1}{2}}={\displaystyle \frac{5\cdot 4.3301}{2}}\doteq 10.8253{\text{cm}}^2 \\ {h}_2^2=h^2+(h_1\mathrm{/}3{)}^2 \\ {h}_2=\sqrt{h^2+(h_1\mathrm{/}3{)}^2}=\sqrt{8^2+(4.3301\mathrm{/}3{)}^2}\doteq 8.1292\text{ cm } \\ {S}_2={\displaystyle \frac{a\cdot h_2}{2}}={\displaystyle \frac{5\cdot 8.1292}{2}}\doteq 20.3229{\text{cm}}^2 \\ S=S_1+3\cdot S_2=10.8253+3\cdot 20.3229\doteq 71.7941{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ {V}_2={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 10.8253\cdot 8\doteq 28.8675{\text{cm}}^3 \\ {V}_2=V \end{gathered}$$

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