Regular quadrilateral pyramid

Find the surface area of a regular quadrilateral pyramid if for its volume V and body height v and the base edge a applies:
V = 2.8 m ^ 3, v = 2.1 m

Correct answer:

S =  13.3038 m²

Step-by-step explanation:

$$\begin{gathered} V=2.8{\text{m}}^3 \\ v=2.1\text{m} \\ V=\frac{1}{3}\cdot S_1\cdot v \\ {S}_1=3\cdot V/v=3\cdot 2.8/2.1=4{\text{m}}^2 \\ {S}_1=a^2 \\ a=\sqrt{S_1}=\sqrt{4}=2\text{m} \\ {h}_2=\sqrt{v^2+(a/2{)}^2}=\sqrt{2.1^2+(2/2{)}^2}\doteq 2.3259\text{m} \\ {S}_2=\frac{a\cdot h_2}{2}=\frac{2\cdot 2.3259}{2}\doteq 2.3259{\text{m}}^2 \\ S=S_1+4\cdot S_2=4+4\cdot 2.3259=13.3038{\text{m}}^2 \end{gathered}$$

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