Quadrilateral 44561

The regular quadrilateral pyramid has a volume of 212 m³ and a base edge a = 7.2 m. Calculate the surface area and height of the pyramid.

Correct answer:

h =  12.2685 m
S =  235.9554 m²

Step-by-step explanation:

$$\begin{gathered} V=212{\text{m}}^3 \\ a=7.2\text{m} \\ {S}_1=a^2=7.2^2=\frac{1296}{25}=51.84{\text{m}}^2 \\ V=\frac{1}{3}{S}_{1}h \\ h=\frac{3\cdot V}{S_1}=\frac{3\cdot 212}{\frac{1296}{25}}=\frac{1325}{108}\text{m}=12.2685\text{m} \end{gathered}$$

$$\begin{gathered} h_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{12.2685^2+(7.2/2{)}^2}\doteq 12.7858{\text{m}}^2 \\ {S}_2=\frac{a\cdot h_2}{2}=\frac{7.2\cdot 12.7858}{2}\doteq 46.0289{\text{m}}^2 \\ S=S_1+4\cdot S_2=51.84+4\cdot 46.0289=235.9554{\text{m}}^2 \end{gathered}$$

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