Quadrilateral pyramid

The volume of a regular quadrilateral pyramid is 72 cm³. Its height is equal to the length of the base edge. Calculate the length of the base and the surface of the pyramid.

Final Answer:

a =  6 cm
S =  116.4984 cm²

Step-by-step explanation:

$$\begin{gathered} V=72{\text{cm}}^3 \\ h=a \\ V=\frac{1}{3}\cdot S_1\cdot h \\ V=\frac{1}{3}\cdot a^2\cdot a \\ V=\frac{1}{3}\cdot a^3 \\ {a}^3=3V \\ a=\sqrt[3]{3\cdot V}=\sqrt[3]{3\cdot 72}=6\text{cm} \end{gathered}$$

$$\begin{gathered} S_1=a^2=6^2=36{\text{cm}}^2 \\ h=a=6\text{cm} \\ {h}_2^2=h^2+(a/2{)}^2 \\ {h}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{6^2+(6/2{)}^2}=3\sqrt{5}\text{cm}\doteq 6.7082\text{cm} \\ {S}_2=\frac{a\cdot h_2}{2}=\frac{6\cdot 6.7082}{2}=9\sqrt{5}{\text{cm}}^2\doteq 20.1246{\text{cm}}^2 \\ S=S_1+4\cdot S_2=36+4\cdot 20.1246=116.4984{\text{cm}}^2 \end{gathered}$$

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