Quadrilateral 46431

Calculate the volume V and the surface S of a regular quadrilateral pyramid, the base edge and height of which are the same size as the edge of a cube with a volume V1 = 27m³

Correct answer:

V =  9 m³
S =  29.1246 m²

Step-by-step explanation:

$$\begin{gathered} V_1=27{\text{m}}^3 \\ {V}_1=a^3 \\ a=\sqrt[ \\ 3]{V_1}=\sqrt[ \\ 3]{27}=3\text{m} \\ h=a=3\text{m} \\ {S}_1=a^2=3^2=9{\text{m}}^2 \\ V=S_1\cdot h/3=9\cdot 3/3=9{\text{m}}^3 \end{gathered}$$

$$\begin{gathered} h_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{3^2+(3/2{)}^2}\doteq 3.3541\text{m} \\ {S}_2=a\cdot h_2/2=3\cdot 3.3541/2\doteq 5.0312{\text{m}}^2 \\ S=S_1+4\cdot S_2=9+4\cdot 5.0312=29.1246{\text{m}}^2 \end{gathered}$$

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