The perimeter

The perimeter of the base of a regular quadrilateral pyramid is the same as its height. The pyramid has a volume of 288 dm³. Calculate its surface area round the result to the whole dm².

Final Answer:

S =  326 dm²

Step-by-step explanation:

$$\begin{gathered} V=288{\text{dm}}^3 \\ 4a=h \\ V=\frac{1}{3}\cdot a^2\cdot h \\ V=\frac{1}{3}\cdot a^2\cdot 4\cdot a \\ a=\sqrt[3]{\frac{3}{4}\cdot V}=\sqrt[3]{\frac{3}{4}\cdot 288}=6\text{dm} \\ h=4\cdot a=4\cdot 6=24\text{dm} \\ {S}_1=a^2=6^2=36{\text{dm}}^2 \\ {h}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{24^2+(6/2{)}^2}=3\sqrt{65}\text{dm}\doteq 24.1868\text{dm} \\ {S}_2=\frac{a\cdot h_2}{2}=\frac{6\cdot 24.1868}{2}=9\sqrt{65}{\text{dm}}^2\doteq 72.5603{\text{dm}}^2 \\ S=S_1+4\cdot S_2=36+4\cdot 72.5603\doteq 326.2413{\text{dm}}^2 \\ \text{ Verifying Solution: } \\ {V}_2=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 36\cdot 24=288{\text{dm}}^3 \end{gathered}$$

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