Find the

Find the surface area of a regular quadrilateral pyramid which has a volume of 24 dm³ and a height of 45 cm.

Correct answer:

S =  5539.5431 cm²

Step-by-step explanation:

$$\begin{gathered} V=24{\text{dm}}^3\to {\text{cm}}^3=24\cdot 1000{\text{cm}}^3=24000{\text{cm}}^3 \\ h=45\text{ cm } \\ V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h \\ {S}_1=3\cdot V\mathrm{/}h=3\cdot 24000\mathrm{/}45=1600{\text{cm}}^2 \\ {S}_1=a^2 \\ a=\sqrt{S_1}=\sqrt{1600}=40\text{ cm } \\ (a\mathrm{/}2{)}^2+h^2=h_2^2 \\ {h}_2=\sqrt{(a\mathrm{/}2{)}^2+h^2}=\sqrt{(40\mathrm{/}2{)}^2+45^2}=5\sqrt{97}\text{ cm}\doteq 49.2443\text{ cm } \\ {S}_2={\displaystyle \frac{a\cdot h_2}{2}}={\displaystyle \frac{40\cdot 49.2443}{2}}=100\sqrt{97}{\text{cm}}^2\doteq 984.8858{\text{cm}}^2 \\ S=4\cdot S_2+S_1=4\cdot 984.8858+1600=5539.5431{\text{cm}}^2 \end{gathered}$$

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