Quadrilateral pyramid

A regular quadrilateral pyramid has a volume of 24 dm³ and a base edge a = 4 dm.
Calculate:
a/height of the pyramid
b/sidewall height
c/surface of the pyramid

Correct result:

h =  4.5 dm
h₂ =  4.9244 dm
S =  55.3954 dm²

Solution:

$$\begin{gathered} V=24{\text{dm}}^3 \\ a=4\text{ dm } \\ {S}_1=a^2=4^2=16{\text{dm}}^2 \\ V={\displaystyle \frac{1}{3}}{S}_{1}h \\ h={\displaystyle \frac{3\cdot V}{S_1}}={\displaystyle \frac{3\cdot 24}{16}}={\displaystyle \frac{9}{2}}=4\frac{1}{2}={\displaystyle \frac{9}{2}}\text{ dm}=4.5\text{ dm} \end{gathered}$$

$h_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{4.5^2+(4\mathrm{/}2{)}^2}=4.9244\text{ dm}$

$$\begin{gathered} S_2={\displaystyle \frac{a\cdot h_2}{2}}={\displaystyle \frac{4\cdot 4.9244}{2}}=\sqrt{97}{\text{dm}}^2\doteq 9.8489{\text{dm}}^2 \\ S=S_1+4\cdot S_2=16+4\cdot 9.8489=55.3954{\text{dm}}^2 \end{gathered}$$

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