The length approximately equals width. The length is three times the height. The volume of the building is about $0.009 km^3$.
This question is supposed to be solved using a polynomial. I got the answer, but I did not use a polynomial equation. I know I'm supposed to convert $km^3$ to $m^3$, which is $9,000,000 m^3$, but I'm not sure what to do after that. Please help.
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The volume, $V$, is the product of the length, $\ell$, width, $w$, and height, $h$.
$$ V = \ell \cdot w \cdot h \text{.} $$
We are given $\ell \approx w$ and $\ell = 3h$, so we can approximately replace $w$ by $\ell$ and replace $h$ by $\ell/3$.
$$ V = \ell \cdot \ell \cdot \frac{\ell}{3} $$
From this, \begin{align*} \ell^3 &= 3V \\ &= 3 \cdot (0.009 \,\mathrm{km}^3) \\ &= 3 \cdot (9 \cdot 0.001 \,\mathrm{km}^3) \\ &= 3 \cdot (3^2 \cdot 10^{-3} \,\mathrm{km}^3) \\ &= 3^3 \cdot 10^{-3} \,\mathrm{km}^3 \\ &= (3 \cdot 10^{-1} \,\mathrm{km})^3 \text{.} \end{align*} Having written the right-hand side this way, it is easy to read out the (approximate) solution: $\ell = 0.3 \,\mathrm{km}$, which is the same as $300 \,\mathrm{m}$. Then the width is approximately $300 \,\mathrm{m}$ and the height is approximately $100 \,\mathrm{m}$.
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To make it a shot as possible, express everything in function of the height $h$: the volume is $V=lwh$ and $w\approx l=3h$, so $V\approx 9h^3$. Now the volume is given as $$9\cdot 10^{-3}\,\mathrm{km}^3=9\cdot10^{-3}\cdot 10^9 =9\cdot10^6\,\mathrm m^3,$$ so we obtain the equation $$9h^3=9\cdot 10^6\iff h^3=10^6=100^3\iff h=100\,\mathrm m.$$