What might be the dimensions 0f a pentagonal prism if its surface area is 180cm
The surface area of a regular pentagonal prism is given by
$\displaystyle TSA = 2A_{\textrm{pentagon}} + 5lh$, where $\displaystyle l$ is one of the side lengths and $\displaystyle h$ is the height.
The area of a regular polygon is given by
$\displaystyle A = \frac{l^2n}{4\tan{\frac{\pi}{n}}}$, where $\displaystyle n$ is the number of sides, so the pentagon's area is
$\displaystyle \frac{5l^2}{4\tan{\frac{\pi}{5}}}$
$\displaystyle = \frac{5l^2}{4\sqrt{5-2\sqrt{5}}}$.
Therefore
$\displaystyle TSA = 2\left(\frac{5l^2}{4\sqrt{5-2\sqrt{5}}}\right) + 5lh$
$\displaystyle = \frac{5l^2}{2\sqrt{5 - 2\sqrt{5}}} + 5lh$.
Choose some values of $\displaystyle l$ and $\displaystyle h$ which will make this value $\displaystyle 180\,\textrm{cm}^2$.