Okay, here we have this:
Considering the provided information, and that the area of a rectangle is the product of its length and width, we obtain the following:
[tex]\begin{gathered} Width=\frac{Area}{Length} \\ Width=\frac{9h^2+45h+36}{3h+12} \\ =\frac{9\left(h+1\right)\left(h+4\right)}{3h+12} \\ =\frac{9\left(h+1\right)\left(h+4\right)}{3\left(h+4\right)} \\ =3\mleft(h+1\mright) \\ =3h+3 \end{gathered}[/tex]
Now, to know if the width is greater or less than the length, we will subtract it, then we have:
[tex]\begin{gathered} (3h+12)-(3h+3) \\ =3h+12-3h-3 \\ =12-3 \\ =9 \end{gathered}[/tex]
Finally we obtain that the width is feet 9 shorter than the length, so the correct answer is the option B
