The perimeter of the rectangle below is $p$ inches and the area of the rectangle is 36 square inches. If $l$ and $w$ are integers, what is one possible value of $p$ ?

(A) 24
(B) 28
(C) 32
(D) 48

Hint:- We will use the formulae of the area and perimeter of a rectangle. Then we will find out the possible values of the length and breadth of the rectangle using the fact that they are integers.

Complete step-by-step answer:

From the given figure, the length of the rectangle is $l$ inches and the breadth of the rectangle is $w$ inches.
Also, given is area of the rectangle is 36 square inches
We know that area of a rectangle is equal to (length) $ \times $ (breadth)
Thus,
$lw = 36$
Now we also know that it has been given that both $l$ and $w$ are integers.
Using this information, we can find out the possible pairs of $\left( {l,w} \right)$
The possible values of $\left( {l,w} \right)$= (1,36), (2,18), (3,12), (4,9), (6,6).
For any rectangle, we know that its perimeter is equal to 2$ \times $ (length + breadth)
Thus, for the given rectangle, its perimeter $p = 2\left( {l + w} \right)$
For the different possible values of $\left( {l,w} \right)$, we get
$\begin{array}{l}
p = 2\left( {1 + 36} \right) = 74\\
p = 2\left( {2 + 18} \right) = 40\\
p = 2\left( {3 + 12} \right) = 30\\
p = 2\left( {4 + 9} \right) = 26\\
p = 2\left( {6 + 6} \right) = 24
\end{array}$
Out the given options only $p = 24$ matches the above possible set of values of $p$ .

Hence the answer is 24.
Note:- The perimeter P of a rectangle is given by the formula, P=2l + 2w , where l is the length and w is the width of the rectangle. The area A of a rectangle is given by the formula, A= lw , where l is the length and w is the width.