A point is selected at random inside a rectangle and perpendiculars are drawn on each side of the rectangle from the point. If the sum of these perpendiculars is 32 cm, then the perimeter of the rectangle will be

The correct answer is (a).

Let $ABCD$ be the rectangle, and let $P$ be the point inside the rectangle. Let $AP$, $BP$, $CP$, and $DP$ be the perpendiculars drawn from $P$ to the sides of the rectangle.

Since $AP$ and $BP$ are perpendicular to $AB$, and $CP$ and $DP$ are perpendicular to $CD$, the quadrilateral $APBC$ is a trapezoid. The sum of the lengths of the bases of a trapezoid is equal to the sum of the lengths of the diagonals. Therefore, $AB + CD = AP + CP = BP + DP = 32$.

Since the perimeter of a rectangle is equal to twice the sum of its sides, the perimeter of $ABCD$ is $2(AB + CD) = 2(32) = \boxed{48}$ cm.

Option (b) is incorrect because $64$ is not a multiple of $32$. Option (c) is incorrect because $96$ is greater than the sum of the lengths of the sides of a rectangle. Option (d) is incorrect because the perimeter of a rectangle is greater than the sum of the perpendiculars drawn from a point inside the rectangle to the sides of the rectangle.