The correct answer is (d).
Let $l$ be the length of the shorter side of the rectangle and $3l$ be the length of the longer side. The area of the rectangle is $l(3l) = 3l^2$.
The perpendiculars drawn from the point inside the rectangle to the sides form four right triangles. The hypotenuses of these triangles are all equal to the length of the diagonal of the rectangle, which is $\sqrt{l^2 + (3l)^2} = 2\sqrt{10}l$.
The sum of the lengths of the perpendiculars is $40$ cm. This means that the sum of the lengths of the legs of the four right triangles is $40$ cm.
The legs of a right triangle can be any two numbers that add up to $2\sqrt{10}l$. There are infinitely many such pairs of numbers, so there are infinitely many possible rectangles that satisfy the given conditions.
The area of any of these rectangles is $3l^2$. Therefore, the area of the rectangle could be any number greater than or equal to $0$.
The options (a), (b), and (c) are all specific values for the area of the rectangle. However, as we have seen, the area of the rectangle could be any number greater than or equal to $0$, so none of these options is correct.