In the given figure, PQRS is a square of side 2 cm and PLMN is a rectangle.…

Let the square have coordinates P = (0, 2), Q = (0, 0), R = (2, 0), and S = (2, 2). Since L lies on QR, write L = (x, 0), where 0 <= x <= 2.

One side of rectangle PLMN is vector PL = (x, -2). The adjacent side PN must be perpendicular to PL, so take PN = k(2, x) for some scale factor k.

The side MN is parallel to PL and passes through S. Hence S can be written as P + PN + t PL for some t between 0 and 1.

Using coordinates:
(0, 2) + k(2, x) + t(x, -2) = (2, 2).

So,
2k + tx = 2,
xk - 2t = 0.

From xk - 2t = 0, t = xk/2. Substituting in 2k + tx = 2 gives
k(2 + x^2/2) = 2,
so k = 4/(x^2 + 4).

Area of rectangle PLMN = |PL| × |PN|
= sqrt(x^2 + 4) × k sqrt(x^2 + 4)
= k(x^2 + 4)
= 4.

Therefore, the area of rectangle PLMN is 4 cm^2.