The given figure shows a parallelogram ABCD in which X and Y are the mid-points of sides AB and AD respectively.

If the area of ΔCXY = 24 cm², then what is the area of parallelogram ABCD?

Join BD and AC such that they intersect at point O.

O is the mid-point of BD.

X, Y, and O are the mid-points of sides AB, AD, and BD respectively.

We know that Area (ΔABD) = $\frac{1}{2}$area (ABCD)

In ΔABC, X is the mid-point of AB i.e., CX is the median of ΔABC with respect to side AB.

∴ Area (ΔBCX) = $\frac{1}{2}$area (ΔABC)

⇒ Area (ΔBCX) = $\frac{1}{2}$ x $\frac{1}{2}$area (ABCD)

⇒ Area (ΔBCX) = $\frac{1}{4}$area (ABCD)        ...........(1)

Similarly, It can be proved that, Area (ΔABD) = 4area (ΔAXY)

∴ $\frac{1}{2}$area (ABCD) = 4area (ΔAXY)

⇒ Area (ΔAXY) = $\frac{1}{8}$area (ABCD)       .............(2)

Similarly, it can be shown that, Area (ΔCDY) = $\frac{1}{4}$area (ABCD)    ............(3)

Therefore, from equation (1), (2) and (3), we get:

Area (ABCD) = Area (ΔCXY) + Area (ΔAXY) + Area (ΔBCX) + Area (ΔCDY)

= 24 cm² + $\frac{1}{8}$area (ABCD) + $\frac{1}{4}$area (ABCD) + $\frac{1}{4}$area (ABCD)

= 24 cm² + $\frac{5}{8}$area (ABCD)

∴ Area (ABCD) −$\frac{5}{8}$area (ABCD) = 24 cm²

⇒ $\left(1-\frac{5}{8}\right)$area (ABCD) = 24 cm²

⇒ $\frac{3}{8}$area (ABCD) = 24 cm²

⇒ Area (ABCD) = $\frac{8}{3}\times 24=64 {\mathrm{cm}}^2$ 

Thus, the area of quadrilateral ABCD is 64 cm².