The given figure shows a rectangle PQRS in which A, B, and C are the mid-points of sides PQ, QR, and diagonal PR respectively. Also, AC ⊥ BC. If the area of ΔABC is 25 cm², then what is the area of rectangle PQRS?

In the figure, A and C are the mid-points of PQ and PR respectively.

Mid-point theorem states, "The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it."

Therefore,

$\mathrm{AC}=\frac{1}{2}\mathrm{QR} \mathrm{and} \mathrm{CB}=\frac{1}{2}\mathrm{PQ}$

The area of ΔABC is 25 cm².

$\therefore \frac{1}{2}\times \mathrm{AC}\times \mathrm{CB}=25 {\mathrm{cm}}^2$

Hence, AC × CB = 50 cm²

$\Rightarrow \left(\frac{1}{2}\mathrm{QR}\right)\left(\frac{1}{2}\mathrm{PQ}\right)=50 {\mathrm{cm}}^2$

PQ × QR = 200 cm²

Thus, the area of the rectangle PQRS is 200 cm².