If ABCD be a rectangle and P,Q,R,S be the mid points of AB, BC, CD, and DA respectively,, then the area of the quadrilateral PQRS is equal to:

Using the theorem that a line joining the mid points of 2 sides of the triangle is parallel to the third side and equal to half of its length.

=> $$PQ = \frac{1}{2}AC$$ and PQ | | AC

Similarly, $$RS = \frac{1}{2}AC$$ and RS | | AC

=> PQRS is a parallelogram.

Also, $$\triangle$$PQB $$\sim \triangle$$ABC

=> $$\frac{ar (\triangle PQB)}{ar (\triangle ABC)} = \frac{PQ^2}{BC^2} = \frac{1}{4}$$

=> $$ar(\triangle PQB) = \frac{1}{4} ar(\triangle ABC)$$

SImilarly, $$ar (\triangle SDR) = \frac{1}{4} ar (\triangle ADC)$$

$$ar (\triangle CRQ) = \frac{1}{4} ar (\triangle CDB)$$

$$ar (\triangle ASP) = \frac{1}{4} ar (\triangle ADB)$$

=> $$ar(PQRS) = ar(ABCD) - ar (\triangle PQB) - ar(\triangle SDR) - ar (\triangle CRQ) - ar(\triangle ASP)$$

=> $$ar(PQRS) = ar(ABCD) - \frac{1}{4} * [ar(\triangle ABC) + ar(\triangle ADC) + ar(\triangle CDB) + ar(\triangle ADB)]$$

=> $$ar(PQRS) = ar(ABCD) - \frac{1}{4} * [2 * ar(ABCD)]$$

=> $$ar(PQRS) = \frac{1}{2} ar(ABCD)$$