In figure (3) given below, E and F are midpoints of sides AB and CD, respectively, of parallelogram ABCD. If the area of parallelogram ABCD is 36 cm²,

(i) state the area of ∆APD.

(ii) Name the parallelogram whose area is equal to the area of ∆APD.

Join the diagonals AC and BD as shown below:

(i) ∆APD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC.

Area of ∆APD = $\dfrac{1}{2}$ Area of ||gm ABCD

= $\dfrac{1}{2}$ × 36

= 18 cm².

Hence, area of ∆APD = 18 cm².

(ii) Let diagonals AC and BD meet at point O.

In ∆ABC,

Since, O is mid-point of AC (as diagonals bisect each other) and E is mid-point of AB so by mid-point theorem,

EO || BC

∴ EF || BC.

Since, BC || AD so,

⇒ EF || AD.

AB || DC (ABCD is a parallelogram)

⇒ AE || DF

Since, EF || AD and AE || DF.

∴ AEFD is a parallelogram.

EF bisects the parallelogram ABCD in two equal halves as E and F are mid-points of AB and CD and EF || BC || AD.

∴ Area of || gm AEFD = $\dfrac{1}{2}$ Area of || gm ABCD = $\dfrac{1}{2}$ × 36 = 18 cm².

∴ Area of ∆APD = Area of || gm AEFD.

Hence, AEFD is the required parallelogram which has area equal to the area of ∆APD.