Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O. ∴ O is the mid - point of AC as well as BD. Now, in △ADB , AO is its median ∴ ar(△ADB) = 2 ar(△AOD) [ ∵ median divides a triangle into two triangles of equal areas] So, (△ADB) = 2 × 4 = 8 cm2 Now, △ADB and ||gm ABCD lie on the same base AB and lie between same parallel AB and CD . ∴ ar(ABCD) = 2 ar(△ADB) = 2 × 8 = 16 cm2