If in the following figure (not to scale), ∠DAB + ∠CBA = 90°, BC = AD, AB = 20 cm, CD = 10 cm then the area of the quadrilateral ABCD is:

75 cm²

Now trapezium ABCD

AD = BC 

CD = PQ = 10

AP = QB 

AP + PQ + QB = AB

= AP + 10 + AP = 20

= 2AP = 20 - 10

= AP = 10/2 = 5 cm

∠DAB + ∠CBA = 90° (Given)

Let ∠DAB = ∠CBA = 45°

Now in right angled triangle APD

∠DAP + ∠APD + ∠PDA = 180°

= 45° + 90° + ∠PDA = 180°

= ∠PDA = 180° - 135°

= ∠PDA = 45°

If ∠PDA = ∠DAP = 45, then

AP = PD = 5 cm

Area of trapezium = \(\frac{1}{2}\)× 30 × 5 = 75 cm²