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:
Answer & explanation
Correct answer: option 4
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