Question

In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD) = ?
(a) 24 cm²
(b) 40 cm²
(c) 55 cm²
(d) 27.5 cm²

Answer 1

(b) 40 cm²

In right angled triangle MBC, we have:
MC = $\sqrt{5^2-4^2}=\sqrt{9}=3\mathrm{cm}$
In right angled triangle ADL, we have:
DL = $\sqrt{5^2-4^2}=\sqrt{9}=3\mathrm{cm}$

Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
∴ Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
= $\frac{1}{2}$ ×​ ( 13 + 7) ​× 4
= 40 cm²