In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠A = 60°. Find : (i) length of AB (ii) distance between AB and DC.

(i) Draw two perpendicular lines, DP and CM, from points D and C to AB, respectively. Since AB is parallel to CD, PMCD will form a rectangle. In Δ APD, cos 60° = ⇒ ⇒ ⇒ AP = 10 Now, Δ APD and Δ BMC, DP = CM (Height of same quadrilateral) AD = CB (Both are 20 cm) ∠ DPA = ∠ CMB So, Δ APD ≅ Δ BMC (∵ RHS congruency) And, by using corresponding sides of congruent triangle, AP = MB = 10 cm PM = DC = 20 cm AB = AP \+ PM \+ MB = 10 \+ 20 \+ 10 cm = 40 cm Hence, AB = 40 cm. (ii) In Δ APD, sin 60° = ⇒ ⇒ ⇒ PD = 10 = 17.32 cm Hence, PD = 17.32 cm.