Question

ab || CD in trapezium ABCD if < A = Y + 60° < b = X + 60 ° < C = 3 x - 40° and < d = 3 Y - 80° then find all the angles of ABCD

Answer 1

Join DY and exted it to meet AB produced at P. ∠BYP = ∠CYD [Vertically opposite angles]. ∠DCY = ∠PBY [Since DC || AP]. and BY = CY (Since Y is the mid-point of BC). So, according to ASA congruence criterion, we get ΔBYP ≅ ΔCYD ⇒ DY = YP and DC = BP Also, X is the mid-point of AD ∴ XY || AP and XY = ½AP ⇒ XY = ½ (AB + BP) ⇒ XY = ½ (AB + DC) ⇒ XY = ½ (50 + 30) = ½ × 80 = 40 cm. Also, We have XY || AP ⇒ XY || AB and AB || DC ⇒ XY || DC ⇒ DCYX is a trapezium. Since X and Y are the mid-points of AD and BC respectively. Therefore trapezium DCYX and ABYX are of same height and assuming it as h cm. Ar(Trap. DCYX) = ½ (DC + XY) × h = ½ (30 + 40) h = 35h cm2

Ar(Trap. ABYX) = ½ (AB + XY) × h = ½ (50 + 40)h = 45h cm2

So, Ar(trap. DCYX) / Ar(trap. ABYX) = 35h / 45h = 7/ 9

⇒ Ar(trap. DCYX) = 7/9 Ar(trap. ABXY)