In trapezium ABCD, AB ‖ DC and L is the midpoint of BC. Through L, a line PQ ‖ AD has been drawn which meets AB in Point P and DC produced in Q. Prove that ar(trap. ABCD) = ar( ‖ gm APQD).

Question

Philoid · Accepted Answer

Given: AB‖DC and L is the midpoint of BC, PQ ‖ AD

Construction: Drop a perpendicular DM from D onto AP

Consider ΔPBL and ΔCQL

Here,

∠LPB = ∠LQC (Alternate interior angles, AB|| DQ)

BL = LC (L is midpoint of BC)

∠PLB = ∠QLC (vertically opposite angles)

∴ By AAS congruency

ΔPBL ΔCQL

∴ PB = CQ (C.P.C.T)

Area (||gm APQD) = base× height = AP × DM –1

Area (Trap.ABCD) = 1/2 × (sum of parallel sides) × height = 1/2 × (AB + DC) × DM

Area (Trap.ABCD) = 1/2 × (AB + DC) × DM = 1/2 × (AP + PB + DC) × DM (∵ AB = AP + PB)

Area (Trap.ABCD) = 1/2 × (AP + CQ + DC) × DM (∵ PB = CQ)

Area (Trap.ABCD) = 1/2 × (AP + DQ) × DM (∵ DC + CQ = DQ)

Area (Trap.ABCD) = 1/2 × (2× AP) × DM (∵ AP = DQ)

Area (Trap.ABCD) = AP × DM –2

From –1 and –2

Area (Trap.ABCD) = Area (||gm APQD)

In trapezium ABCD, AB‖DC and L is the midpoint of BC. Through L, a line PQ ‖ AD has been drawn which meets AB in Point P and DC produced in Q.Prove that ar(trap. ABCD) = ar(‖gm APQD).

In trapezium ABCD, AB‖DC and L is the midpoint of BC. Through L, a line PQ ‖ AD has been drawn which meets AB in Point P and DC produced in Q.
Prove that ar(trap. ABCD) = ar(‖gm APQD).