Question

ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are respectively the mid-points of AD and BC, then prove that
ar (DCYX) = 7/9 ar (XYBA).

Answer 1

Given: ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm.

Also, X and Y are respectively the mid-points of AD and BC.

To prove: ar (DCYX) = \(\frac { 7 }{ 9 }\) ar (XYBA).

Proof: Now, in ∆DCY and ∆PBY

CY = BY (given)

∠C = ∠B (alternate interior angles)

and ∠3 = ∠2 (vertically interior angles)

∆DCY = ∆PBY (by ASA congruence rule)

Then DC = BP (by c.p.c.t)

⇒ DC = PB = 30 cm

AP = BA + BP = 50 + 30 = 80 cm

Now, in ∆ADP, using mid-point theorem

XY = \(\frac { 1 }{ 2 }\) AP = \(\frac { 1 }{ 2 }\) x 80 = 40 cm

Let the distance between AB, XY and DC be h cm.

Now, in trapezium DCYX,

ar (trapezium DCYX)

Hence proved.