If there is a trapezium ABCD …

Ans.

Given : ABCD is a trapezium with AB || DC. AB = 50 cm and CD = 30 cm.  X and Y are Mid-point of AD and BC respectively.

Construction : Join DY and produce it to meet AB produced at P.

Proof: In \(\triangle BYP \) and \( \triangle CYD \)

\(\angle BYP = \angle CYD \)       [vertically opposite angles]

\(\angle DCY = \angle PBY\)     [Alternate opposite angles as DC || AP and BC is the transversal]

 BY = CY              [Y is the mid point of BC]

 Thus \( \triangle BYP\)  \(\cong\)  \(\triangle CYD \)    [by ASA cogence criterion]

=> DY = YP and DC = BP  [By CPCT]

Also, X is the mid-point of AD

=> \(XY \parallel AP\)

=> XY  = \({1\over 2} AP \)  [By Mid-point theorem ]

=> \(XY = {1\over 2} [AB + BP]\)

=> \(XY = {1\over 2} [AB + CD ] \)

=> \(XY = {1\over 2} [50+30] = {1\over2} \times 80 = 40cm\)

As X and Y are the mid points of AD and BC respectively.

Trapezium DCYX and ABYX are of same height, say K cm.

Now, 

\({ar(DCYX) \over ar(ABYX) } = {{{1\over2} [DC+XY]\times K} \over {{1\over2} [AB+XY]\times K}} = {[30+40] \over [50+40]} = {70\over90} = {7\over9}\)

=> ar(DCYX) = 7/9 ar(ABYX)

Hence Proved