ABCD is a trapezium in which …

Ans.

Given : ABCD is a trapezium with AB || DC, DC = 30cm , AB = 50cm

To Prove:  \(ar(DCYX)={7\over 9} ar(XYBA)\)
Construction: Join DY and produce it to meet AB produced at P.

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

\(\angle BYP = \angle CYD\)      (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 criterion)

=>  DY = YP and DC = BP    [by CPCT]  ......... (2)

=> Y is the mid point of AD

XY || AP and XY = \({1\over 2}\)AP   [Mid-point theorem]     ........ (2)

=> AP = AB + BP 

=> AP = AB + DC   [From (1)]

=> AP = 50 + 30 = 80 

\(=> XY = {1\over 2} \times 80 = 40 cm\)

Since X and Y are the mid points of AD and BC respectively.
Trapezium DCYX and ABYX are of same height, say h cm
Now, 

\(=> {ar(DCYX) \over ar(XYBA)} = {{1\over 2} (DC+XY)\times h \over {1\over 2} (AB+XY)\times h} = {30+40\over 50+40} = {70\over 90} = {7\over 9}\)

\(=> {ar(DCYX) ={7\over 9} ar(XYBA)} \)

Hence Proved