ABCD is 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, prove that
`ar (DCYX) = (7)/(9) ar (XYBA)`.

Explanation

Given In a trapezium ABCD, `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) = (7)/(9)ar (XYBA)`
Construction Join DY and extend it to meet produced AB at P.
Proof In `DeltaDCY` and `DeltaPBY`,
`" " CY = BY" "` [since, Y is the mid-point of BC]
`" " angleDCY = anglePBY" "` [alternate interior angles]
and `" " angle2 = angle3" "` [vertically opposite angles]
`therefore" "` `DeltaDCY ~= PBY" "` [by ASA congruence rule]
Then, `" " DC = BP" "` [by CPCT]
But `" " DC = 30 cm" "` [given]
`therefore" "` `DC = BP = 30 cm`
Now, `" " AP = AB + BP`
`=50 + 30 = 80 cm`
In `DeltaADP`, by mid-point theorem,
`XY = (1)/(2)AP = (1)/(2) xx 80 = 40 cm`
Let distance between AB, XY and XY, DC is h cm.
Now, area of trapezium `DCYX = (1)/(2)h(30 + 40)`
`[because "area of trapezium " = (1)/(2) "sum of parallel sides" xx "distance between them"]`
`= (1)/(2) h (70) = 35 h cm^(2)`
Similarly, `" "` `"area of trapezium" = (1)/(2) h (40 + 50) = (1)/(2)h xx 90 = 45h cm^(2)`
`therefore" "` `(ar (DCYX))/(ar(XYBA)) = (35h)/(45h) = (7)/(9)`
`rArr" "` `ar (DCYX) = (7)/(9) ar (XYBA)" "` Hence proved.