Question

ABCD is a trapezium in which AB || DC, DC = 30cm and AB = 50cm. If X and Y are,
respectively the mid-points of AD and BC, prove that

ar( DCYX) = 7/9 ar( XYBA).

Answer 1

hope it helps you........

Answer 2

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

To prove : ar( DCYX) = 7/9 ar( XYBA).

Construction : join DY and extend it to meet produced AB at P

Proof :

In triangle DCY and BPY

CY = BY (y is the mid point)

∠DCY = ∠PBY (alternate angles)

∠2 = ∠3 ( vertical opp angles)

by ASA congruency  rule

$\bigtriangleup DCY \cong \bigtriangleup PBY$

Thus ,

DC = BP (by cpct)

DC = BP = 30 cm

AP = AB+BP

AP = 50+30 = 80 cm

now ,

In triangle ADP  using mid point theorem

$XY = \frac{1}{2}AP= \frac{1}{2}BD = 40$

now , let DC = h cm

then

area of trapezium DCYX =

$\frac{1}{2}h(30+40)\\\\= 35 h$

similarly

area of trapezium XYBA

$\frac{1}{2}h(40+50)\\\\= 45 h$

now ,

ratio is

$\frac{ar(DCYX)}{ar(XYBA)}= \frac{35h}{45h}= \frac{7}{9}\\\\therefore \\\\ar(DCYX)= \frac{7}{9}ar(XYBA)$

hence proved