Question

Question 5
ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. if x and y are, respectivelythe mid - points of AD and BC, prove that $ar\left(DCYX\right)=\frac{7}{9}ar\left(XYBA\right)$

Answer 1

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

Proof:
In $\Delta DCY$ and $\Delta PBY$
CY = BY [since, Y is the mid - point of BC]
$\angle DCY=\angle PBY$ [alternate interior angles]
And $\angle 2=\angle 3$ [vertically opposite angles]
$\therefore \Delta DCY\cong \Delta PBY$ [by ASA congruence rule]
Then, DC = BP [by CPCT]
DC = 30 cm [given]
AP = AB + BP
= 50 + 30 = 80 cm

In $\Delta ADP$, by mid - point theorem,
$XY=\frac{1}{2}AP=80=40cm$

Let the distance between AB, XY and XY, DC be h cm.
Now, area of trapezium DCYX
$=\frac{1}{2}h(30+40)$
[$\because$ area of trapezium = $\frac{1}{2}$ sum of parallel sides × distance between them]
=$\frac{1}{2}h\left(70\right)=35hc{m}^2$

Similarly,
Area of trapezium ABYX
$=\frac{1}{2}h(40+50)$
$=45hc{m}^2$
$\therefore \frac{ar\left(DCYX\right)}{ar\left(XYBA\right)}=\frac{35h}{45h}=\frac{7}{9}$

$\Rightarrow ar\left(DCYX\right)=\frac{7}{9}ar\left(XYBA\right)$