Question

ABCD is a trapezium in which $AB\parallel DC,AB=50$ cm, $DC=30$ cm. If $X$ and $Y$ are respectively the mid-points of AD and BC, prove that
$ar\left(\text{trap.}DCYX\right)=\frac{7}{9}ar\left(\text{trap.}XYBA\right).$

Answer 1

$ABCD$ is a trapezium.

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

In $△BYP$ and $△CYD$

$\Rightarrow$ $\angle BYD=\angle CYD$ [ Vertically opposite angles ]

$DC\parallel AP$ and $BC$ is the transversal.

$\Rightarrow$ $\angle DCY=\angle PBY$ [ Alternate angles ]

$\therefore$ $△BYP\cong △CYD$ [ By ASA congruence rule ]

$\Rightarrow$ $DY=YP$ and $DC=BP$ [ CPCT ]

$Y$ is the mid point of $AD$

$\therefore$ $XY\parallel AP$ and $XY=\frac{1}{2}\times AP$ [ Mid -point theorem ]

$\Rightarrow$ $XY=\frac{1}{2}\times (AB+BP)$

$\Rightarrow$ $XY=\frac{1}{2}\times (AB+DC)$

$\Rightarrow$ $XY=\frac{1}{2}\times (50+30)$

$\Rightarrow$ $XY=\frac{1}{2}\times 80$

$\therefore$ $XY=40cm$

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

$\therefore$ Trapezium $DCYX$ and $XYBA$ are of same height, say $hcm$

Now,

$\Rightarrow$ $\frac{ar(trap.DCYX)}{ar(trap.XYBA)}=\frac{\frac{1}{2}(DC+XY)\times h}{\frac{1}{2}(AB+XY)\times h}=\frac{30+40}{50+40}=\frac{70}{90}=\frac{7}{9}$

$\Rightarrow$ $9ar(trap.DCXY)=7ar(trap.XYBA)$

$\therefore$ $ar(trap.DCYX)=\frac{7}{9}ar(trap.XYBA)$