Question

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

Answer 1

Proving that $\mathrm{ar}\left(\mathrm{DCYX}\right)=\frac{7}{9}\mathrm{ar}\left(\mathrm{XYBA}\right)$:

Given:

$\mathrm{ABCD}$is a trapezium with $\mathrm{AB}\left|\right|\mathrm{DC}$

To prove:

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

Construction:

Join $\mathrm{DY}$ and produce it to meet $\mathrm{AB}$ produced at $\mathrm{M}$.

Proof

In$∆\mathrm{BYM}$ and $∆\mathrm{CYD}$

$\angle \mathrm{BYM}=\angle \mathrm{CYD}$ (vertically opposite angles)

$\angle \mathrm{DCY}=\angle \mathrm{MBY}$ (alternate interior angles of $\mathrm{DC}\left|\right|\mathrm{AM}$ and $\mathrm{BC}$ is the transversal)

$\mathrm{BY}=\mathrm{CY}$ ($\mathrm{Y}$is the midpoint of $\mathrm{BC}$)

Thus,

$∆\mathrm{BYM}\cong ∆\mathrm{CYD}$ (by $\mathrm{ASA}$ congruence criterion)

So, $\mathrm{DY}=\mathrm{YM}$ and$\mathrm{DC}=\mathrm{BM}$ (by CPCT)

Therefore $\mathrm{Y}$is the midpoint of $\mathrm{DM}$

now, $\mathrm{X}$ is the midpoint of $\mathrm{AD}$ given

∴ $\mathrm{XY}\left|\right|\mathrm{AM}$ and $\mathrm{XY}=\frac{1}{2}\mathrm{AM}$ (by midpoint theorem)

$\begin{array}{rcl}& \Rightarrow & \mathrm{XY}=\frac{1}{2}\mathrm{AM}\\ & =& \frac{1}{2}(\mathrm{AB}+\mathrm{BM})\\ & =& \frac{1}{2}(\mathrm{AB}+\mathrm{DC})\\ & =& \frac{1}{2}(50+30)\\ & =& \frac{1}{2}\times 80\mathrm{cm}\\ & =& 40\mathrm{cm}\end{array}$

Since $\mathrm{X}$ is the midpoint of $\mathrm{AD}$

And $\mathrm{Y}$ is the midpoint of $\mathrm{BC}$

Hence, trapezium $\mathrm{DCYX}$ and $\mathrm{ABYX}$ are of same height, $\mathrm{h}\mathrm{cm}$

$\begin{array}{rcl}\frac{\mathrm{ar}\mathrm{DCYX}}{\mathrm{ar}\mathrm{ABYX}}& =& \frac{{\displaystyle \frac{1}{2}}\left(DC+XY\right)\times h}{{\displaystyle \frac{1}{2}}\left(AB+XY\right)\times h}\\ & =& \frac{\left(30+40\right)}{\left(50+40\right)}\\ & =& \frac{70}{90}\\ \frac{\mathrm{ar}\mathrm{DCYX}}{\mathrm{ar}\mathrm{ABYX}}& =& \frac{7}{9}\end{array}$

⇒ $9\mathrm{ar}\left(\mathrm{DCXY}\right)=7\mathrm{ar}\left(\mathrm{XYBA}\right)$

⇒$\mathrm{ar}\left(\mathrm{DCXY}\right)=\frac{7}{9}\mathrm{ar}\left(\mathrm{XYBA}\right)$

Hence,proved that $\mathrm{ar}\left(\mathrm{DCYX}\right)=\frac{7}{9}\mathrm{ar}\left(\mathrm{XYBA}\right)$