Question

In the given figure, ABCD is a parallelogram with x : y = 1 : 1 and OE || AD. Then, (area of $\Delta ADC$ + area of $\Delta BOC$) : area of $\Delta ODC$ is

• A. 4 : 1
• B. 9 : 1
• C. 16 : 1
• D. None of the above

Answer 1

The correct option is D

None of the above

Given, ABCD is a parallelogram with x : y = 1 : 1
i.e. x = y
Also, OE || AD
$\Rightarrow$ OE || BC
Now, in $\Delta CAD$ and $\Delta COE$,
OE || AD
$\therefore \angle COE=\angle CAD$
and $\angle CEO=\angle CDA$ [corresponding angles]
$\Rightarrow \Delta CAD\sim \Delta COE$ [by AA similarity]
$\Rightarrow \frac{CE}{CD}=\frac{CO}{CA}$
$\Rightarrow \frac{y}{x+y}=\frac{CO}{CA}$ ...(i)
Similarly, $\Delta DOE\sim \Delta DBC$
$\Rightarrow \frac{x}{x+y}=\frac{DO}{DB}$ ...(ii)
From Eq. (i),
$\frac{Areaof\Delta COE}{Areaof\Delta CAD}=\frac{(y{)}^2}{(x+y{)}^2}\Rightarrow \frac{Areaof\Delta CAD}{Areaof\Delta COE}=\frac{(x+y{)}^2}{(y{)}^2}\Rightarrow \frac{Areaof\Delta CAD}{Areaof\Delta COE}-1=\frac{(x+y{)}^2}{(y{)}^2}-1\Rightarrow \frac{AreaoftrapeziumOEDA}{Areaof\Delta COE}=\frac{x^2+2xy}{y^2}\Rightarrow \frac{AreaoftrapeziumOEDA}{Areaof\Delta COE}=\frac{x^2+2x^2}{x^2}[\because x=y]=\frac{3x^2}{x^2}=\frac{3}{1}\Rightarrow AreaoftrapeziumOEDA=3Areaof\Delta COE...\left(iii\right)$
From Eq. (ii),
$AreaoftrapeziumOECB=3Areaof\Delta OED...\left(iv\right)$
On adding Eqs. (iii) and (iv), we get
$AreaoftrapeziumOEDA+AreaoftrapeziumOECB=3(Areaof\Delta COE+Areaof\Delta OED)\Rightarrow Areaof\Delta ADC+Areaof\Delta BOC=3Areaof\Delta ODC$
$\Rightarrow$ $\frac{Areaof\Delta ADC+Areaof\Delta BOC}{Areaof\Delta ODC}$ = $\frac{3}{1}$