Question

In the given figure, O is a point on side CD of a parallelogram ABCD such that the ratio of areas of $\Delta$OAD and $\Delta$ OBC is 1:2. If area of parallelogram is $60c{m}^2$, then the area of $\Delta$OBC will be

• A. $10c{m}^2$
• B. $15c{m}^2$
• C. $20c{m}^2$
• D. $30c{m}^2$

Answer 1

The correct option is C $20c{m}^2$

$ar\left(\Delta AOB\right)=ar\left(\Delta AOD\right)+ar\left(\Delta BOC\right)$
$=\frac{1}{2}ar\left(\text{parallelogram}ABCD\right)$
$=\frac{1}{2}\times 60c{m}^2$
$=30c{m}^2$
Let $ar\left(\Delta AOD\right)=x$
and $ar\left(\Delta BOC\right)=2x$
$\Rightarrow x+2x=30c{m}^2$
$\Rightarrow x=10c{m}^2$
$\therefore$ Area of $\Delta OBC=2x$
$=2\times 10c{m}^2$
$=20c{m}^2$
Hence, the correct answer is option (3)