Question

In parallelogram ABCD, E is mid-point of side AB and CE meets the diagonal BD at point O.
If area of $\Delta EOB=80c{m}^2$. Calculate the area of parallelogram $ABCD$.

Answer 1

From the figure,

$△EOB\sim △COD$

The corresponding sides are proportional.

$\Rightarrow \frac{EB}{CB}=\frac{OE}{OC}=\frac{1}{2}$

$\Rightarrow \frac{ar\left(EOB\right)}{ar\left(COD\right)}={\left(\frac{EO}{CO}\right)}^2$

$\therefore ar\left(COD\right)=80\left(4\right)=240c{m}^2$

The area of triangle $BOC$ be $y.$

The area of two triangle between the two parallel lines are equal. So,

$\therefore ar\left(EDC\right)=ar\left(BDC\right)$

$\Rightarrow ar\left(EOD\right)+240=240+y$

$\therefore ar\left(EOD\right)=y$

Similarly, $ar\left(△ADE\right)=ar\left(△BCE\right)$

$=80+y$

$\Rightarrow 2ar\left(BCD\right)=ar\left(ABCD\right)$

$\Rightarrow 2(240+y)=(240+4)+(80+y)+(80+y)$

$\Rightarrow y=80$

$\therefore ar\left(ABC\right)=ar\left(BCD\right)$

$\Rightarrow 240+80=320c{m}^2$

$\therefore ar\left(ABCD\right)=3ar\left(ABC\right)=960c{m}^2$

Hence, the answer is $960c{m}^2.$