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 $\Delta EBC\left(c{m}^2\right)$

Answer 1

From the figure,

$△EOB$ and $△COD$ are similar

Corresponding sides are proportional

$\frac{EB}{CD}=\frac{OE}{OC}$

$⟹\frac{OE}{OC}=\frac{1}{2}$ ----------as $P$ is the mid point of $AB$

We know that the ratio of the two similar triangles is equal to the square of the ratio of the corresponding sides.

$\frac{Areaof△EOB}{Areaof△COD}=(\frac{EO}{CO}{)}^2$

$\frac{80}{Areaof△COD}=(\frac{1}{2}{)}^2=\frac{1}{4}$

$Areaof△COD=80\times 4=240sq.cm$

Let the area of $△BOC$ be y

As we know that area of two triangles between the two parallel lines are equal. So,

$Areaof△EDC=Areaof△BDC$

$Areaof△EOD+240=240+y$

$Areaof△=y$

Similarly,

$Areaof△ADE=Areaof△BCE$

$Areaof△ADE=80+y$

Now,

$2Areaof△BCD=AreaofparallelogramABCD$

$2(240+y)=(240+y)+(80+y)+(80+y)$

$y=80$

So,

$Areaof△BOC=80sq.cm$

$Areaof△EBC=160sq.cm$