Question

In parallelogram $ABCD$, $E$ is mid-point of side $AB$ and $CE$ meets the diagonal $BD$ at point $O$, then find area of $△EBC$. Given, Area of $△EOB=80{cm}^2$.

• A. $240c{m}^2$
• B. $140c{m}^2$
• C. $200c{m}^2$
• D. $220c{m}^2$

Answer 1

The correct option is A $240c{m}^2$

Given, Parallelogram $ABCD$ with $E$ as a mid point of $AB,CE$ meets diagonal $BD$ at $O$.

Now, In $△EOB$ and $△DOC$

$\angle DOC=\angle EOB$ (Vertically opposite angles are equal.)
$\angle EBO=\angle ODC$ (Alternate angles, $\because AB\parallel CD$ and $BD$ intersect it)
$\angle OCD=\angle OEB$ (Alternate angles, $\because AB\parallel CD$ and $EC$ intersect it)
Using $A-A-A$ property, $△EOB\sim △DOC$
Using similarity property,
$\frac{OE}{OC}=\frac{EB}{DC}$

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

(As $AB=DC$,opposite sides of parallelogram)

$AB=2EB$

(As $E$ is mid point of AB)
$\therefore \frac{OE}{OC}=\frac{1}{2}$

So, $OE:OC=1:2$

Therefore,$OC=2OE$

$EC=OC+OE$

or $EC=2OE+OE$

or $EC=3OE$

We draw a perpendicular from $B$ to $EC$ which intersects it at $F$.

In $△EBC$ Area of triangle $=\frac{1}{2}\times base\times altitude$

$=\frac{1}{2}\times BF\times EC$

Now, In $△EOB$ Area of triangle $=\frac{1}{2}\times base\times altitude$

$=\frac{1}{2}\times BF\times OE$

Now taking ratio of area of $△EBC$ and $△EOB$,

$\frac{Areaof△EBC}{Areaof△EOB}=\frac{\frac{1}{2}\times EC\times BF}{\frac{1}{2}\times OE\times BF}=\frac{EC}{OE}=\frac{3}{1}$ ....(As $EC=3OE$)

So, an Area of $△EBC$ is $3$ times Area of $△EOB$.

Given, Area of $△EOB=80{cm}^2$

So, Area of $△EBC=3\times 80{cm}^2=240{cm}^2$