Question

In the above figure, $D$ and $E$ are the mid-points of the sides $AC$ and $BC$ respectively of $△ABC$. If $ar\left(△BED\right)=12c{m}^2$, then $ar\left(ABED\right)=$

• A. $36c{m}^2$
• B. $48c{m}^2$
• C. $24c{m}^2$
• D. None of these

Answer 1

The correct option is A $36c{m}^2$

Given-

D &E are the mid points of the sides AC & BC respectively of $\Delta ABC.$

$ar.\Delta BED=12{cm}^2$

F, the mid point of the side AB, is joined to D.

D is the mid point of AC and E is the mid point of BC.

So, by mid point theorem,

$DE\parallel AB.$.

Again, F is the mid point of AB.

So $DF\parallel BC.$

i.e BEDF is a parallelogram and BD is its diagonal.

So BD divides BEDF into two equal areas.

$\therefore ar.\Delta BED=ar.\Delta BDF=12{cm}^2$

Again, in $\Delta ADB$

DF is the median since F is the mid point of AB.

$\therefore ar.\Delta BDF=ar.\Delta AFD=12{cm}^2$\\

(median divides a triangle into two equal areas).

Now $ar.ABED=ar.\Delta AFD+ar.\Delta BDF+ar.\Delta BED=12{cm}^2+12{cm}^2+12{cm}^2=36{cm}^2.$

Ans- Option A.