TRIANGLE ABC IS AN EQUILATERAL TRIANGLE. D AND E ARE THE MID-POINTS OF SIDES BC AND AB RESPECTIVELY. IF BC= 4cm,FIND ar(BED).

Answer :

We form our diagram from given information , As  :

Here D and E are mid points of BC and AB , We join DE and from " converse of mid-point theorem "  we get AC | | ED 

And BC  =  4 cm , SO

AB  = BC  = CA  =  4 cm (  As given ABC is a equilateral triangle )
And
D and E are mid points of BC and AB , So

BE  =  EA  =  BD  =  DC  =  2 cm 

Now In $∆$ BAC and $∆$ BED  
$\angle$ ABC  =  $\angle$ EBD                                      ( Same angles )

$\angle$ BAC  =  $\angle$ BED                                      (  As we know AC  | | ED and take AB as transversal line , So these angles are Corresponding angles )

And

$\angle$ BCA  =  $\angle$ BDE                                      (  As we know AC  | | ED and take CB as transversal line , So these angles are Corresponding angles )

Hence $∆$  BAC $~$ $∆$  BED                     ( By AAA rule )

So we know

$$\begin{gathered} \frac{BA}{BE} = \frac{AC}{ED} \\ \Rightarrow \frac{4}{2} = \frac{4}{ED} \\ \Rightarrow ED = 2 \end{gathered}$$

And

$$\begin{gathered} \frac{Area of BAC}{Area of BED} = \frac{A{C}^2}{E{D}^2} \\ \Rightarrow \frac{Area of BAC}{Area of BED} = \frac{4^2}{2^2} \\ \Rightarrow \frac{Area of BAC}{Area of BED} = \frac{16}{4} \end{gathered}$$
So,
Area of $∆$ BED   =  4 cm²                                  ( Ans )