Question

anyone please solve this question​

Answer 1

Given :

• ABCD is a parallelogram
• Side BC is produced to E such that CE = BC.
• AE intersects CD at F.
• A (Δ BDF) = 3 cm²

To Find :

• Area of parallelogram ABCD.

Solution :

AD = BC = EC ___(1)

We have Δ ADF and Δ ECF.

Let's try to show them congruent.

In Δ ADF and Δ ECF,

➣ $\angle{ADF}$ = $\angle{ECF}$ $\sf{\big[{Alternate\:\angle\big]}}$

➣ $\mathtt{AD\:=\:EC}$ $\sf{\big[{From(1)\big]}}$

➣ $\angle{DFA}$ = $\angle{CFE}$$\sf{\big[{V. O\:\angle\big]}}$

•°• Δ ADF $\congruent$ Δ ECF. $\sf{\big[{By\:ASA\:rule\big]}}$

So, we have :

• AD = EC
• DF = CF
• AF = EF

We have, BF as a median in △BDC.

➣ $\mathtt{A\:(\triangle\:BDC)}$ = $\mathtt{2\:A\:<strong> </strong>(\triangle\:BDF)}$

We have, A(Δ BDF) = 3 cm²

➣ $\mathtt{A\:(\triangle\:BDC)}$ = $\mathtt{2\:\:\times\:3}$

$\mathtt{A\:(\triangle\:BDC)}$ = $\mathtt{6m^2}$ ___(2)

$\large{\boxed{\sf{\red{A\:(\triangle\:BDC)\:=\:6\:cm^2}}}}$

Area of parallelogram ABCD :

We can see that the parallelogram ABCD is formed by twice of Δ BDC.

➣ $\mathtt{Area_{ABCD}\:=\:2\:\times\:A\:(\triangle\:BDC)}$

➣ $\mathtt{Area_{ABCD}\:=\:2\:\times\:6}$

➣ $\mathtt{Area_{ABCD}\:=\:12}$

$\large{\boxed{\sf{\red{Area_{ABCD}}\:=\:12\:cm^2}}}$