Question

Abcd is parallelogram in which Bc is produced to E such that CE=BC AE intersect CD at F if area DFB=3 cmsquare and area of parallelogram ABCD

Answer 1

In △ADF and △ECF , we have

∠ADF = ∠ECF   [alt.int.∠s]

AD = EC [∵ AD = BC and BC = EC]

∠DFA = ∠CFE   [vert. opp. ∠s]

∴ By AAS congruence rule ,

△ADF ≅ △ECF

 ⇒ DF = CF   [c.p.c.t.]

⇒  ar(△ADF) = ar(△ECF) 

Now, DF = CF

 ⇒ BF is a median in △BDC 

⇒  ar(△BDC) = 2 ar(△DFB)  

= 2 × 3 = 6 cm2   [∵ar(△DFB) = 3 cm2] 

Thus, ar(||gm  ABCD) = 2 ar(△BDC) 

= 2 × 6 = 12 cm2  

Answer 2

Answer:

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