ABCD is a parallelogram in which BC is produced to E such that CE = BC (figure). AE intersects … 3 cm2 , find the area of the…
Answers:
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
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