The area of triangle ABC is 27cm² and AB= 6cm
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ar △ABC=27cm
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ABCF is a ∥gm in which EF∥BC. Also, AF∥BC
Now, in △ABC and △ACF
(i) ∠BAC=∠ACF [Alternate angles]
(ii) AC = AC [common]
(iii) AF = BC [∵AF∥BC]
Therefore, △ABC≅△ACF [By SAS congreuncy]
Hence, ar △ABC=△ACF [By CPCT]
∴ar△ABC=ar△ACF=27cm
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And ar ∥gmABCF=ar△ABC+ar△ACF=27+27=54cm
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