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Question

Triangle ABC and parallelogram ABEF are on the same base, AC in between the same parallels AB and EF. Prove that ar(ABC)=12ar(||gmABEF)

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Solution

Through B draw BH||AC to meet FE produced at H
ABHC$ is a parallelogram
Diagonal BC divides it into two congruent triangles
area(ABC)=area(BCH)
=12area(||gmABHC)
But || gm $ABHC and || gm ABEF are on the same base AB and between same parallels AB and EF
area(||gmABHC)=area(||gmABEF)
Hence area(ABC)=12area(||gmABEF)
From the result, we say that "the area of a triangle is equal to half the area of the parallelogram on the same base and between the same parallels".
706412_569094_ans_6ff937957bf24dbeb0f04af42b18376c.png

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