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".