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$is a parallelogramDiagonal BC divides it into two congruent triangles∴area(△ABC)=area(△BCH)=12area(||gmABHC)But || gm$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".

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