in the adjoining figure, two parallelograms ABCD and AEFB are drawn on opposite sides of AB.
Prove that: ar (Ilgm ABCD) + ar (Ilgm AEFB) = ar (11gm EFCD).
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Area of DCFE = Area of ABCD + Area of AEFB
Step-by-step explanation:
Compare Δ ADE & ΔBCF
DE = CE (opposite sides of parallelogram EFCD)
AD = BC (opposite sides of parallelogram ABCD)
AE = BF (opposite sides of parallelogram AFEB)
=> Δ ADE ≅ ΔBCF
=> Area of Δ ADE = Area of ΔBCF
Area of DCFE = Area of ABCD + Area of AEFB - Area of Δ ADE + Area of ΔBCF
=> Area of DCFE = Area of ABCD + Area of AEFB
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