Question

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).
C​

Answer 1

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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