Mathematics
In the adjoining figure, two parallelograms ABCD and AEFB are drawn on opposite sides of AB. Prove that : ar (∥ gm ABCD) + ar (∥ gm AEFB) = ar (∥ gm EFCD).
Theorems on Area
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Answer
Parallelogram ABCD and Parallelogram AEFB are on opposite sides of the common base AB.
Let h1 be the distance between parallel lines AB and CD, h2 be the distance between parallel lines AB and EF.
Area of ||gm = Base × Height
ar (∥ gm ABCD) = AB × h1
ar (∥ gm AEFB) = AB × h2
ar (∥ gm ABCD) + ar(∥ gm AEFB) = AB × h1 + AB × h2
ar (∥ gm ABCD) + ar(∥ gm AEFB) = AB × (h1 + h2) ……(1)
From figure,
In ∥ gm EFCD,
Height = Height of ||gm ABCD + Height of ||gm AEFB = h1 + h2
ar (∥ gm EFCD) = DC × (h1 + h2)
Since, AB = DC [Opposite sides of parallelogram are equal]
ar (∥ gm EFCD) = AB × (h1 + h2)…..(2)
From equation (1) and (2),
ar(∥ gm ABCD) + ar(∥ gm AEFB) = ar (∥ gm EFCD)
Hence, proved that ar(∥ gm ABCD) + ar(∥ gm AEFB) = ar (∥ gm EFCD).
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