Question

Parallelogram ABCD and rectange ABEF are on the same base AB and also have equal area show the the perimeter of the parallelogram is greater than of the rectangle.

Answer 1

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Answer 2

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Here's your answer!!!
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ɢɪᴠᴇɴ: ||gm. ABCD and rect. ABEF are on the same base AB and ar (||gm ABCD = ar (rect. ABEF)

ᴛᴏ ᴘʀᴏᴠᴇ : Perimeter (|| gm ABCD) > perimeter (rect.ABEF)

ᴄᴏɴsᴛ : Draw DL perpendicular AB

ᴘʀᴏᴏғ : In rt ∆ ALD, hyp. AD > side LD

Now,

ar (rect. ABEF) = AB×BE

ar (rect. ABCD) = AB×LD

But

ar (rect. ABEF) = ar(|| gm ABCD)

=> AB×BE = AB×LD => BE = LD

Perimeter (rect. ABEF) = 2 (AB×BE) = 2 (AB+LD)

Perimeter (|| gm ABCD) = 2(AB+AD)

∵ AD > LD

∴ 2(AB + AD) > 2(AB +LD)

=> Perimeter (|| gm ABCD) > Perimeter (rect. ABEF)
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