Math, asked by sindhusharavuri, 1 year ago

parallelogram abcd and rectangle abef are on same base ab and have equal areas. Show that perimeter of the parallelogram is greater than that of the rectangle.

Answers

Answered by Anonymous
16
Given parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.
To prove :  The perimeter of the parallelogram ABCD is greater than that of rectangle ABEF.
Proof : Let ABCD parallelogram and ABEF rectangle are on the same base AB and between the same parallels AB and FC.
Then perimeterof the parallelogram ABCD = 2(AB+AD)
and perimeterof the rectangle ABEF = 2(AB+AF)
In ΔADF ∠ AFD = 90
∴ ∠ ADF is an acute angle
∴ ∠ AFD > ∠ ADF
∴ AB + AD >  AB + AF
∴ 2( AB + AD) >  2(AB + AF )
∴ perimeter of the parallelogram is greater than that of the rectangle.
Answered by Anonymous
1

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