prove perimeter of square is less than that of an equivalent parallelogram on same base
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I have uploaded a diagram regarding the answer of this question.
Let PQCB be a square and ABCD be a parallelogram (please have a close look at the figure).
If you look at the figure you can see that AB and CD hypotenuses which are certainly greater in length than PB and QC. So the perimeter of square PQCB is less than that of parallelogram ABCD on same base.
Hence proved.
I have uploaded a diagram regarding the answer of this question.
Let PQCB be a square and ABCD be a parallelogram (please have a close look at the figure).
If you look at the figure you can see that AB and CD hypotenuses which are certainly greater in length than PB and QC. So the perimeter of square PQCB is less than that of parallelogram ABCD on same base.
Hence proved.
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