A parallelogram and rectangle are on the same base and between the same parallel lines. Then perimeter of rectangle is:

equal to the perimeter of the parallelogram.

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greater than the perimeter of the parallelogram.

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less than the perimeter of the parallelogram.

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Twice the perimeter of the parallelogram.

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Solution

The correct option is C
less than the perimeter of the parallelogram.

$△$ EDC $≅$ $△$ FAB

So, ED = AF (CPCT) ------------------------------ 1

Since, $E C > C D$ and $B F > A B$ (Hypotenuse of right angle triangle is the greatest side)

Therefore, EC + BF $>$ CD + AB ---------------------------------- II

EC + BF + CB + ED + DF = perimeter of EFCB = EC + BF + CB + DF + AF (from I)

$\Rightarrow$ EC + BF + CB + ED + DF $>$ CD + AB + CB + DF + AF (from II)

$\Rightarrow$ Perimeter of EFBC $>$ Perimeter of ABCD

Therefore, perimeter of rectangle is less than that of parallelogram.

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