# Parallelogram $\mathrm{ABCD}$ and rectangle $\mathrm{ABEF}$ are on the same base $\mathrm{AB}$ and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

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Given:

Parallelogram $\mathrm{ABCD}$ and rectangle $\mathrm{ABEF}$ are on the same base $\mathrm{AB}$ and have equal areas.

To do:

We have to show that the perimeter of the parallelogram is greater than that of the rectangle.

Solution:

Parallelogram $\mathrm{ABCD}$ and rectangle $\mathrm{ABEF}$ are on the same base $\mathrm{AB}$ and have equal areas.

This implies,

Parallelogram $\mathrm{ABCD}$ and rectangle $\mathrm{ABEF}$ lie between the same parallels $AB$ and $CF$.

We know that,

The opposite sides of a rectangle are equal.

Therefore,

$AB = EF$

Similarly,

The opposite sides of a parallelogram are equal.

This implies,

$AB = CD$

$\Rightarrow CD = EF$

$AB + CD = AB + EF$............(i)

In the right-angled triangle $AFD$, $AD$ is the hypotenuse.

This implies,

$AF Similarly, In the right-angled triangle$EBC$,$EB$is the altitude and$BC$is the hypotenuse. This implies,$BE

Adding (ii) and (iii), we get,

$AF + BE From (i) and (iv), we get,$AB + EF + AF + BE

Perimeter of rectangle \$ABEF

This implies,

The perimeter of the parallelogram is greater than that of the rectangle.

Hence proved.

Updated on 10-Oct-2022 13:42:14