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# 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.

**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.

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