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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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