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Volume and Its Measurement

A rectangle a...

Question

A rectangle and a parallelogram of equal area and equal base are given. If $r$ and $p$ denote their respective perimeters, then

r = p

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r = 2 p

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r > 2 p

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r < p

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Solution

The correct option is D $r < p$
Perimeter of the rectangle ABCD
$= (A B + A D) + (D C + B C) = r$ ........(i)

Perimeter of parallelogram ABFE
$= (A B + A E) + (E F + B F) = p$ ........(ii)

in right angled $△ A D E$
$A E > A D$ ...[since being the hypotenuse side]

in right angled $△ B C F$
$=> B F > B C$ ...[since being the hypotenuse side]

Also, $A B = D C$ and $A B = E F$ ( $∵$ opposite sides of a rectangle and a parllelogram are equal )
$=> D C = E F$

Comparing (i) & (ii), we get
$P e r i m e t e r o f r e c t a n g l e < P e r i m e t e r o f P a r a l l e l o g r a m$ $(∵ b o t h B F a n d A E a r e g r e a t e r t h a n B C a n d A D r e s p e c t i v e l y)$

i.e. $r < p$

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