Question

A square and a rectangle are of equal area. If the perimeter of rectangle is $p$ and that of a square is $q$, then

• A. $p=q$
• B. $p<q$
• C. $p>q$
• D. $p\le q$

Answer 1

The correct option is C

$p>q$

Let the length and breadth of a rectangle be $l$and $b$respectively.

So, Area of rectangle $=l\times b$.

Let the side of the square be $a$.

Area of square $=a^2$.

Perimeter of rectangle $=2(l+b)$.

Perimeter of square $=4\times a$.

By the given condition,

A square and a rectangle are of equal area.

So, $l\times b=a^2$………………….(1)

Let us assume perimeter of square is greater.

which means, $2(l+b)<4a$

Dividing both sides by $2$.

$\frac{2(l+b)}{2}<\frac{4a}{2}$

$l+b<2a$.

Squaring both sides,

${(l+b)}^2<{\left(2a\right)}^2$

$l^2+2lb+b^2<4a^2$

But from (1),

$l\times b=a^2$

So, $l^2+2lb+b^2<4lb$

$l^2+2lb-4lb+b^2<0$

$l^2-2lb+b^2<0$.

${(l-b)}^2<0$ (${(a-b)}^2=a^2-2ab+b^2$)

$l-b<0$

$l<b$ which is not possible

As, the length of a rectangle is always greater than the breadth.

So, our assumption is wrong.

So, the perimeter of a rectangle is greater than that of a square.

Perimeter of rectangle is $p$ and that of a square is $q$.

Hence, $p>q$.

Option C is the correct option.