Question

A square and a rectangle have equal areas.If there perimeters are p1 and p2 respectively, then:

Answer 1

The relation between $p_{1}$ and $p_{2}$ = $p_{1}$ < $p_{2}$

Step-by-step explanation:

Let the side of square = a and

The length and breadth of rectangle = l and b

To find, the relation between $p_{1}$ and $p_{2}$ = ?

According to question,

The area of the square = The area of the rectangle

⇒ $a^{2} =lb$

⇒ $a=\sqrt{lb}$               ............. (1)

The perimeter of the square($p_{1}$) = 4a and

The perimeter of the rectangle($p_{2}$) = 2(l + b)

⇒ $p_{1}$ = $p_{2}$

⇒ 4a = 2(l+b)

⇒ 2a = l+b

⇒ a = $\dfrac{l+b}{2}$

From equation (1), we get

⇒ $\sqrt{lb}$ = $\dfrac{l+b}{2}$

⇒ G.M. < A.M.

⇒ $p_{1}$ < $p_{2}$

Thus, the relation between $p_{1}$ and $p_{2}$ = $p_{1}$ < $p_{2}$