Question

If the areas of a square and a rectangle is equal then compare their perimeters

Answer 1

Yup. take the following measures :

♦ L = length of rectangle 
♦ B = breadth of rectangle

♣ S = side of a square 
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→ Areas of both the figure equal => LB = S²
 => S = √( LB )

=> 2( L + B ) : 4S = ( L + B ) : 2S
=> ( L + B ) : 2√( LB )
=> Ratio of perimeter =  Arithmetic Mean of Sides of Rectangle : G.M. of Sides
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 ♦ Applying A.M. G.M. inequality : 
→ Ratio is = 1:1 if and only if  L = B
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Answer 2

Hey friend , Harish here.

Here is your answer:

Given that,

Area of Square = Area of Rectangle .

Objective:

To compare their perimeters.

Solution:

We know that,

Area of square = Area of rectangle.

Let Side of Square be ' a ' and Length & breadth of the rectangle be ' l ' , ' b ' respectively.

Then,

a² = l × b

a = √(l×b)

So, Perimeter of square = 4 × a = 4 √(l × b)

      Perimeter of  rectangle = 2 (l + b)

Then, 

Ratio of  perimeters = 4a : 2(l+b)

                                = 2a : ( l+b)
 
                                = 2√(lb) : (l+b)

  Now it is in the form of:   G.M of sides : A.M of sides.

(A.M - Arithmetic mean, G.M - Geometric Mean ).

We know that, 

A.M  ≥ G.M 

So, If AM = GM.

Then, The ratios Of perimeter is 1 : 1 And is Only possible when, ( l = b = a).

And Next case if AM > GM Then . The ratio is less than one and hence the perimeter of rectangles is larger than that of the square.
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Hope my answer is helpful to you.