If the area of a square and a rectangle is equal then compare their perimeters
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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.
__________________________________________________
Hope my answer is helpful to you.
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.
__________________________________________________
Hope my answer is helpful to you.
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