Question

A wire of fixed length is bent to form a square and the perimeter is noted.
The same wire is now bent in the shape of a circle and again the perimeter is
noted. What do we observe? and why?

Next, for the same square, derive the relationship between its perimeter and area using their formulas.

Answer 1

Answer:

$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$$\sf \dfrac{\cancel{450}{×6×7}}{\cancel{100}}$

Answer 2

if the wire is of fixed length then the sum of all the sides of the square which means the perimeter of the square will be equal to the the perimeter of the circle that is circumference of the circle which means if we consider side of the square as A, then

4 x A = 2πR