Q:

Area of a quarter circle C1 equals the area of an inner circle C2 where C2.diameter = C1.radius

Say we have a circle C1 with radius 2. Inside of that we draw circle C2 going from the centre point of C1 to the perimeter of C1 (making it diameter = 2)
Let A = a quarter of C1
Let B = area of C2 (inner circle)

Is it true that the the areas are the same?

Prove that A = B
A = (pi * r ^ 2 ) / 4
  = (pi * 4) / 4
  = pi

B = pi * r ^ 2
  = pi * (2 / 2) ^ 2
  = pi

Is this correct?  
If this is the case, then with flexible walls, 4 circles of diameter equaling the radius of the outer circle, would fit in the larger circle (Assumption#1)
So this means that the perimeter of A and B should also be the same, since when we squeeze the 4 circles inside the big one the shape will change from a little circle to a quarter circle
Perim A = (arc / 4) + r + r
        = (2 * r * pi ) / 4 + 2 r
        = (2 * 2 * pi ) / 4 + 4
        = pi + 4

Permin B = 2 * r * pi 
         = 2 * (2/2) * pi
         = 2 * pi

pi + 4 != 2 * pi
This shows the the circumference of A (quarter of big circle) doesn't equal B (smaller circle) 
I am practicing and relearning maths after a long break, so not sure if there is something wrong with the numbers, or is assumption#1 wrong?

A:

Equality of areas does not imply equality of perimeters, and you provided a counterexample yourself. Your assumption is invalid because you do not have a good argument to base it on.
Think about a surface such as an $A4$ paper. We can find a ruler which has the same surface area as that of the sheet's, but this doesn't meant that we could fit the ruler on the sheet's surface in some way. However, if we break it into small pieces, then we can do so. This is the same as your situation. We can fit $4$ small circles into the larger one, but we'll have to deform them in some way.

