increasing area of circle,square &rectangle Recently our teacher asked us to arrange a circle, a square, a rectangle in the order of their increasing areas(perimeters are equal).
I found that circle has much area then a square.
But I was unable to find relation of third one.
Please help me to arrange them (with proof).
 A: Let the radius of the circle be $r$, the side length of the square $s$ and the length and breadth of the rectangle $l$ and $b$ respectively. So, we have, $$2\pi r = 4s = 2\left(l+b\right)$$  Now, we have to arrange $\pi r^2, s^2$ and $lb$ in increasing order.   
$(1)$ Circle and square: We have $\pi r = 2s$. So, $$\text{area of circle}= \pi r^2 = \pi \left(\frac{2s}{\pi}\right)^2 = \frac{4s^2}{\pi} > s^2 = \text{area of square}$$    
$(2)$ Square and rectangle: We have $2s = (l+b)$. So, $$\text{area of square} = s^2 = \frac{(l+b)^2}{4} \geq \text{area of rectangle}$$ 
Note: The areas are equal if both the length and breadth are same otherwise the area of the square is greater. We can also see that proving $(3)$ is not necessary but...  
$(3)$ Circle and rectangle: We have $\pi r =(l+b)$. So, $$\text{area of circle} = \pi r^2 = \pi\left(\frac{l+b}{\pi}\right)^2 = \frac{(l+b)^2}{\pi} > \text{area of rectangle}$$ Note: The area of the circle is always greater than that of the rectangle. We can easily deduce that there can be no equality.     
So, our inference is 
$$\text{area of rectangle} \leq \text{area of square} < \text{area of circle}$$
Hope it helps.
A: The relation is the following : Area of circle > Area of rectangle = Area of square.
Since perimeter is same (let it be P), the radius of circle is $\frac{P}{2\pi}$, side of square is $\frac{P}{4}$ and sum of sides of the rectangle is $\frac{P}{2}$. This gives us area of circle being $\frac{P^2}{4\pi}$ and area of square being $\frac{P^2}{16}$. For rectangle we have to solve the maximizing problem that is given sum of sides find the rectangle of maximum area which will come out to be $\frac{P^2}{4}$. To see this note we have to maximize $l\times b$ given $l+b=P/2$ which reduces to maximizing $l(l-P/2)$ where $0\le l\le \frac{P}{2}$. You can take derivative wrt $l$ and you will find that max occur at $l=\frac{P}{4}$. 
A: Let's say the perimeter is 40. 
Case 1-
As for the square, one side = 10. 
Area of square $10^2$ = 100. 
Case 2-
As for the rectangle -
a) Let's say the L is 2 and W is 18. 
Area of rectangle 2 * 18 = 36. 
b) If L is 10 and W is 10.
Area of rectangle = 10 * 10 = 100. 
c) If L is 14 and W is 6, 
Area of rectangle 14 * 6 = 84. 
It will ALWAYS be smaller than or equal to the 100 Area of a square in above sub-cases.
So you can say Area of rectangle either smaller than or equal to the Area of square.
Case 3-
As for the circle -
2πr = 40
r = $\frac{140}{11}$
Area of circle $π * \frac{140}{11} * \frac{140}{11}$ =  127.27
So area of circle is greater than both.

Edit - 

The area of a circle is $πr^2$, where r is the radius of the circle. 
But r = $\frac{P}{2π}$, where P is the perimeter of the circle. 
So the area of the circle is $\frac{P^2}{4π}$
The area of a square is $s^2$, where s is the side length. 
But s = $\frac{P}{4}$
So the area of a square is $\frac{P^2}{16}$. 
Since $\frac{1}{4π} > \frac{1}{16}$, the circle has more area than the square.

Edit - Rectangle and Square.

For Square perimeter = Rectangle perimeter
4S = 2(L + B) 
S = 0.5(L + B) 
Area of Square = $S^2 = 0.25[(L + B)^2]$
Area of rectangle = L*B
Area of Square - Area of Rectangle 
= $0.25[(L + B)^2] - L*B$
= $0.25(L)^2 + 0.25(B)^2 + 0.5 LB - LB$
= $0.25(L)^2 + 0.25(B)^2 - 0.5 LB$
= $0.25[L^2 + B^2 - 2LB]$
= $0.25[(L - B)^2]$
Since (L-B)^2 is always greater than zero i.e. positive 
then a square's area is greater than or equal a rectangle's area with a given perimeter.
