Which Areas are closer? 
Each circle-and-square pair share the same center. Which areas are closer to being equal?

Choices:
$A$ and $B$
$C$ and $D$
The difference of $B$ and $A$ is equal to the difference of $C$ and $D$.

My answer is $C$ and $D$ because judging the diagram, I can assume that $35^2 - 400\pi$ are more closer than $36^2 - 400\pi.$ My question is, how do I show my work to prove that my answer is correct? 
 A: You want to say that, considering the internal areas and using symmetry, 


*

*$400\pi - 4A = 36^2-4B$ so $B-A = \dfrac{36^2-400\pi}{4}$  

*$400\pi - 4C = 35^2-4D$ so $C-D = \dfrac{400\pi-35^2}{4}$
and then use the calculations as to which difference is closer to $0$ in absolute terms
A: 
My question is, how do I show my work to prove that my answer is correct?

$4(B - A)  = 36^2 - \pi \left( \dfrac{40}{2} \right)^2$
$4(B - A)  = 1296 - 400\pi$
$B - A = 324 - 100\pi$
$B - A ≈ 9.84$
$4(D - C)  = 35^2 - \pi \left( \dfrac{40}{2} \right)^2$
$4(D - C)  = 1225 - 400\pi$
$D - C = \dfrac{1225}{4} - 100\pi$
$D - C ≈ 7.91$
Since $C - D < B - A,$ $C$ and $D$ are closer to being equal.
A: First note that:
$$35^2<400\pi<36^2$$
as
$$\frac{35^2}{400}=\frac{1225}{400} = 3.0625<\pi<3.24=\frac{1296}{400}=\frac{36^2}{400}\ \ \ \ \ \cdots(i)$$
So we will compare the positive differences $400\pi-35^2$ and $36^2-400\pi$. One way to do this is to observe from equation $(i)$ that $3.0625$ is closer to $\pi=3.14159...$ than $3.24$ is. A more systematic way woul be to compare $400\pi-35^2$ and $36^2-400\pi$ with an unknown comparing operator $\square$ which we will try to find out as either $>$ or $<$.
So
$$400\pi-35^2\ \square\ 36^2-400\pi$$
$$800\pi\ \square\ 36^2+35^2=1296+1225=2521$$
$$\pi\ \square\ \frac{2521}{800}=3.15125$$
Which gives us a sign $<$. So again $400\pi-35^2<36^2-400\pi$
