Proving that $\mu(C(x,\delta))\ge \delta^2/2 $ where $C(x,\delta)=B(x,\delta)\cap C$ and $C$ is a square. Let $\mu$ denote Lebesgue measure. Consider a closed square $C\subset \mathbb{R}^2$. Let $d$ be the diameter of $C$ and $$C(x,\delta)=B(x,\delta)\cap C,$$
where $B(x,\delta)=\{y\in \mathbb{R}^2: |y-x|<\delta\}$. I can prove that for $x\in C$ and $\delta< d$, the quantity $$\frac{\mu(C(x,\delta))}{\delta^2},$$
is bounded below by some postive constant, independently of $x$ and $\delta$. However, I fail to see why this constant can be takne to be $1/2$. Any idea is appreciated.
 A: First note that if $x$ is in the corner of the square, we have that $$\mu(C(x,\delta))\le \mu(C(y,\delta)),$$
for any $y\in C$, therefore, we can prove the estimate only for $x$ in the corner. Assume without loss of generality that the square is given by $[0,t]\times [0,t]$.
If $\delta\le t$, we are done, so assume that $\delta> t$ and let $P_1=(\sqrt{\delta^2-t^2},t)$ and $P_2=(t,\sqrt{\delta^2-t^2})$ be the two points in the intersection of $C$ and $\partial B(x,\delta)$. We see that the desired area is composed of the area of two triangles and a circular sector. 
The area of the two triangles is $t\sqrt{\delta^2-t^2}$, while the area of the circular sector is given by $(\pi/4-\arctan({\sqrt{\delta^2-t^2}/t}))\delta^2$. Therefore, $$\frac{\mu(C(x,\delta))}{\delta^2}=\frac{t\sqrt{\delta^2-t^2}+(\pi/4-\arctan({\sqrt{\delta^2-t^2}/t}))\delta^2}{\delta^2}, \ \delta\in(t,\sqrt{2}t). \tag{1}$$
From $(1)$ we see that if $\delta=t$ then , $\mu(C(x,\delta))/\delta^2=\pi/4$, while if $\delta=\sqrt{2}t$ then, $\mu(C(x,\delta))/\delta^2=1/2$. Moreover, by differentiating the function $\mu(C(x,\delta))/\delta^2$, with respect to $\delta$, we see that it is a decreasing function in the interval $(t,\sqrt{2}t)$, therefore, the estimates is proved.
