Equation of parabola

$y=x^2-4$ and $2y=4-x^2$

Let point $B=\left(h,\frac{4-h^2}{2}\right)$

$A=\left(-h,\frac{4-h^2}{2}\right)$

$C=\left(h_1{h}^2-4\right)$

$D=\left(-h,h^2-4\right)$

Area of rectangle

$ABCD=AB\times BC=2h\times \left(\frac{4-h^2}{2}-h^2+4\right)$

$\Rightarrow A=12h-3h^3\Rightarrow \frac{dA}{dh}=12-9h^2$

For maxima or minima

$\frac{dA}{dh}=0\Rightarrow 12-9h^2=0$

$h=\pm \frac{2}{\sqrt{3}}$ maximum at $h=\frac{2}{\sqrt{3}}$

$\therefore A=12\left(\frac{2}{\sqrt{3}}\right)-3{\left(\frac{2}{\sqrt{3}}\right)}^3=\frac{24}{\sqrt{3}}-\frac{8}{\sqrt{3}}$

$=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}$

$A-\frac{16\times 173}{3}=9.22$

$\left[A\right]=\left[9.22\right]=9$