A disc of mass  $\mathrm{m}$ and radius  $\mathrm{R}$ has a concentric hole of radius $\mathrm{r}$  . Its moment of inertia about an axis through centre and normal to its plane is

Moment of Inertia of remaining disc is

${\mathrm{I}}_0=\frac{1}{2}{ \mathrm{m}}_1{\mathrm{R}}^2-\frac{1}{2}{ \mathrm{m}}_2{\mathrm{r}}^2$

Mass of remaining disc is m.

${\mathrm{m}}_1-{\mathrm{m}}_2=\mathrm{m}$

Mass is directly proportional to its area. So,

${\mathrm{m}}_1=\frac{{\mathrm{mR}}^2}{{\mathrm{R}}^2-{\mathrm{r}}^2}$

${\mathrm{m}}_2=\frac{{\mathrm{mr}}^2}{{\mathrm{R}}^2-{\mathrm{r}}^2}$

$\Rightarrow {\mathrm{I}}_0=\frac{\mathrm{m}}{2}\left({\mathrm{R}}^2+{\mathrm{r}}^2\right)$