The moment of inertia of semicircular plate of radius R and mass M about axis AA' in its plane passing through its centre is

Moment of inertia of disk w.r.t. principle axis (1) $=\frac{M{R}^2}{2}$

Moment of inertia of disk w.r.t. diametrical axis (2)

From perpendicular axes theorem

${\mathrm{I}}_{\mathrm{z}}={\mathrm{I}}_{\mathrm{x}}+{\mathrm{I}}_{\mathrm{y}}\dots \left(\mathrm{i}\right)$

here $I_x=I_y$, because both are diametrical axis,

So here  $I_{\mathit{z}}\mathit{=}\frac{\mathit{M}{\mathit{R}}^{\mathit{2}}}{\mathit{2}}$and ${\mathrm{I}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{y}}=\mathrm{I}\left(\text{ Let }\right)$

So from eq. (i)

$\Rightarrow \frac{{\mathrm{MR}}^2}{2}=\mathrm{I}+\mathrm{I}\Rightarrow \mathrm{I}=\frac{{\mathrm{MR}}^2}{4}$

For half disk also, it is applicable, So

$\mathrm{I}=\frac{{\mathrm{MR}}^2}{4}$

(It does not depend on $\theta$ given in diagram, because given axis is also similar to diametrical axis)