A metallic sheet of mass 2 kg has the shape of an isosceles triangle OAB as shown in figure. Then its moment of inertia about an axis passing through O and perpendicular to its plane is 

If four such triangles are assembled a square ABCD is formed as shown in figure. Whose side $\mathrm{l}=2\times 2\cos {45}^{\mathrm{o}}\mathrm{m}=2\sqrt{2}\mathrm{m}$.

$m^l=$Mass of the square $\text{=4×2\hspace{0.33em}Kg=8\hspace{0.33em}Kg}$

Moment of inertia of the square about an axis passing through O and perpendicular to its plane is given by 

${\text{I}}_0^1=\frac{1}{12}{m}^1{l}^2+\frac{1}{12}{m}^1{l}^2=\frac{1}{6}\times 8\times {\left(2\sqrt{2}\right)}^{2}Kg-m^2=\frac{32}{3}Kg-m^2$

$\therefore$ Moment of inertia of the triangle ABC, $I_0=\frac{I_0^1}{4}=\frac{8}{3}Kg-m^2$