State and prove perpendicular axes theorem.

Statement: Sum of moments of inertia of the plane lamina about any two mutually perpendicular axes in its plane is equal to moment of inertia of the plane lamina about a perpendicular axis passing through point of intersection of first two axes
$\therefore {\mathrm{I}}_{\mathrm{z}}={\mathrm{I}}_{\mathrm{x}}+{\mathrm{I}}_{\mathrm{y}}$
Proof: Consider a plane lamina revolving about the z-axis. Let ‘o’ be the origin of the axis. Imagine a particle of mass ‘m’ lying at a distance ‘r’ from point ‘o’ on the plane. 
Let x,y be the coordinates of the point P.

Thus ${\mathrm{r}}^2={\mathrm{x}}^2+{\mathrm{y}}^2$

Then the moment inertia of the body about x-axis ${\mathrm{I}}_{\mathrm{x}}={\mathrm{\Sigma my}}^2$

The moment inertia of the body about y-axis
${\mathrm{I}}_{\mathrm{y}}={\mathrm{\Sigma mx}}^2$
Then the moment of inertia of the body about z-axis
$\begin{array}{l}{\mathrm{I}}_{\mathrm{z}}={\mathrm{\Sigma mr}}^2\\ {\mathrm{I}}_{\mathrm{z}}=\mathrm{\Sigma m}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\\ {\mathrm{I}}_{\mathrm{z}}={\mathrm{\Sigma mx}}^2+{\mathrm{\Sigma my}}^2={\mathrm{I}}_{\mathrm{y}}+{\mathrm{I}}_{\mathrm{x}}\\ \therefore {\mathrm{I}}_{\mathrm{z}}={\mathrm{I}}_{\mathrm{y}}+{\mathrm{I}}_{\mathrm{x}}\end{array}$
Hence perpendicular axes is proved.