(N/A) The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.
Consider a physical body with centre $O$ and a point mass $m$ in the $x-y$ plane at $(x, y)$.
Moment of inertia about $x$-axis,$I_{x} = m y^{2}$.
Moment of inertia about $y$-axis,$I_{y} = m x^{2}$.
Moment of inertia about $z$-axis,$I_{z} = m(x^{2} + y^{2})$.
Thus,$I_{x} + I_{y} = m y^{2} + m x^{2} = m(x^{2} + y^{2}) = I_{z}$.
Hence,the theorem is proved.
$(b)$ The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
Suppose a rigid body is made up of $n$ particles,having masses $m_{1}, m_{2}, \dots, m_{n}$ at perpendicular distances $r_{1}, r_{2}, \dots, r_{n}$ respectively from the centre of mass $O$ of the rigid body.
The moment of inertia about axis $RS$ passing through the point $O$ is $I_{RS} = \sum m_{i} r_{i}^{2}$.
The perpendicular distance of mass $m_{i}$ from the axis $QP$ is $(a + r_{i})$.
Hence,the moment of inertia about axis $QP$ is $I_{QP} = \sum m_{i}(a + r_{i})^{2} = \sum m_{i}(a^{2} + r_{i}^{2} + 2 a r_{i}) = \sum m_{i} a^{2} + \sum m_{i} r_{i}^{2} + 2 a \sum m_{i} r_{i}$.
Since the origin is the centre of mass,$\sum m_{i} r_{i} = 0$.
Also,$\sum m_{i} = M$,where $M$ is the total mass of the rigid body.
Therefore,$I_{QP} = M a^{2} + I_{RS}$.
Hence,the theorem is proved.