(i) Moment of inertia of a triangular lamina about one of its side:

Let us consider a triangular lamina ABC of mass M, is rotating about one of its sides BC, where the length of the side BC is a. If $h$ be the height of this lamina, then the area of this lamina is $\left(\frac{1}{2}\cdot{a}\cdot{h}\right)$ and the mass density i.e., mass per unit area is $\frac{M}{\frac{1}{2}ah}$ = $\left(\frac{2M}{ah}\right)$.

In order to calculate the moment of inertia of this lamina about the axis of rotation BC, let us consider a small elementary area $EF$ of thickness $dz$ and length $l$ at a distance $z$ from the axis of rotation BC. So the area of $EF$ is $l\cdot{dz}$ and the mass is $\left(\frac{2M}{ah}\right)l\ dz$. Now the moment of inertia of this element about the axis of rotation is

$dI=\left(\frac{2Ml}{ah}\right)z^2\ dz\tag{1}$

From Fig. 1, we can write that

$\left(\frac{l}{h-z}\right)=\left(\frac{a}{h}\right)$

$or,\ \displaystyle{l=\frac{a(h-z)}{h}}$

Putting the value of $l$ in equation (1) we get,

$dI=\left(\frac{2M}{ah}\right)\frac{a(h-z)}{h}z^2\ dz$

$or,\ dI=\frac{2M}{h^2}(h-z)z^2\ dz$

So the moment of inertia of the whole triangular lamina is

$I=\frac{2M}{h^2}\displaystyle{\int_0^h}(h-z)z^2\ dz$

$or,\ I=\frac{2M}{h^2}\displaystyle{\int_0^h}(hz^2-z^3)\ dz$

$or,\ I=\frac{2M}{h^2}{\left[\frac{1}{3}hz^3-\frac{1}{4}z^4\right]}_0^h$

$or,\ I=\frac{2M}{h^2}{\left[\frac{1}{3}h^4-\frac{1}{4}h^4\right]}$

$or,\ I=\frac{2M}{h^2}\frac{1}{12}h^4$

$or,\ \displaystyle{I=\frac{1}{6}Mh^2}\tag{2}$

(ii) About an axis passing through the centre of gravity and parallel to one side of the triangular lamina:

Let us consider the rectangular lamina ABC is rotating about an axis PQ passing through the centre of gravity G of the lamina and parallel to the side BC. We know that the perpendicular distance of G from each side is $\left(\frac{h}{3}\right)$, where h is the height of the triangular lamina.

So the perpendicular distance between PQ and BC is $(\frac{h}{3})$

If $I_G$ be the moment of inertia of this lamina about the axis PQ, then by applying the theorem of parallel axes, we can write that

$I=I_G+M{\left(\frac{h}{3}\right)}^2$

$or,\ I_G=I-\frac{1}{9}Mh^2$

$or,\ I_G=\frac{1}{6}Mh^2-\frac{1}{9}Mh^2$
[putting the value of $I$ from equation (2)]

$or,\ \displaystyle{I_G=\frac{1}{18}Mh^2}\tag{3}$

(iii) Moment of inertia of the rectangular lamina about an axis passing through one of its vertices and parallel to opposite side:

Let us consider the rectangular lamina ABC is rotating about an axis RS passing through vertex A and parallel to the side BC. We know that the distance of each vertex from the centre of gravity is $\left(\frac{2h}{3}\right)$, where h is the height of the rectangular lamina. So the perpendicular distance between PQ and RS is $\left(\frac{2h}{3}\right)$.

By applying the theorem of parallel axes, we can write that the moment of inertia of this rectangular lamina about the axis of rotation $RS$ is

$I_V=I_G+M{\left(\frac{2h}{3}\right)}^2$

$or,\ I_V=\frac{1}{18}Mh^2+\frac{4}{9}h^2$
[putting the value of $I_G$ from equation (3)]

$or,\ \displaystyle{I_V=\frac{1}{2}Mh^2}\tag{4}$