What is the moment of inertia of the triangle ?

Dear Student,

Let the side of the triangle along y-axis is AB and is of height a and the side along x-axis is the base of length b. Since moment of inertia is to be determined about an axis of rotation and from the provided diagram, it seems that the student is interested in finding the moment of inertia about the side (AB) along y-axis.
Considering an element DE parallel to y-axis at a distance x from origin and width dx.
Let the mass of the triangle be M. Then mass per unit area of the triangle would be 
$\sigma =\frac{M}{A}=\frac{M}{{\displaystyle \frac{1}{2}}ab}=\frac{2M}{ab}$
$$\begin{gathered} \Delta ABC and \Delta CDE are similar, so \\ \frac{AB}{DE}=\frac{BC}{DC} \\ \frac{a}{DE}=\frac{b}{(b-x)} \\ DE=\frac{a(b-x)}{b} \\ Area of the element DE = \frac{a(b-x)}{b}\times dx \\ mass the element, dm = \sigma \frac{a(b-x)}{b}dx \\ Moment of inertia about AB, \\ I={\int }_0^b{x}^{2}dm={\int }_0^b{x}^2\sigma \frac{a(b-x)}{b}dx=\sigma \frac{a}{b}{\int }_0^b\left(b{x}^2-x^3\right)dx=\sigma \frac{a}{b}\left(\frac{b^4}{3}-\frac{b^4}{4}\right) \\ I=\sigma \frac{a}{b}\left(\frac{b^4}{12}\right)=\sigma \frac{a{b}^3}{12} \\ Putting the value of \sigma \\ I=\frac{a{b}^3}{12}\times \frac{2M}{ab}=\frac{M{b}^2}{6}. \end{gathered}$$

Regards