Moment of inertia of a triangle of base width ‘b’ and height ‘h’ with respect to the centroidal axis parallel to its base is equal to: 

Explanation:

Moment of inertia of a triangle with base b and height h

\({I_{base}} = \frac{{b{h^3}}}{{12}}\)

Parallel axis theorem

Parallel axis theorem is used to find a moment of inertia about an axis which is at some distance from the centroidal axis and parallel to centroidal.

Suppose an axis is at distance d from the centroidal axis and parallel to the centroidal axis.

according to the parallel axis theorem, the moment of inertia about that axis is given by,

\({I_{about~axis~parallel~to~centroida~laxis}} = {I_{about~centroidal~axis}} + A{d^2}\)

Moment of inertia about the centroidal axis of the triangle

Applying parallel axis theorem we get

\({I_{xx}} = {I_{cc}} + A{\left( {\frac{h}{3}} \right)^2}\)

\({I_{cc}} = {I_{xx}} - A{\left( {\frac{h}{3}} \right)^2}\)

\(\frac{{b{h^3}}}{{12}} - \frac{{bh}}{2} \times \frac{{{h^2}}}{9} = \frac{{b{h^3}}}{{36}}\)

\({I_{centroid}} = \frac{{b{h^3}}}{{36}}\)