Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is

Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is Correct Answer $$\frac{{{\text{b}}{{\text{h}}^3}}}{{36}}$$

Related Questions

According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $$I$$P) is given by (where, A = Area of the section, $$I$$G = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)
According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $${I_{\text{P}}}$$) is given by (where, A = Area of the section, $${I_{\text{G}}}$$ = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)