The moment of inertia of a triangular section (height h, base b) about its base, is
A. $$\frac{{{\text{b}}{{\text{h}}^2}}}{{12}}$$
B. $$\frac{{{{\text{b}}^2}{\text{h}}}}{{12}}$$
C. $$\frac{{{\text{b}}{{\text{h}}^3}}}{{12}}$$
D. $$\frac{{{{\text{b}}^3}{\text{h}}}}{{12}}$$
Answer: Option C

I for triangle =bh^3/36
At base
I+Ay^2=bh^3/36+1/2*b*h(h/3)^2
=bh^3/12
About its centroid bh^3/36
And
Base bh^3/12
Wrong
BH^3/36 for triangular section