Question

Moment of inertia of isosceles triangle of angle theta

Answer 1

$\mathfrak{The\:Answer\:is}$


Mathematically,

\(I_{0}=\frac{bh^{3}}{36}\)

———————————————————–

\(Iy_{0}= \frac{hb^{3}}{48}\)


$\boxed{Hope\:This\:Helps}$

Answer 2

The moment of inertia of a triangle with respect to an axis passing through its centroid, parallel to its base, is given by the following expression: where b is the base width, and specifically the triangle side parallel to the axis, and h is the triangle height (perpendicular to the axis and the base).