 Solution

Measure the axial coordinate $z$ from the vertex. The slice radius is $z\tan \alpha$ and its mass is $dm=3m{z}^2{H}^{-3}dz$, since cross-sectional area is proportional to $z^2$. Its axial moment is $\frac{1}{2}(z\tan \alpha {)}^{2}dm$. ThereforeFor an axis through the vertex perpendicular to the cone axis, a thin disc contributes $[z^2+\frac{1}{4}{z}^2{\tan }^2\alpha ]dm$: the second term is its diametral moment. Integration gives the two equal transverse principal momentsRotational symmetry eliminates the mixed moments and makes these the principal axes. The centre of mass is at $3H/4$, so the parallel-axis theorem would give $A_G=3m{H}^2/80+3m{H}^2{\tan }^2\alpha /20$ and $C_G=C_O$.