= Solution
The <parallel axis theorem> relates the <moment of inertia> about an axis through the <centre of mass> to that about a parallel axis a perpendicular distance $d$ away:
$$
\boxed{I=I_{\mathrm{CM}}+M_{\mathrm{tot}}d^2.}
$$
To see the cross term vanish, write each perpendicular displacement as its displacement from the <centre of mass> plus the fixed offset. In the defining mass <integral>, the first moment about the <centre of mass> is zero.
For a thin planar <rigid body> in the $xy$ plane, the <perpendicular axis theorem> says
$$
\boxed{I_z=I_x+I_y,}
$$
where all three axes pass through the same point, $x,y$ lie in the plane, and $z$ is perpendicular to it. This follows directly from $x^2+y^2$ being the squared distance from the $z$ axis. The thin-planar condition and common intersection point are essential; the formula is not a general identity for three-dimensional bodies.
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