= Solution
Take the square to have uniform <surface mass density> $\sigma=M/a^2$, and put its centre at $x=y=0$. For an in-plane axis through the centre parallel to a side, direct <integration> gives
$$
I_x=\frac M{a^2}\int_{-a/2}^{a/2}\int_{-a/2}^{a/2}y^2\,dx\,dy=\frac{Ma^2}{12}.
$$
The same holds for $I_y$, and the mixed <integral> $\int xy\,dm$ vanishes by reflection symmetry. The squared perpendicular distance to an axis making angle $\psi$ with the $x$ axis is $(-x\sin\psi+y\cos\psi)^2$. Its <integral> therefore gives
$$
I_\psi=\frac{Ma^2}{12}(\sin^2\psi+\cos^2\psi)=\frac{Ma^2}{12}.
$$
Thus a diagonal through opposite corners and its perpendicular \b[through the centre] have
$$
\boxed{I_{\rm diagonal}=I_{\rm central\ perpendicular}=\frac{Ma^2}{12}.}
$$
This is the <in-plane inertia isotropy of a square plate>.
\b[Axis-location qualification:] perpendicularity fixes a direction, not a line's position. If the second axis is instead shifted by distance $d$ from the centre, the <parallel axis theorem> gives $I_2=Ma^2/12+Md^2$. A perpendicular through a corner on that diagonal has $d=a/\sqrt2$ and hence $I_2=7Ma^2/12$, a concrete different value satisfying the stated directional description. If “through the corners” means an edge rather than a diagonal, that edge has $I=Ma^2/3$; two adjacent edges both have that value. The general displaced-axis formula and these choices make the location convention explicit rather than silently assigning a unique value to an unspecified second axis.
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