$A$ uniform square plate $S$ (side $c$) and a uniform rectangular plate $R$ (sides $b, a$) have identical areas and masses. Show that:
$(i) \frac{I_{xR}}{I_{xS}} < 1$
$(ii) \frac{I_{yR}}{I_{yS}} > 1$
$(iii) \frac{I_{zR}}{I_{zS}} > 1$

(N/A) Given that the area of the square plate $S$ is equal to the area of the rectangular plate $R$.
Therefore,$c^2 = a \times b$,which implies $c^2 = ab$.
Let $M$ be the mass of both plates.
The moment of inertia of a rectangular plate of sides $a$ and $b$ about the $x$-axis (passing through the center and parallel to side $a$) is $I_{xR} = \frac{Mb^2}{12}$.
The moment of inertia of a square plate of side $c$ about the $x$-axis is $I_{xS} = \frac{Mc^2}{12}$.
$(i)$ $\frac{I_{xR}}{I_{xS}} = \frac{Mb^2/12}{Mc^2/12} = \frac{b^2}{c^2} = \frac{b^2}{ab} = \frac{b}{a}$. Since $a > b$ for a rectangle,$\frac{b}{a} < 1$,so $\frac{I_{xR}}{I_{xS}} < 1$.
$(ii)$ Similarly,$I_{yR} = \frac{Ma^2}{12}$ and $I_{yS} = \frac{Mc^2}{12}$.
$\frac{I_{yR}}{I_{yS}} = \frac{a^2}{c^2} = \frac{a^2}{ab} = \frac{a}{b}$. Since $a > b$,$\frac{a}{b} > 1$,so $\frac{I_{yR}}{I_{yS}} > 1$.
$(iii)$ By the perpendicular axis theorem,$I_z = I_x + I_y$.
$I_{zR} = \frac{M(a^2 + b^2)}{12}$ and $I_{zS} = \frac{M(c^2 + c^2)}{12} = \frac{2Mc^2}{12}$.
$\frac{I_{zR}}{I_{zS}} = \frac{a^2 + b^2}{2c^2} = \frac{a^2 + b^2}{2ab}$.
Since $(a - b)^2 > 0$,we have $a^2 + b^2 > 2ab$,therefore $\frac{a^2 + b^2}{2ab} > 1$,which implies $\frac{I_{zR}}{I_{zS}} > 1$.