According to the problem, Area of square = Area of rectangular plate ⇒ c2 = a × b ⇒ c2 = ab - $\frac{\text{I}_\text{XR}}{\text{I}_\text{XS}}=\frac{\text{b}^2}{\text{c}^2}$
It is clear from diagram that b < c.
$\Rightarrow\ \frac{\text{I}_\text{XR}}{\text{I}_\text{XS}}=\Big(\frac{\text{b}}{\text{c}}\Big)^2<1$
$\Rightarrow{\text{I}_\text{XR}}<{\text{I}_\text{XS}}$
- $\frac{\text{I}_\text{YR}}{\text{I}_\text{RS}}=\frac{\text{a}^2}{\text{c}^2}$ (It is clear that a > c)
$\Rightarrow\ \frac{\text{I}_\text{YR}}{\text{I}_\text{RS}}=\Big(\frac{\text{a}}{\text{c}}\Big)^2>1$
- $\text{I}_\text{zR}=\frac{1}{12}\text{M}(\text{a}^2+\text{b}^2)$
$\Rightarrow\ \text{I}_\text{zS}=\frac{1}{12}\text{M}(\text{c}^2+\text{c}^2)$
Now, $\text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big[\text{a}^2+\text{b}^2-2\text{c}^2\big]$
$\Rightarrow\ \text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big(\text{a}^2+\text{b}^2-2\text{ab}\big)$
$\Rightarrow\ \text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big(\text{a}-\text{b}\big)^2>0$
$\Rightarrow\ \frac{\text{I}_\text{zR}}{\text{I}_\text{zS}}>1$
