A uniform square plate S(side c) and a uniform rectangular plate R(sides b, a) have identical areas and masses: 
Show that:
Show that:
- $\frac{\text{I}_\text{xR}}{\text{I}_\text{xS}}<1$
- $\frac{\text{I}_\text{ys}}{\text{I}_\text{ys}}>1$
- $\frac{\text{I}_{2\text{R}}}{\text{I}_{2\text{s}}}>1$
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According to the problem, Area of square = Area of rectangular plate $\Rightarrow c^2 = a \times b \Rightarrow c^2 = ab$
$\Rightarrow\ \frac{\text{I}_\text{XR}}{\text{I}_\text{XS}}=\Big(\frac{\text{b}}{\text{c}}\Big)^2<1$
$\Rightarrow{\text{I}_\text{XR}}<{\text{I}_\text{XS}}$
Now, $\text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big[\text{a}^2+\text{b}^2-2\text{c}^2\big]$
$\Rightarrow\ \text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big(\text{a}^2+\text{b}^2-2\text{ab}\big)$
$\Rightarrow\ \text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big(\text{a}-\text{b}\big)^2>0$
$\Rightarrow\ \frac{\text{I}_\text{zR}}{\text{I}_\text{zS}}>1$
- $\frac{\text{I}_\text{XR}}{\text{I}_\text{XS}}=\frac{\text{b}^2}{\text{c}^2}$
$\Rightarrow\ \frac{\text{I}_\text{XR}}{\text{I}_\text{XS}}=\Big(\frac{\text{b}}{\text{c}}\Big)^2<1$
$\Rightarrow{\text{I}_\text{XR}}<{\text{I}_\text{XS}}$
- $\frac{\text{I}_\text{YR}}{\text{I}_\text{RS}}=\frac{\text{a}^2}{\text{c}^2}$ (It is clear that a > c)
- $\text{I}_\text{zR}=\frac{1}{12}\text{M}(\text{a}^2+\text{b}^2)$
Now, $\text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big[\text{a}^2+\text{b}^2-2\text{c}^2\big]$
$\Rightarrow\ \text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big(\text{a}^2+\text{b}^2-2\text{ab}\big)$
$\Rightarrow\ \text{I}_\text{zR}-\text{I}_\text{zS}=\frac{1}{12}\text{M}\big(\text{a}-\text{b}\big)^2>0$
$\Rightarrow\ \frac{\text{I}_\text{zR}}{\text{I}_\text{zS}}>1$
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