A uniform square plate S (side c) and a unifo | vedclass

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(C) Given that the areas are identical,$ab = c^2$. Since $a > b$,it follows that $a > c > b$.For the rectangular plate $R$: $I_{xR} = \frac{Mb^2}{12}$

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$(a)$,$(b)$,and $(c)$

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(C) Given that the areas are identical,$ab = c^2$. Since $a > b$,it follows that $a > c > b$.For the rectangular plate $R$: $I_{xR} = \frac{Mb^2}{12}$,$I_{yR} = \frac{Ma^2}{12}$,$I_{zR} = \frac{M(a^2 + b^2)}{12}$.For the square plate $S$: $I_{xS} = \frac{Mc^2}{12}$,$I_{yS} = \frac{Mc^2}{12}$,$I_{zS} = \frac{Mc^2}{6}$.Check $(a)$: $I_{xR}/I_{xS} = b^2/c^2$. Since $b Check $(b)$: $I_{yR}/I_{yS} = a^2/c^2$. Since $a > c$,$a^2/c^2 > 1$. Thus,$(b)$ is correct.Check $(c)$: $I_{zR}/I_{zS} = \frac{M(a^2 + b^2)/12}{Mc^2/6} = \frac{a^2 + b^2}{2c^2} = \frac{1}{2} \left( \frac{a^2}{c^2} + \frac{b^2}{c^2} \right)$.Since $ab = c^2$,we have $b^2/c^2 = c^2/a^2$. So,$I_{zR}/I_{zS} = \frac{1}{2} \left( \frac{a^2}{c^2} + \frac{c^2}{a^2} \right)$.Using the property $x + 1/x > 2$ for $x > 0$ (where $x = a^2/c^2$),it follows that $I_{zR}/I_{zS} > 1$. Thus,$(c)$ is correct.Therefore,all statements $(a)$,$(b)$,and $(c)$ are correct.

$A$ uniform square plate $S$ (side $c$) and a uniform rectangular plate $R$ (sides $b$,$a$) $(a > b)$ have identical areas and masses. Based on the figure,which of the following is correct?
$(a)\ I_{xR}/I_{xS} < 1$
$(b)\ I_{yR}/I_{yS} > 1$
$(c)\ I_{zR}/I_{zS} > 1$

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$A$ uniform square plate $S$ (side $c$) and a uniform rectangular plate $R$ (sides $b$,$a$) $(a > b)$ have identical areas and masses. Based on the figure,which of the following is correct?$(a)\ I_{xR}/I_{xS} < 1$$(b)\ I_{yR}/I_{yS} > 1$$(c)\ I_{zR}/I_{zS} > 1$