Explanation

(b) Let the each side of square lamina is $d$.
So, $I_{EF} = I_{GH}$ (due to symmetry)
And $I_{AC} = I_{BD}$ (due to symmetry)
Now, according to theorem of perpendicular axis,
[image:7_image84.emf]
$I_{AC} + I_{BD} = I_{0}$
$\Rightarrow$ $2I_{AC} = I_{0}$ …(i)
and $I_{EF} + I_{GH} = I_{0}$
$\Rightarrow$ $2I_{EF} = I_{0}$ …(ii)
From Eqs. (i) and (ii), we get
$I_{AC} = I_{EF}$
$\therefore$ $I_{AD} = I_{EF} + \frac{md^{2}}{4}$
$= \frac{md^{2}}{12} + \frac{md^{2}}{4}\ \ \ \ \ \ \ \ \left( as\ I_{EF} = \frac{md^{2}}{12} \right)$
So, $I_{AD} = \frac{md^{2}}{3} = 4I_{EF}$