Moment of inertia of a square sheet about one side is I. Then the moment of inertia of the same square sheet about one of its diagonals is 

(3)

$\begin{array}{l}Itlbethelengthofeachside,ThenI=\frac{1}{3}m{l}^2\\ \therefore I_{x\text{'}}=I_{AB}-m.{\left(\frac{a}{2}\right)}^2\\ =\frac{1}{3}m{l}^2-\frac{1}{4}m{a}^2=\frac{1}{12}m{a}^2\\ Byperpendicularaxistheorem\\ I_o=I_{xi}+I_{yj}=2I_x\\ Also{I}_o=I_{AC}+I_{BO}=2I_{AC}\\ \therefore I_{AC}=I_x=\frac{1}{12}m{x}^2=\frac{1}{12}\times 3I=\frac{1}{4}I.\\ \end{array}$