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(B) Consider a square plate of side $l$ and mass $2M$ formed by joining two such triangular sheets $ABC$ along the hypotenuse $AC.$Let the square be $

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$\frac{Ml^2}{12}$

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(B) Consider a square plate of side $l$ and mass $2M$ formed by joining two such triangular sheets $ABC$ along the hypotenuse $AC.$Let the square be $PQRS$ where $AC$ is a diagonal.The moment of inertia of a square plate of mass $M_{total} = 2M$ and side $l$ about its diagonal is given by $I_{diagonal} = \frac{1}{12} M_{total} l^2.$Substituting $M_{total} = 2M$,we get $I_{diagonal} = \frac{1}{12} (2M) l^2 = \frac{Ml^2}{6}.$Since the square is composed of two identical triangular sheets $ABC$ of mass $M$ each,the moment of inertia of one triangular sheet about the diagonal $AC$ is half of the moment of inertia of the square about the diagonal.Therefore,$I_{AC} = \frac{1}{2} 	imes I_{diagonal} = \frac{1}{2} 	imes \frac{Ml^2}{6} = \frac{Ml^2}{12}.$

The figure shows a thin metallic triangular sheet $ABC.$ The mass of the sheet is $M.$ The moment of inertia of the sheet about side $AC$ is

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