The given triangular sheet has an area equal to half of a square sheet having same side length as the length of the base of the triangle.Let MI of the triangular sheet about AC be I.I is half of the MI of the square plate about an axis passing through the centre O and being in the plane of the plate.Let the MI of the square about an axis XY be \\\\(I_{XY}\\\\).Due to symmetry\\\\(I_{AB} = I_{CD}\\\\) AND \\\\(I_{PQ} = I_{RS}\\\\)Let \\\\(I_{square}\\\\) be the MI a bout an axis passing through the centre of the square and perpendicular to the plane of the square plate.Then using Perpendicular Axis Theorem\\\\(I_{square} = I_{AB} + I_{CD}=I_{PQ} + I_{RS} \\\\)\\\\( \ herefore I_{square} = 2 I_{AB} = I_{PQ} \\\\)But, \\\\(I_{AB}= 2I\\\\)\\\\( \ herefore I = \\dfrac{1}{2}I_{AB} = \\dfrac{1}{2} *(\\dfrac{1}{2}\\dfrac{ML^2}{6})\\\\) ....(1) where M is the mass of the square plate.But the triangular sheet has half the mass of the square plate. Writing I in terms of the mass of the plate.\\\\( \ herefore I =\\dfrac{M}{2}*L^2/12=\\dfrac{1}{12}M_{triangle}L^2\\\\)