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For a any plane figure having x and y axis on the figure plane and z axis perpendicular to the figure plane (see figure attached) it can be shown that
`I_x =I_y =(1/2)*I_z`
regardless where is located the intersection of the x,y and z axis, and regardless how x and y axis are rotated in the plane of the figure (see the figure).
For a thin square sheet having sides of length l, the momentum of inertia with respect to a perpendicular axis through its center is
Therefore, relative to the x and y axis parallel to square sides in the figure plane we have
`I_x =I_y = (M*l^2)/12` .
Since the diagonals in a square are also perpendicular to each other than we can rotate x and y axis to become x' and y' axis and the momentum above stay the same.
`I_(x') =I_(y') =(M*l^2)/12`
Now the momentum of the right triangle PQR in the figure with respect to the axis PR is just half of the above momentum (from symmetry considerations).
`I_("triangle") =(1/2)*I_(y') =(M*l^2)/24`
The correct answer is A) `(M*l^2)/24`
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