For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is:

Moment of inertia of a rectangular sheet of mass $$M$$ and dimensions $$a\times b $$ about an axis passing through its center of mass $$O$$ and perpendicular to its plane:

$$I_O = \frac{1}{12}M(a^2 + b^2)$$

$$I_O = \frac{1}{12}M(60^2 + 80^2) = \frac{1}{12}M(3600 + 6400) = \frac{10000}{12}M = \frac{2500}{3}M$$

$$\text{By the parallel axis theorem}, I_{O'} = I_O + Md^2$$

$$\text{where } d \text{ is the distance from the center of mass } O\left(\frac{a}{2}, \frac{b}{2}\right) \text{ to the corner point } O'(a, b)\text{:}$$

$$d^2 = \left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4} = \frac{60^2 + 80^2}{4} = \frac{10000}{4} = 2500\text{ cm}^2$$

$$I_{O'} = \frac{2500}{3}M + 2500M = 2500M\left(\frac{1}{3} + 1\right) = \frac{4}{3}(2500)M = \frac{10000}{3}M$$

$$\text{Ratio of the moments of inertia: } \frac{I_O}{I_{O'}} = \frac{\frac{2500}{3}M}{\frac{10000}{3}M} = \frac{2500}{10000} = \frac{1}{4}$$