1. Formula for Moment of Inertia The moment of inertia of a distributed mass about an axis is: \[ I = \int r_{\perp}^2 dm \] Here, \( r_{\perp} = y \), the vertical distance from the axis PQ.

2. Mass Element Consider a strip at height \( y \) with thickness \( dy \). The width of the triangle at this height, using similar triangles, is: \[ w(y) = b \left( 1 – \frac{y}{h} \right) \] where \( b \) is the length of base AB and \( h \) is the total height.
The mass of this strip is: \[ dm = \rho \times (\text{Thickness } t) \times (\text{Area } w(y) dy) \] \[ dm = \rho t b \left( 1 – \frac{y}{h} \right) dy \]

3. Integration Substitute \( dm \) into the integral: \[ I = \int_0^h y^2 \left[ \rho t b \left( 1 – \frac{y}{h} \right) \right] dy \] \[ I = \rho t b \int_0^h \left( y^2 – \frac{y^3}{h} \right) dy \] \[ I = \rho t b \left[ \frac{y^3}{3} – \frac{y^4}{4h} \right]_0^h \] \[ I = \rho t b \left( \frac{h^3}{3} – \frac{h^3}{4} \right) = \rho t b \left( \frac{h^3}{12} \right) \]

4. Expressing in terms of Mass Total Mass \( M = \text{Density} \times \text{Volume} = \rho \times t \times (\frac{1}{2}bh) \). Substituting \( \rho t b h = 2M \) into the equation for I: \[ I = (2M) \frac{h^2}{12} = \frac{M h^2}{6} \]

5. Comparison The result \( I = \frac{M h^2}{6} \) depends only on the total mass and the height. Since both plates have the same base, height, thickness, and material, they have the same Mass \( M \) and Height \( h \). Thus, their moments of inertia are equal.