\includegraphics{figure_4}
A uniform square lamina \(ABCD\) has sides of length \(10\text{cm}\). The point \(E\) is on \(BC\) with \(EC = 7.5\text{cm}\), and the point \(F\) is on \(DC\) with \(CF = x\text{cm}\). The triangle \(EFC\) is removed from \(ABCD\) (see diagram). The centre of mass of the resulting shape \(ABEFD\) is a distance \(\bar{x}\text{cm}\) from \(CB\) and a distance \(\bar{y}\text{cm}\) from \(CD\).
- Show that \(\bar{x} = \frac{400 - x^2}{80 - 3x}\) and find a corresponding expression for \(\bar{y}\). [4]
The shape \(ABEFD\) is in equilibrium in a vertical plane with the edge \(DF\) resting on a smooth horizontal surface.
- Find the greatest possible value of \(x\), giving your answer in the form \(a + b\sqrt{2}\), where \(a\) and \(b\) are constants to be determined. [3]