6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{266c4f52-f35f-459c-9184-836b0f3baf5b-20_570_608_287_669}
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\caption{Figure 3}
\end{figure}
The uniform lamina \(A B C D\) is a square with sides of length \(6 a\). The point \(E\) is the midpoint of side \(A D\). The triangle \(C D E\) is removed from the square to form the uniform lamina \(L\), shown in Figure 3. The centre of mass of \(L\) is at the point \(G\).
- Show that the distance of \(G\) from the side \(A B\) is \(\frac { 7 } { 3 } a\).
- Find the distance of \(G\) from the side \(A E\).
The mass of \(L\) is \(M\). A particle of mass \(k M\) is attached to \(L\) at the point \(E\). The lamina, with the particle attached, is freely suspended from \(A\) and hangs in equilibrium with the diagonal \(A C\) vertical.
- Find the value of \(k\).