4.

\begin{figure}[h]
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  \includegraphics[alt={},max width=\textwidth]{1dea68fe-7916-41ed-894e-6b48f8d989bb-12_662_716_255_614}
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\caption{Figure 2}
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\end{figure}

The uniform lamina $L$, shown shaded in Figure 2, is formed by removing the square $P Q R V$, of side $2 a$, and the square $R S T U$, of side $4 a$, from a uniform square lamina $A B C D$, of side $8 a$. The lines $Q R U$ and $V R S$ are straight. The side $A D$ is parallel to $P V$ and the side $A B$ is parallel to $P Q$. The distance between $A D$ and $P V$ is $a$ and the distance between $A B$ and $P Q$ is $a$. The centre of mass of $L$ is at the point $G$.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of $G$ from the side $A D$ is $\frac { 42 } { 11 } a$

The mass of $L$ is $M$. A particle of mass $k M$ is attached to $L$ at $C$.

The lamina, with the attached particle, is freely suspended from $B$ and hangs in equilibrium with $B C$ making an angle of $45 ^ { \circ }$ with the horizontal.
\item Find the value of $k$.

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\begin{tabular}{|l|l|l|}
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2020 Q4 [9]}}