\begin{enumerate}
  \item \hspace{0pt} [The centre of mass of a semicircular arc of radius $r$ is $\frac { 2 r } { \pi }$ from the centre.]
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \begin{tikzpicture}[>=Stealth, scale=0.18]
  % Define coordinates
  % A is at top, ADE goes down-left, ABC goes down-right
  % angle AEC = 90°, EC = 7a, ADE = 24a, ABC = 25a
  % ABD is a semicircular arc of radius 7a
  
  % Place E at origin, with AE vertical and EC horizontal (angle AEC = 90°)
  % AE = 24a, EC = 7a
  % Then AC = 25a (which checks out: 24² + 7² = 576 + 49 = 625 = 25²)
  
  \coordinate (E) at (0, 0);
  \coordinate (A) at (0, 24);
  \coordinate (C) at (7, 0);
  
  % ABC is straight from A to C, length 25a
  % B is on AC such that ABD is a semicircular arc of radius 7a
  % The semicircle has diameter AD where D is on AE
  % Center of semicircle is midpoint of AD
  % AB = arc starts at A, goes through B, ends at D
  % Since ABD is a semicircular arc of radius 7a, the diameter AD = 14a
  % D is on AE at distance 14a from A (since AD = 14a, ADE = 24a, so DE = 10a)
  
  \coordinate (D) at (0, 10); % 24 - 14 = 10 from E, i.e., y=10
  
  % Center of semicircle is midpoint of AD
  \coordinate (O) at (0, 17); % midpoint of A(0,24) and D(0,10)
  
  % B is on line AC. Direction from A to C: (7, -24)/25
  % B is at distance from A along AC such that |OB| = 7 (radius)
  % Parameterize: point on AC = A + t*(C - A) = (7t, 24 - 24t)
  % Distance from O(0,17): sqrt(49t² + (7-24t)²) = 7
  % 49t² + 49 - 336t + 576t² = 49
  % 625t² - 336t = 0
  % t(625t - 336) = 0 => t = 336/625
  
  \coordinate (B) at ({7*336/625}, {24 - 24*336/625});
  % B = (2352/625, 24 - 8064/625) = (3.7632, 11.0976)
  
  % Draw ADE (straight line from A to E)
  \draw[thick] (A) -- (E);
  
  % Draw EC (straight line from E to C)
  \draw[thick] (E) -- (C);
  
  % Draw ABC (straight line from A through B to C)
  \draw[thick] (A) -- (C);
  
  % Draw semicircular arc ABD
  % Center O at (0,17), radius 7a
  % A is at angle 90° from O, D is at angle 270° (or -90°)
  % The semicircle goes from A through B (on the right side) to D
  \draw[thick] (A) arc[start angle=90, end angle=-90, radius=7];
  
  % Labels
  \node[above] at (A) {$A$};
  \node[right] at (B) {$B$};
  \node[right] at (C) {$C$};
  \node[left] at (D) {$D$};
  \node[below left] at (E) {$E$};
  
  % Dimension label for ADE: 24a on the left
  \draw[<->] (-3, 0) -- (-3, 24);
  \node[left] at (-3, 12) {$24a$};
  
  % Dimension label for EC: 7a at bottom
  \draw[<->] (0, -3) -- (7, -3);
  \node[below] at (3.5, -3) {$7a$};
  
  % Right angle mark at E
  \draw (0, 1.5) -- (1.5, 1.5) -- (1.5, 0);
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Uniform wire is used to form the framework shown in Figure 2.\\
In the framework,

\begin{itemize}
  \item $A B C$ is straight and has length $25 a$
  \item $A D E$ is straight and has length $24 a$
  \item $A B D$ is a semicircular arc of radius $7 a$
  \item $E C = 7 a$
  \item angle $A E C = 90 ^ { \circ }$
  \item the points $A , B , C , D$ and $E$ all lie in the same plane
\end{itemize}

The distance of the centre of mass of the framework from $A E$ is $d$.\\
(a) Show that $d = \frac { 53 } { 2 ( 7 + \pi ) } a$

The framework is freely suspended from $A$ and hangs in equilibrium with $A C$ at angle $\alpha ^ { \circ }$ to the downward vertical.\\
(b) Find the value of $\alpha$.\\

\hfill \mbox{\textit{Edexcel M2 2022 Q6 [11]}}