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\begin{questionparts}
\item
A uniform lamina $OXYZ$ is in the shape of the trapezium shown in the diagram. It is right-angled at $O$ and $Z$, and $OX$ is parallel to $YZ$. The lengths of the sides are given by $OX=9\,$cm, $XY=41\,$cm, $YZ=18\,$cm and $ZO=40\,$cm. Show that its centre of mass is a distance $7\,$cm from the edge $OZ$.
\begin{center}
\begin{tikzpicture}
% Define coordinates based on the PSTricks code
\coordinate (O) at (3.1,0.87);
\coordinate (X) at (4.77,0.87);
\coordinate (Z) at (3.1,4.85);
\coordinate (Y) at (6.04,4.85);
% \coordinate (W) at (8,6);
% \coordinate (V) at (10.71,6);
% \coordinate (T) at (8,2);
% \coordinate (U) at (9.45,1.99);
% Draw the main quadrilateral OZYX
\draw (Z) -- (Y);
\draw (Y) -- (X);
\draw (O) -- (Z);
\draw (O) -- (X);
% Draw the upper part
% \draw (Z) -- (W);
% \draw (W) -- (V);
% \draw (V) -- (Y);
% \draw (V) -- (U);
% \draw (U) -- (X);
% % Draw dashed lines for hidden edges
% \draw[dashed] (W) -- (T);
% \draw[dashed] (T) -- (O);
% \draw[dashed] (T) -- (U);
% Draw the small vertical line near W
% \draw (W) -- (8,5.33);
% Add vertex labels
\node[below left] at (O) {$O$};
\node[below right] at (X) {$X$};
\node[above left] at (Z) {$Z$};
\node[above right] at (Y) {$Y$};
% \node[above] at (W) {$W$};
% \node[above] at (V) {$V$};
% \node[above right] at (T) {$T$};
% \node[below right] at (U) {$U$};
% Add measurements
\node[below] at ($(Z)!0.5!(Y)$) {$18$};
\node[right] at ($(X)!0.5!(Y)$) {$41$};
\node[below] at ($(O)!0.5!(X)$) {$9$};
\node[left] at ($(Z)!0.5!(O)$) {$40$};
% \node[above] at ($(Z)!0.5!(W)$) {$d$};
% \node[below right] at ($(X)!0.5!(U)$)  {$d$};
\end{tikzpicture}
\end{center}
\item
The diagram shows a tank with no lid made of thin sheet metal. The base $OXUT$, the back $OTWZ$ and the front $XUVY$ are rectangular, and each end is a trapezium as in part \textbf{(i)}. The width of the tank is $d\,$cm.
\begin{center}
\begin{tikzpicture}
% Define coordinates based on the PSTricks code
\coordinate (O) at (3.1,0.87);
\coordinate (X) at (4.77,0.87);
\coordinate (Z) at (3.1,4.85);
\coordinate (Y) at (6.04,4.85);
\coordinate (W) at (8,6);
\coordinate (V) at (10.71,6);
\coordinate (T) at (8,2);
\coordinate (U) at (9.45,1.99);
% Draw the main quadrilateral OZYX
\draw (Z) -- (Y);
\draw (Y) -- (X);
\draw (O) -- (Z);
\draw (O) -- (X);
% Draw the upper part
\draw (Z) -- (W);
\draw (W) -- (V);
\draw (V) -- (Y);
\draw (V) -- (U);
\draw (U) -- (X);
% Draw dashed lines for hidden edges
\draw[dashed] (W) -- (T);
\draw[dashed] (T) -- (O);
\draw[dashed] (T) -- (U);
% Draw the small vertical line near W
\draw (W) -- (8,5.33);
% Add vertex labels
\node[below left] at (O) {$O$};
\node[below right] at (X) {$X$};
\node[above left] at (Z) {$Z$};
\node[above right] at (Y) {$Y$};
\node[above] at (W) {$W$};
\node[above] at (V) {$V$};
\node[above right] at (T) {$T$};
\node[below right] at (U) {$U$};
% Add measurements
\node[below] at ($(Z)!0.5!(Y)$) {$18$};
\node[right] at ($(X)!0.5!(Y)$) {$41$};
\node[below] at ($(O)!0.5!(X)$) {$9$};
\node[left] at ($(Z)!0.5!(O)$) {$40$};
\node[above] at ($(Z)!0.5!(W)$) {$d$};
\node[below right] at ($(X)!0.5!(U)$)  {$d$};
\end{tikzpicture}
\end{center}
 Show that the centre of mass of the tank, when empty,
is a distance 
\[
\frac {3(140+11d)}{5(12+d)}\,\text{cm}
\]
from the back of the tank.
The tank is then filled with a liquid.  The mass per unit volume of this liquid  is $k$ times the mass per unit area of the sheet metal. In the case $d=20$, find an expression for the distance of the centre of mass of the filled tank from the back of the tank.
\end{questionparts}

Solution (Optional)

\begin{questionparts}
\item
\begin{center}
\begin{tikzpicture}

% Define coordinates based on the PSTricks code
\coordinate (O) at (3.1,0.87);
\coordinate (X) at (4.77,0.87);
\coordinate (Z) at (3.1,4.85);
\coordinate (Y) at (6.04,4.85);
\coordinate (x) at (4.77,4.85);
% \coordinate (W) at (8,6);
% \coordinate (V) at (10.71,6);
% \coordinate (T) at (8,2);
% \coordinate (U) at (9.45,1.99);

% Draw the main quadrilateral OZYX
\draw (Z) -- (Y);
\draw (Y) -- (X);
\draw (O) -- (Z);
\draw (O) -- (X);

\draw[dashed] (X) -- (x);

% Draw the upper part
% \draw (Z) -- (W);
% \draw (W) -- (V);
% \draw (V) -- (Y);
% \draw (V) -- (U);
% \draw (U) -- (X);

% % Draw dashed lines for hidden edges
% \draw[dashed] (W) -- (T);
% \draw[dashed] (T) -- (O);
% \draw[dashed] (T) -- (U);

% Draw the small vertical line near W
% \draw (W) -- (8,5.33);

% Add vertex labels
\node[below left] at (O) {$O$};
\node[below right] at (X) {$X$};
\node[above] at (x) {$X'$};
\node[above left] at (Z) {$Z$};
\node[above right] at (Y) {$Y$};
% \node[above] at (W) {$W$};
% \node[above] at (V) {$V$};
% \node[above right] at (T) {$T$};
% \node[below right] at (U) {$U$};

% Add measurements
\node[below] at ($(Z)!0.5!(Y)$) {$18$};
\node[right] at ($(X)!0.5!(Y)$) {$41$};
\node[below] at ($(O)!0.5!(X)$) {$9$};
\node[left] at ($(Z)!0.5!(O)$) {$40$};
% \node[above] at ($(Z)!0.5!(W)$) {$d$};
% \node[below right] at ($(X)!0.5!(U)$)  {$d$};

\end{tikzpicture}
\end{center}

\begin{array}{c|c|c|c}
& OXX'Z & XX'Y & OXYZ \\ \hline
\text{Area} & 360 & 180 & 540\\
\text{COM} & \binom{4.5}{20} & \binom{12}{\frac{80}{3}} & \binom{\overline{x}}{\overline{y}}
\end{array}

\begin{align*}
&& 2 \binom{3}{20} + \binom{12}{\frac{80}{3}} &= 3 \binom{\overline{x}}{\overline{y}} \\
\Rightarrow &&  \binom{\overline{x}}{\overline{y}} &= \frac13  \binom{21}{\frac{200}{3}} \\
&&&= \binom{7}{\frac{200}{9}}
\end{align*}

ie, the centre of mass is $7\text{ cm}$ from $OZ$

\item \begin{align*}
&& \underbrace{540 \cdot 7}_{OXYZ} + \underbrace{540 \cdot 7}_{TUVW} + \underbrace{40d\cdot 0}_{OTWZ} + \underbrace{9d\cdot 4.5}_{OXUT} + \underbrace{41d \cdot 13.5}_{XUVY} &= (540+540+40d+9d+41d) \overline{x} \\
\Rightarrow && \overline{x} &= \frac{540\cdot 14 + 50d \cdot 4.5 + 41d \cdot 9}{1080 + 90d} \\
&&&= \frac{90 \cdot 84 + 225d + 369d}{1080+90d} \\
&&&=  \frac{90 \cdot 84 + 594d}{1080+90d} \\
&&&= \frac{54(140+11d)}{90(12+d)} \\
&&&= \frac{3(140+11d)}{5(12+d)}
\end{align*}

The volume of the prizm is $540d$, it's center of mass is $7$.

For the tank, it COM is $\frac{3(140+11\cdot20)}{5(12+20)} = \frac{27}4$ and area is $2880$

Therefore for the combined shape we have:

\begin{align*}
&& 540dk \cdot 7 + 2880 \cdot \frac{27}{4} &= (540 \cdot20 k+2880) \overline{x} \\
\Rightarrow && \overline{x} &= \frac{720(150k+27)}{720(15k + 4)} \\
&&&= \frac{3(50k+9)}{15k+4}
\end{align*}

\begin{align*}
&& 
\end{align*}

\end{questionparts}

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