5 In the diagram, ABED is a trapezium with right angles at E and D , and CED is a straight line. The lengths of AB and BC are $$ 2 d $$ and $$ (2 \sqrt{3}) d $$ respectively, and angles BAD and CBE are $$ 30^{\circ} $$ and $$ 60^{\circ} $$ respectively. (i) Find the length of CD in terms of $$ d $$.

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To find the length of $$ CD $$ in terms of $$ d $$, we can break down the problem step by step using the given information.

1. **Identify the trapezium and angles:**
   - We have trapezium $$ ABED $$ with right angles at $$ E $$ and $$ D $$.
   - $$ AB = 2d $$ and $$ BC = (2\sqrt{3})d $$.
   - Angle $$ BAD = 30^\circ $$ and angle $$ CBE = 60^\circ $$.

2. **Determine the coordinates of points:**
   - Let's place point $$ A $$ at the origin, $$ A(0, 0) $$.
   - Since $$ AB = 2d $$ and is horizontal, point $$ B $$ will be at $$ B(2d, 0) $$.
   - To find point $$ D $$, we need to use the angle $$ BAD $$ and the right angle at $$ D $$.

3. **Calculate the coordinates of point $$ D $$:**
   - The angle $$ BAD = 30^\circ $$ means that the line $$ AD $$ makes an angle of $$ 30^\circ $$ with the horizontal.
   - The vertical component of $$ AD $$ can be calculated using $$ AD = AB \cdot \tan(30^\circ) $$:
     $$
     AD = 2d \cdot \tan(30^\circ) = 2d \cdot \frac{1}{\sqrt{3}} = \frac{2d}{\sqrt{3}}
     $$
   - Therefore, the coordinates of point $$ D $$ are $$ D(0, \frac{2d}{\sqrt{3}}) $$.

4. **Calculate the coordinates of point $$ C $$:**
   - Since $$ CED $$ is a straight line and $$ E $$ is directly below $$ B $$ (due to the right angle at $$ E $$), the coordinates of point $$ E $$ are $$ E(2d, \frac{2d}{\sqrt{3}}) $$.
   - The angle $$ CBE = 60^\circ $$ means that the line $$ BC $$ makes an angle of $$ 60^\circ $$ with the horizontal.
   - The vertical component of $$ BC $$ can be calculated using $$ BC = (2\sqrt{3})d \cdot \sin(60^\circ) $$:
     $$
     BC_y = (2\sqrt{3})d \cdot \sin(60^\circ) = (2\sqrt{3})d \cdot \frac{\sqrt{3}}{2} = 3d
     $$
   - Therefore, the coordinates of point $$ C $$ are $$ C(2d, 0 + 3d) = C(2d, 3d) $$.

5. **Calculate the length of $$ CD $$:**
   - The coordinates of $$ C $$ are $$ (2d, 3d) $$ and the coordinates of $$ D $$ are $$ (0, \frac{2d}{\sqrt{3}}) $$.
   - The length $$ CD $$ can be calculated using the distance formula:
     $$
     CD = \sqrt{(x_C - x_D)^2 + (y_C - y_D)^2}
     $$
     $$
     CD = \sqrt{(2d - 0)^2 + (3d - \frac{2d}{\sqrt{3}})^2}
     $$
     $$
     = \sqrt{(2d)^2 + \left(3d - \frac{2d}{\sqrt{3}}\right)^2}
     $$

6. **Simplify the expression:**
   - First, simplify $$ 3d - \frac{2d}{\sqrt{3}} $$:
     $$
     3d - \frac{2d}{\sqrt{3}} = d\left(3 - \frac{2}{\sqrt{3}}\right) = d\left(\frac{3\sqrt{3} - 2}{\sqrt{3}}\right)
     $$
   - Now, substitute back into the distance formula:
     $$
     CD = \sqrt{(2d)^2 + \left(d\left(\frac{3\sqrt{3} - 2}{\sqrt{3}}\right)\right)^2}
     $$
     $$
     = \sqrt{4d^2 + d^2\left(\frac{(3\sqrt{3} - 2)^2}{3}\right)}
     $$
     $$
     = d\sqrt{4 + \frac{(3\sqrt{3} - 2)^2}{3}}
     $$

7. **Final expression for $$ CD $$:**
   - Thus, the length of $$ CD $$ in terms of $$ d $$ is:
     $$
     CD = d\sqrt{4 + \frac{(3\sqrt{3} - 2)^2}{3}}
     $$

This is the final answer for the length of $$ CD $$ in terms of $$ d $$.
The length of $$ CD $$ is $$ d\sqrt{4 + \frac{(3\sqrt{3} - 2)^2}{3}} $$.

5 In the diagram, ABED is a trapezium with right angles at E and D , and CED is a straight line. The lengths of AB and BC are \( 2 d \) and \( (2 \sqrt{3}) d \) respectively, and angles BAD and CBE are \( 30^{\circ} \) and \( 60^{\circ} \) respectively. (i) Find the length of CD in terms of \( d \).

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