Find the values of $x$ and $y$ which satisfy the equations $2^x \div 4^y = \frac{1}{8},$ $9^y (\sqrt{3})^{2x} = 3\sqrt{3}.$ $ABCD$ is a trapezium with an area of $12 + 11\sqrt{3}\text{ cm}^2$, $AB = 8 - \sqrt{48}\text{ cm}$, $DC = 12 - \sqrt{12}\text{ cm}$ and $AD = BC$. Without using a calculator, obtain an expression for the perpendicular distance between $AB$ and $DC$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers.

1. Express the first equation in base 2

$2^x \div (2^2)^y = 2^{-3}$
$2^{x - 2y} = 2^{-3}$

2. Form the first linear equation

$x - 2y = -3 \quad \text{--- (1)}$

3. Express the second equation in base 3

$(3^2)^y (3^{\frac{1}{2}})^{2x} = 3^1 \cdot 3^{\frac{1}{2}}$
$3^{2y} \cdot 3^x = 3^{\frac{3}{2}}$
$3^{x + 2y} = 3^{\frac{3}{2}}$

4. Form the second linear equation

$x + 2y = \frac{3}{2} \quad \text{--- (2)}$

5. Solve the simultaneous linear equations

Adding (1) and (2):
$2x = -3 + \frac{3}{2} = -\frac{3}{2}$
$x = -\frac{3}{4}$

Substitute $x = -\frac{3}{4}$ into (2):
$-\frac{3}{4} + 2y = \frac{3}{2}$
$2y = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$
$y = \frac{9}{8}$

1. Simplify the given lengths of the parallel sides

$AB = 8 - \sqrt{48} = 8 - \sqrt{16 \times 3} = 8 - 4\sqrt{3}$
$DC = 12 - \sqrt{12} = 12 - \sqrt{4 \times 3} = 12 - 2\sqrt{3}$

2. Set up the equation for the area of the trapezium

Area $= \frac{1}{2}(AB + DC) \times h$
$12 + 11\sqrt{3} = \frac{1}{2}(8 - 4\sqrt{3} + 12 - 2\sqrt{3}) \times h$
$12 + 11\sqrt{3} = \frac{1}{2}(20 - 6\sqrt{3}) \times h$
$12 + 11\sqrt{3} = (10 - 3\sqrt{3}) \times h$

3. Express $h$ and rationalise the denominator

$h = \frac{12 + 11\sqrt{3}}{10 - 3\sqrt{3}}$
$h = \frac{12 + 11\sqrt{3}}{10 - 3\sqrt{3}} \times \frac{10 + 3\sqrt{3}}{10 + 3\sqrt{3}}$

4. Expand and simplify to find the final expression for $h$

$h = \frac{120 + 36\sqrt{3} + 110\sqrt{3} + 33(3)}{10^2 - (3\sqrt{3})^2}$
$h = \frac{120 + 146\sqrt{3} + 99}{100 - 27}$
$h = \frac{219 + 146\sqrt{3}}{73}$
$h = 3 + 2\sqrt{3}$