Determine BC, giving your answer in the form $k\sqrt 3$, $k \in {\mathbb{Z}^ + }$ .

The point D lies on (BC), and (AD) is perpendicular to (BC). Determine AD.

${\text{B}}{{\text{C}}^2} = 12 + 81 + 2 \times 2\sqrt 3  \times 9 \times \frac{{\sqrt 3 }}{2} = 147$     M1A1

${\text{BC}} = 7\sqrt 3$     A1

[3 marks]

area of triangle ${\text{ABC}} = \frac{1}{2} \times 9 \times 2\sqrt 3  \times \frac{1}{2}{\text{ }}\left( { = \frac{{9\sqrt 3 }}{2}} \right)$     M1A1

therefore $\frac{1}{2} \times {\text{AD}} \times 7\sqrt 3  = \frac{{9\sqrt 3 }}{2}$     M1

${\text{AD}} = \frac{9}{7}$     A1

[4 marks]