B4 (i) The length of each side of an equilateral triangle is \( 6(\sqrt{3}-1) \mathrm{cm} \). Without using a calculator, express the area of the triangle in the form \( (a+b \sqrt{3}) \mathrm{cm}^{2} \), where \( a \) and \( b \) are integers. (ii) The triangle in part (i) is the base of a prism. Given that the volume of the prism is \( 9(\sqrt{3}-1) \mathrm{cm}^{3} \), show that the height of the prism can be expressed in the form \( (m+n \sqrt{3}) \mathrm{cm} \), where \( m \) and \( n \) are fractions in the simplest form.

Let the side length be \[ s=6(\sqrt{3}-1) \text{ cm}. \] **(i) Finding the area of the equilateral triangle** The area of an equilateral triangle is given by \[ A=\frac{\sqrt{3}}{4}s^2. \] First, compute \(s^2\): \[ s^2=[6(\sqrt{3}-1)]^2=36(\sqrt{3}-1)^2. \] Now, calculate \((\sqrt{3}-1)^2\): \[ (\sqrt{3}-1)^2= (\sqrt{3})^2 - 2\sqrt{3}\cdot 1 + 1^2 = 3-2\sqrt{3}+1=4-2\sqrt{3}. \] Thus, \[ s^2=36(4-2\sqrt{3})=144-72\sqrt{3}. \] Substitute into the area formula: \[ A=\frac{\sqrt{3}}{4}(144-72\sqrt{3})=\frac{144\sqrt{3}}{4}-\frac{72\sqrt{3}\cdot\sqrt{3}}{4}. \] Since \(\sqrt{3}\cdot\sqrt{3}=3\), \[ A=36\sqrt{3}-\frac{216}{4}=36\sqrt{3}-54. \] Thus, the area of the triangle is \[ A=(-54+36\sqrt{3}) \text{ cm}^2. \] **(ii) Finding the height of the prism** The volume \(V\) of the prism is given by the product of the base area and its height \(h\): \[ V=A\cdot h. \] We are given \[ V=9(\sqrt{3}-1) \text{ cm}^3, \] and from part (i) we have \[ A=36\sqrt{3}-54. \] Thus, \[ h=\frac{V}{A}=\frac{9(\sqrt{3}-1)}{36\sqrt{3}-54}. \] Factor the denominator: \[ 36\sqrt{3}-54=18(2\sqrt{3}-3). \] Then, \[ h=\frac{9(\sqrt{3}-1)}{18(2\sqrt{3}-3)}=\frac{\sqrt{3}-1}{2(2\sqrt{3}-3)}. \] To simplify, multiply the numerator and the denominator by the conjugate \((2\sqrt{3}+3)\): \[ h=\frac{(\sqrt{3}-1)(2\sqrt{3}+3)}{2(2\sqrt{3}-3)(2\sqrt{3}+3)}. \] Compute the denominator: \[ (2\sqrt{3}-3)(2\sqrt{3}+3)=(2\sqrt{3})^2-3^2=4\cdot3-9=12-9=3, \] so the full denominator is: \[ 2\cdot 3=6. \] Now expand the numerator: \[ (\sqrt{3}-1)(2\sqrt{3}+3)=\sqrt{3}\cdot2\sqrt{3}+\sqrt{3}\cdot3-1\cdot2\sqrt{3}-1\cdot3, \] \[ =2\cdot3+3\sqrt{3}-2\sqrt{3}-3=6+ \sqrt{3}-3=3+\sqrt{3}. \] Thus, \[ h=\frac{3+\sqrt{3}}{6}=\frac{3}{6}+\frac{\sqrt{3}}{6}=\frac{1}{2}+\frac{1}{6}\sqrt{3}. \] Therefore, the height of the prism is \[ h=\left(\frac{1}{2}+\frac{1}{6}\sqrt{3}\right) \text{ cm}. \]