Question 2: The diagram shows right angled \(\triangle ABC\) in which the length of \(\overline{AC}\) is \(\left(\sqrt{3}+\sqrt{5}\right)cm\). The area of \(\triangle ABC\) is \(\left(1+\sqrt{15}\right)cm^2\). Find the length \(\overline{AB}\) in the form \(\left(a\sqrt{3}+b\sqrt{5}\right)cm\), where \(a\) and \(b\) are integers.
Solution:
\[
\begin{align*}
Area&=\frac{1}{2}\left(\overline{AB}\right)\left(\overline{AC}\right)\\
\overline{AB}&=\frac{2 \times A}{\overline{AC}}\\
&=\frac{2 \times (1+\sqrt{15})}{\sqrt{3}+\sqrt{5}}\\
&=\frac{2\left( 1+\sqrt{15}\right)}{\sqrt{3} +\sqrt{5}} \times \frac{\sqrt{3} -\sqrt{5}}{\sqrt{3} -\sqrt{5}}\\
&=\frac{2\left( 1+\sqrt{15}\right)\left(\sqrt{3} -\sqrt{5}\right)}{\left(\sqrt{3}\right)^{2} -\left(\sqrt{5}\right)^{2}}\\
\end{align*}
\]
\[\small{=\frac{2\left(\sqrt{3} -\sqrt{5} +\sqrt{3}\sqrt{15} -\sqrt{5}\sqrt{15}\right)}{3-5}}\]
\[\scriptsize{=\frac{2\left(\sqrt{3} -\sqrt{5} +\sqrt{3}\sqrt{3\times 5} -\sqrt{5}\sqrt{3\times 5}\right)}{-2}}\]
\[\scriptsize{=\frac{2\left(\sqrt{3} -\sqrt{5} +\sqrt{3\times 3\times 5} -\sqrt{5\times 3\times 5}\right)}{-2}}\]
\[=\frac{\sqrt{3} -\sqrt{5} +\sqrt{3^{2} \times 5} -\sqrt{5^{2} \times 3}}{-1}\]
\[=-\left(\sqrt{3} -\sqrt{5} +3\sqrt{5} -5\sqrt{3}\right)\]
\[=-\left( -4\sqrt{3} +2\sqrt{5}\right)\]
\[=4\sqrt{3} -2\sqrt{5}\]