What is a surd? \begin{solutionordottedlines}[2cm] A surd is an irrational number that can be expressed in the form of a root, such as \(\sqrt{2}\), which cannot be simplified to a rational number. \end{solutionordottedlines} You must realise that a surd is an \textbf{exact value}, and anything that your calculator spits back at you is only an \emph{approximation}. In your calculator $\sqrt{2} = 1.414213562$, but in truth $\sqrt{2}$ never terminates, it continues forever as it is \emph{irrational}! Let's round some surds to 3 decimal places: \begin{examplebox} \section{Examples} \begin{multicols}{4} \begin{questions} \Question[1] \(\sqrt{19}=\)\fillin[4.359]{0.7in} \Question[1] \(\sqrt{37}=\)\fillin[6.083]{0.7in} \Question[1] \(\sqrt{161}=\)\fillin[12.688]{0.65in} \Question[1] \(\sqrt{732}=\)\fillin[27.055]{0.65in} \end{questions} \end{multicols} \end{examplebox} Now we shall find the \emph{exact} lengths of the following sides. Realise that unlike before our side lengths are not whole or even \emph{rational} numbers. \begin{exercisebox} \section{Exercises} \begin{questions} \begin{multicols}{3} \Question[1]\, \drawTriangle{bottom left}{5.83}{3}{5}{$3$}{$5$}{$x$}{0.5} \begin{solutionordottedlines}[1in] \( x = \sqrt{5.83^2 - 3^2} = \sqrt{34} \approx 5.831 \) \end{solutionordottedlines} \Question[1]\, \drawTriangle{top left}{8.944}{8}{4}{$4$}{$8$}{$y$}{0.5} \begin{solutionordottedlines}[1in] \( y = \sqrt{8^2 - 4^2} = \sqrt{48} \approx 6.928 \) \end{solutionordottedlines} \Question[1]\, \drawTriangle{bottom right}{7}{2}{6.708}{$2$}{$x$}{$7$}{0.5} \begin{solutionordottedlines}[1in] \( x = \sqrt{7^2 - 2^2} = \sqrt{45} \approx 6.708 \) \end{solutionordottedlines} \end{multicols} \Question[2] A door frame has height $1.8$m and width $1$m. Will a square piece of board $2.1$m wide fit through the opening? \begin{solutionordottedlines}[1.5in] The diagonal of the door frame is \(\sqrt{1.8^2 + 1^2} = \sqrt{4.24} \approx 2.059\), which is greater than $2.1$m, so the boar will fit. \end{solutionordottedlines} \Question[3] A signwriter leans his ladder against a wall so that he can paint a sign. The wall is vertical and the ground in front of the wall is horizontal. The signwriter's ladder is $4$m long.\\If the signwriter wants the top of the ladder to be $3.8$m above the grou when leaning against the wall, how far, correct to $1$ decimal place should the foot of the ladder be placed from the wall? \begin{solutionordottedlines}[1.5in] The distance from the wall is \(\sqrt{4^2 - 3.8^2} = \sqrt{0.36} \approx 0.6\)m. \end{solutionordottedlines} \end{questions} \end{exercisebox}