\begin{testbox}
\begin{questions}
    \question[4] For each of these right-angled triangles, find the value of the pronumeral, correct to 1 decimal place.
    \begin{parts}\begin{multicols}{2}
        \part \drawTriangle{bottom right}{8}{12}{14.42}{8}{12}{$a$}{0.2}
        \begin{solutionordottedlines}[1cm]
            $a = \sqrt{8^2 + 12^2} = \sqrt{64 + 144} = \sqrt{208} \approx 14.42$
        \end{solutionordottedlines}
        \part \drawTriangle{bottom left}{6.2}{11.7}{13.241}{6.2}{11.7}{$b$}{0.2}
        \begin{solutionordottedlines}[1cm]
            $b = \sqrt{6.2^2 + 11.7^2} = \sqrt{38.44 + 136.89} = \sqrt{175.33} \approx 13.24$
        \end{solutionordottedlines}
        \part \drawTriangle{top right}{5}{13.07}{14}{5}{$c$}{14}{0.2}
        \begin{solutionordottedlines}[1cm]
            $c = \sqrt{14^2 - 5^2} = \sqrt{196 - 25} = \sqrt{171} \approx 13.07$
        \end{solutionordottedlines}
        \part \drawTriangle{bottom right}{18}{37.947}{42}{18}{$d$}{42}{0.1}
        \begin{solutionordottedlines}[1cm]
            $d = \sqrt{42^2 - 18^2} = \sqrt{1764 - 324} = \sqrt{1440} \approx 37.95$
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] The lengths of the sides of a triangle are \(8.2 \mathrm{~cm}, 11.6 \mathrm{~cm}\) and \(14.3 \mathrm{~cm}\). Is the triangle right-angled?
    \begin{solutionordottedlines}[1in]
        To check if the triangle is right-angled, use Pythagoras' theorem:
        $8.2^2 + 11.6^2 \approx 67.24 + 134.56 = 201.8$
        $14.3^2 \approx 204.49$
        Since $201.8$ is not equal to $204.49$, the triangle is not right-angled.
    \end{solutionordottedlines}
    \question[6] In each part below, the two shorter side lengths of a right-angled triangle are given. State the length of the hypotenuse.
    \begin{parts}\begin{multicols}{2}
        \part \(3 \mathrm{~cm}, 4 \mathrm{~cm}\)
        \begin{solutionordottedlines}[2cm]
            Hypotenuse $= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \mathrm{~cm}$
        \end{solutionordottedlines}
        \part \(5 \mathrm{~cm}, 12 \mathrm{~cm}\)
        \begin{solutionordottedlines}[2cm]
            Hypotenuse $= \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \mathrm{~cm}$
        \end{solutionordottedlines}
        \part \(4 \mathrm{~cm}, 7.5 \mathrm{~cm}\)
        \begin{solutionordottedlines}[2cm]
            Hypotenuse $= \sqrt{4^2 + 7.5^2} = \sqrt{16 + 56.25} = \sqrt{72.25} \approx 8.5 \mathrm{~cm}$
        \end{solutionordottedlines}
        \part \(0.3 \mathrm{~cm}, 0.4 \mathrm{~cm}\)
        \begin{solutionordottedlines}[2cm]
            Hypotenuse $= \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \mathrm{~cm}$
        \end{solutionordottedlines}
        \part \(1 \mathrm{~cm}, 2.4 \mathrm{~cm}\)
        \begin{solutionordottedlines}[2cm]
            Hypotenuse $= \sqrt{1^2 + 2.4^2} = \sqrt{1 + 5.76} = \sqrt{6.76} \approx 2.6 \mathrm{~cm}$
        \end{solutionordottedlines}
        \part \(12 \mathrm{~cm}, 22.5 \mathrm{~cm}\)
        \begin{solutionordottedlines}[2cm]
            Hypotenuse $= \sqrt{12^2 + 22.5^2} = \sqrt{144 + 506.25} = \sqrt{650.25} \approx 25.5 \mathrm{~cm}$
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] A gardener is designing a rectangular lawn \(A B C D\). If \(A B=4.2 \mathrm{~m}\) and \(B C=3.15 \mathrm{~m}\), how far apart should \(A\) and \(C\) be to ensure \(\angle A B C=90^{\circ}\) ?
        \begin{center}
            \begin{tikzpicture}
                \draw (0,0) rectangle (4,2);
                \node at (0,0) [below left] {D};
                \node at (4,0) [below right] {C};
                \node at (0,2) [above left] {A};
                \node at (4,2) [above right] {B};
            \end{tikzpicture}
        \end{center}
        \begin{solutionordottedlines}[1in]
            Diagonal \(AC = \sqrt{AB^2 + BC^2} = \sqrt{4.2^2 + 3.15^2} = \sqrt{17.64 + 9.9225} = \sqrt{27.5625} \approx 5.25 \mathrm{~m}\)
        \end{solutionordottedlines}
    \question[3] A plane takes off and after climbing on a straight line path for a distance of \(1 \mathrm{~km}\), it has flown a horizontal distance of \(900 \mathrm{~m}\). What is the plane's altitude, correct to the nearest metre?
        \begin{solutionordottedlines}[1in]
            Altitude $= \sqrt{1^2 - 0.9^2} = \sqrt{1 - 0.81} = \sqrt{0.19} \approx 435.89 \mathrm{~m} \approx 436 \mathrm{~m}$
        \end{solutionordottedlines}
    \question[6] Simplify each of these surds.
    \begin{parts}\begin{multicols}{2}
        \part \(\sqrt{20}\)
        \begin{solutionordottedlines}[1cm]
            $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$
        \end{solutionordottedlines}
        \part \(\sqrt{75}\)
        \begin{solutionordottedlines}[1cm]
            $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$
        \end{solutionordottedlines}
        \part \(2 \sqrt{18}\)
        \begin{solutionordottedlines}[1cm]
            $2 \sqrt{18} = 2 \sqrt{9 \cdot 2} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}$
        \end{solutionordottedlines}
        \part \(4 \sqrt{50}\)
        \begin{solutionordottedlines}[1cm]
            $4 \sqrt{50} = 4 \sqrt{25 \cdot 2} = 4 \cdot 5\sqrt{2} = 20\sqrt{2}$
        \end{solutionordottedlines}
        \part \(5 \sqrt{108}\)
        \begin{solutionordottedlines}[1cm]
            $5 \sqrt{108} = 5 \sqrt{36 \cdot 3} = 5 \cdot 6\sqrt{3} = 30\sqrt{3}$
        \end{solutionordottedlines}
        \part \(9 \sqrt{27}\)
        \begin{solutionordottedlines}[1cm]
            $9 \sqrt{27} = 9 \sqrt{9 \cdot 3} = 9 \cdot 3\sqrt{3} = 27\sqrt{3}$
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] Write each number as the square root of a whole number.
    \begin{parts}\begin{multicols}{2}
        \part \(2 \sqrt{3}\)
        \begin{solutionordottedlines}[1cm]
            $2 \sqrt{3} = \sqrt{4 \cdot 3} = \sqrt{12}$
        \end{solutionordottedlines}
        \part \(3 \sqrt{2}\)
        \begin{solutionordottedlines}[1cm]
            $3 \sqrt{2} = \sqrt{9 \cdot 2} = \sqrt{18}$
        \end{solutionordottedlines}
        \part \(10 \sqrt{5}\)
        \begin{solutionordottedlines}[1cm]
            $10 \sqrt{5} = \sqrt{100 \cdot 5} = \sqrt{500}$
        \end{solutionordottedlines}
        \part \(4 \sqrt{7}\)
        \begin{solutionordottedlines}[1cm]
            $4 \sqrt{7} = \sqrt{16 \cdot 7} = \sqrt{112}$
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[8] Simplify:
    \begin{parts}\begin{multicols}{2}
        \part \(4 \sqrt{2}+7 \sqrt{2}\)
        \begin{solutionordottedlines}[1cm]
            $4 \sqrt{2}+7 \sqrt{2} = (4+7) \sqrt{2} = 11\sqrt{2}$
        \end{solutionordottedlines}
        \part \(8 \sqrt{3}-5 \sqrt{3}\)
        \begin{solutionordottedlines}[1cm]
            $8 \sqrt{3}-5 \sqrt{3} = (8-5) \sqrt{3} = 3\sqrt{3}$
        \end{solutionordottedlines}
        \part \(4 \sqrt{2} \times 5 \sqrt{3}\)
        \begin{solutionordottedlines}[1cm]
            $4 \sqrt{2} \times 5 \sqrt{3} = 20 \sqrt{2 \cdot 3} = 20\sqrt{6}$
        \end{solutionordottedlines}
        \part \(3 \sqrt{5} \times 4 \sqrt{7}\)
        \begin{solutionordottedlines}[1cm]
            $3 \sqrt{5} \times 4 \sqrt{7} = 12 \sqrt{5 \cdot 7} = 12\sqrt{35}$
        \end{solutionordottedlines}
        \part \(\sqrt{18}+\sqrt{32}\)
        \begin{solutionordottedlines}[1cm]
            $\sqrt{18}+\sqrt{32} = \sqrt{9 \cdot 2} + \sqrt{16 \cdot 2} = 3\sqrt{2} + 4\sqrt{2} = 7\sqrt{2}$
        \end{solutionordottedlines}
        \part \(\sqrt{27}-\sqrt{12}\)
        \begin{solutionordottedlines}[1cm]
            $\sqrt{27}-\sqrt{12} = \sqrt{9 \cdot 3} - \sqrt{4 \cdot 3} = 3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$
        \end{solutionordottedlines}
        \part \(4 \sqrt{12}+3 \sqrt{75}\)
        \begin{solutionordottedlines}[1cm]
            $4 \sqrt{12}+3 \sqrt{75} = 4 \sqrt{4 \cdot 3} + 3 \sqrt{25 \cdot 3} = 8\sqrt{3} + 15\sqrt{3} = 23\sqrt{3}$
        \end{solutionordottedlines}
        \part \(8 \sqrt{50}-2 \sqrt{98}\)
        \begin{solutionordottedlines}[1cm]
            $8 \sqrt{50}-2 \sqrt{98} = 8 \sqrt{25 \cdot 2} - 2 \sqrt{49 \cdot 2} = 40\sqrt{2} - 14\sqrt{2} = 26\sqrt{2}$
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[8] Expand and simplify:
    \begin{parts}\begin{multicols}{2}
        \part \(\sqrt{2}(\sqrt{3}+\sqrt{10})\)
        \begin{solutionordottedlines}[2cm]
        \( \sqrt{2}\sqrt{3} + \sqrt{2}\sqrt{10} = \sqrt{6} + \sqrt{20} = \sqrt{6} + 2\sqrt{5} \)
        \end{solutionordottedlines}
        \part \(\sqrt{3}(4 \sqrt{3}-5)\)
        \begin{solutionordottedlines}[2cm]
        \( \sqrt{3} \cdot 4\sqrt{3} - \sqrt{3} \cdot 5 = 4 \cdot 3 - 5\sqrt{3} = 12 - 5\sqrt{3} \)
        \end{solutionordottedlines}
        \part \(3 \sqrt{5}(2 \sqrt{2}-4 \sqrt{5})\)
        \begin{solutionordottedlines}[2cm]
        \( 3\sqrt{5} \cdot 2\sqrt{2} - 3\sqrt{5} \cdot 4\sqrt{5} = 6\sqrt{10} - 3 \cdot 4 \cdot 5 = 6\sqrt{10} - 60 \)
        \end{solutionordottedlines}
        \part \(2 \sqrt{2}(3 \sqrt{3}+4 \sqrt{2})\)
        \begin{solutionordottedlines}[2cm]
        \( 2\sqrt{2} \cdot 3\sqrt{3} + 2\sqrt{2} \cdot 4\sqrt{2} = 6\sqrt{6} + 2 \cdot 4 \cdot 2 = 6\sqrt{6} + 16 \)
        \end{solutionordottedlines}
        \part \((2 \sqrt{3}+1)(3 \sqrt{3}-2)\)
        \begin{solutionordottedlines}[2cm]
        \( (2\sqrt{3})(3\sqrt{3}) + (2\sqrt{3})(-2) + 1(3\sqrt{3}) - 1(2) = 6 \cdot 3 - 4\sqrt{3} + 3\sqrt{3} - 2 = 18 - \sqrt{3} - 2 \)
        \end{solutionordottedlines}
        \part \((4 \sqrt{2}+3)(5 \sqrt{2}-7)\)
        \begin{solutionordottedlines}[2cm]
        \( (4\sqrt{2})(5\sqrt{2}) + (4\sqrt{2})(-7) + 3(5\sqrt{2}) - 3(7) = 4 \cdot 5 \cdot 2 - 28\sqrt{2} + 15\sqrt{2} - 21 = 40 - 13\sqrt{2} - 21 \)
        \end{solutionordottedlines}
        \part \((3 \sqrt{2}-1)^{2}\)
        \begin{solutionordottedlines}[2cm]
        \( (3\sqrt{2})^2 - 2(3\sqrt{2})(1) + 1^2 = 9 \cdot 2 - 6\sqrt{2} + 1 = 18 - 6\sqrt{2} + 1 \)
        \end{solutionordottedlines}
        \part \((\sqrt{5}+1)^{2}\)
        \begin{solutionordottedlines}[2cm]
        \( (\sqrt{5})^2 + 2(\sqrt{5})(1) + 1^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5} \)
        \end{solutionordottedlines}
        \part \((2 \sqrt{5}+7 \sqrt{2})(2 \sqrt{5}-7 \sqrt{2})\)
        \begin{solutionordottedlines}[2cm]
        \( (2\sqrt{5})^2 - (7\sqrt{2})^2 = 4 \cdot 5 - 7^2 \cdot 2 = 20 - 49 \cdot 2 = 20 - 98 = -78 \)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] Express each number with a rational denominator.
    \begin{parts}\begin{multicols}{2}
        \part \[\frac{3}{\sqrt{3}}\]
        \begin{solutionordottedlines}[2cm]
        \( \frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \)
        \end{solutionordottedlines}
        \part \[\frac{2 \sqrt{5}}{\sqrt{5}}\]
        \begin{solutionordottedlines}[2cm]
        \( \frac{2 \sqrt{5}}{\sqrt{5}} = 2 \)
        \end{solutionordottedlines}
        \part \[\frac{2}{4 \sqrt{3}}\]
        \begin{solutionordottedlines}[2cm]
        \( \frac{2}{4\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{12} = \frac{\sqrt{3}}{6} \)
        \end{solutionordottedlines}
        \part \[\frac{5 \sqrt{3}}{3 \sqrt{2}}\]
        \begin{solutionordottedlines}[2cm]
        \( \frac{5 \sqrt{3}}{3 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{6}}{3 \cdot 2} = \frac{5\sqrt{6}}{6} \)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] Express each number with a rational denominator.
    \begin{parts}\begin{multicols}{2}
        \part \[\frac{3 \sqrt{2}}{\sqrt{5}+2}\]
        \begin{solutionordottedlines}[3cm]
        \( \frac{3 \sqrt{2}}{\sqrt{5}+2} \cdot \frac{\sqrt{5}-2}{\sqrt{5}-2} = \frac{3\sqrt{2}(\sqrt{5}-2)}{5-4} = 3\sqrt{10} - 6\sqrt{2} \)
        \end{solutionordottedlines}
        \part \[\frac{\sqrt{3}}{2 \sqrt{3}-1}\]
        \begin{solutionordottedlines}[3cm]
        \( \frac{\sqrt{3}}{2\sqrt{3}-1} \cdot \frac{2\sqrt{3}+1}{2\sqrt{3}+1} = \frac{\sqrt{3}(2\sqrt{3}+1)}{3-1} = \frac{2 \cdot 3 + \sqrt{3}}{2} = 3 + \frac{\sqrt{3}}{2} \)
        \end{solutionordottedlines}
        \part \[\frac{3 \sqrt{2}+1}{\sqrt{5}+2}\]
        \begin{solutionordottedlines}[3cm]
        \( \frac{3 \sqrt{2}+1}{\sqrt{5}+2} \cdot \frac{\sqrt{5}-2}{\sqrt{5}-2} = \frac{(3\sqrt{2}+1)(\sqrt{5}-2)}{5-4} = 3\sqrt{10} - 6\sqrt{2} + \sqrt{5} - 2 \)
        \end{solutionordottedlines}
        \part \[\frac{\sqrt{2}+1}{\sqrt{3}+\sqrt{2}}\]
        \begin{solutionordottedlines}[3cm]
        \( \frac{\sqrt{2}+1}{\sqrt{3}+\sqrt{2}} \cdot \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{2}(\sqrt{3}-\sqrt{2})+1(\sqrt{3}-\sqrt{2})}{3-2} = \sqrt{6} - 2 + \sqrt{3} - \sqrt{2} \)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[6] For the rectangular prism to the right, calculate the length of each of these intervals. Give your answers as surds in simplest form.
    \begin{center}
    \includegraphics[width=\textwidth]{prism}
    \end{center}
    \begin{parts}\begin{multicols}{2}
        \part \(E G\)
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
        \part \(E C\)
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
        \part \(H C\)
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
        \part \(G M\)
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
        \part \(C M\)
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
        \part \(A M\)
        \begin{solutionordottedlines}[1cm]
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] A vase is in the shape of a cylinder with base radius \(4 \mathrm{~cm}\) and height \(10 \mathrm{~cm}\). What is the length, correct to 1 decimal place, of the longest flower stem that can just fit in the vase?
        \begin{solutionordottedlines}[1in]
        The longest stem that can fit into the vase is the diagonal of the cylinder. This can be calculated using the Pythagorean theorem for the diagonal (d) of the cylinder with radius (r) and height (h):
        \[ d = \sqrt{r^2 + h^2} \]
        \[ d = \sqrt{4^2 + 10^2} \]
        \[ d = \sqrt{16 + 100} \]
        \[ d = \sqrt{116} \]
        \[ d \approx 10.8 \text{ cm} \]
        \end{solutionordottedlines}
\end{questions}
\end{testbox}
