# Surd (Trigonometry)

7DO NOT USE A CALCULATOR IN THIS QUESTION

In this question all lengths are in centimetres.

(CIE 0606/2021 March/12/Q7)
The diagram shows a trapezium $ABCDE$ such that $AB$ is parallel to $EC$ and $ABCD$ is a rectangle. It is given that $BC=\sqrt{17}+1, ED=\sqrt{17}-1$ and

$DC=\sqrt{17}+4$.

(a) Find the perimeter of the trapezium, giving your answer in the form $a+b\sqrt{17},$ where $a$ and $b$ are  integers. [3]

(b) Find the area of the trapezium, giving your answer in the form $c+d\sqrt{17},$ where $c$ and $d$ are  integers.

(c) Find $\tan AED,$ giving your answer in the form $\dfrac{e+f\sqrt{17}}{8},$ where $e$ and $f$ are integers. [2]

(d) Hence show that $\sec^2 AED = \dfrac{81+9\sqrt{17}}{32}.$  [2]

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**********math solution**********

From the diagram,

\begin{aligned} AB & =\sqrt{17}+4,AD=\sqrt{17}+1\\\\ A E^{2} & =(\sqrt{17}-1)^{2}+(\sqrt{17}+1)^{2} \\\\ & =(17-2 \sqrt{17}+1)+(17+2 \sqrt{17}+1) \\\\ & =36 \\\\ A E & =6 \end{aligned}

\begin{aligned} \text { (a)} & \text {Perimeter of trapezium } \\\\ & =6+((\sqrt{17}-1)+(\sqrt{17}+4))+(\sqrt{17}+1)+(\sqrt{17}+4) \\\\ & =4 \sqrt{17}+14\\\\ \text { (b)} & \text {Area of trapezium }\\\\ & =\frac{1}{2}((\sqrt{17}+4)+(\sqrt{17}-1+\sqrt{17}+4))(\sqrt{17}+1) \\\\ & =\frac{1}{2}(3 \sqrt{17}+7)(\sqrt{17}+1) \\\\ & =\frac{1}{2}(51+3 \sqrt{17}+7 \sqrt{17}+7) \\\\ & =\frac{1}{2}(58+10 \sqrt{17}) \\\\ & =29+5 \sqrt{17} \end{aligned}

\begin{aligned} \text {(c)} \tan A E D & =\frac{\sqrt{17}+1}{\sqrt{17}-1} \times \frac{\sqrt{17}+1}{\sqrt{17}+1} \\\\ & =\frac{17+2 \sqrt{17}+1}{17-1} \\\\ & =\frac{18+2 \sqrt{17}}{16} \\\\ & =\frac{9+\sqrt{17}}{8}\\\\ \text {(d)} \sec ^{2} A E D & =1+\tan ^{2} A E D \\\\ & =1+\left(\frac{9+\sqrt{17}}{8}\right)^{2} \\\\ & =1+\frac{81+18 \sqrt{17}+17}{64} \\\\ & =\frac{64+98+18 \sqrt{17}}{64} \\\\ & =\frac{162+18 \sqrt{17}}{64} \\\\ & =\frac{81+9 \sqrt{17}}{32} \end{aligned}

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