Higher November 2023 Unit 2 Q18
18. The diagram below is a sketch of a regular 12-sided polygon, with centre \(O\).
The polygon has sides of length 5 cm.
\(AE\) and \(OE\) are straight lines.
Diagram not drawn to scale
(a) Calculate the length of \(CE\).
You must show all your working. [6]
You must show all your working. [6]
(b) Hence, calculate the area of triangle \(CGE\). [2]
From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Area of triangle \(= \dfrac{1}{2}ab\sin C\)
| Scheme | Marks | Comments |
|---|---|---|
| \(E\hat{C}G\) = 30(°) | B1 | Angles must be clearly stated or seen on the diagram. |
| Sight of \(C\hat{E}G\) = 45(°) AND \(C\hat{G}E\) = 105(°) | B2 | B1 for \(C\hat{E}G\) = 45(°) OR \(C\hat{G}E\) = 105(°) Sight of \(C\hat{E}G\) = 45(°) implies previous B1. FT ‘their \(E\hat{C}G\) = 30’ to calculate:
|
| \(CE = \dfrac{5 \times \sin 105}{\sin 45}\) | M2 | Use of correct angles implies B1B2. FT ‘their \(C\hat{G}E\) = 105’ AND ‘their \(C\hat{E}G\) = 45’. M1 for \(\dfrac{CE}{\sin 105} = \dfrac{5}{\sin 45}\) Award M2 for a correct use of ‘two-step’ trigonometric relationship. M0 otherwise. |
| = 6·8(3…cm) or \(\dfrac{5 + 5\sqrt{3}}{2}\) (cm) | A1 | Allow 7(cm) from correct working. |
18(a) Alternative version to find OC first and then use triangle OCE
| Scheme | Marks | Comments |
|---|---|---|
| \(C\hat{O}E\) = 30(°) | B1 | Angles must be clearly stated or seen on the diagram. |
| Sight of \(C\hat{E}G\) = 45(°) | B1 | Sight of \(C\hat{E}G\) = 45(°) implies previous B1 FT ‘their \(E\hat{C}G\) = 30’ to calculate: \(C\hat{E}G\) = 90 – ‘their 30’ × 3/2 |
| \(\left(OC = \dfrac{2{\cdot}5}{\sin(15)} = \right)\) 9·6(5…cm) or 9·7(cm) or \(\dfrac{5\sqrt{6} + 5\sqrt{2}}{2}\) (cm) | B1 | Or equivalent e.g. using the sine rule in triangle OCG FT ‘sin (½ of ‘their \(C\hat{O}E\) = 30’)’ |
| \(CE = \dfrac{9{\cdot}6(5\ldots) \times \sin 30}{\sin 45}\) | M2 | Use of correct side and angles implies B1B1B1. FT ‘their derived OC’ AND ‘their \(C\hat{O}E\) = 30’ AND ‘their \(C\hat{E}G\) = 45’. M1 for \(\dfrac{CE}{\sin 30} = \dfrac{9{\cdot}6(5\ldots)}{\sin 45}\) Award M2 for a correct use of ‘two-step’ trigonometric relationship. M0 otherwise. |
| = 6·8(3…cm) or \(\dfrac{5 + 5\sqrt{3}}{2}\) (cm) | A1 | Allow an answer of 7(cm) from correct working. |
Corrected by Westie's Workshop
- The board's mark scheme printed “FT ‘sin (½ of ‘their \(C\hat{O}E\) = 30’)”. Changed to “FT ‘sin (½ of ‘their \(C\hat{O}E\) = 30’)’” because a closing quotation mark was missing at the end of the phrase.
Original: see the board's mark scheme PDF (link above).
| Scheme | Marks | Comments |
|---|---|---|
| ½×5×6·8(3…)×sin30 | M1 | FT their solution in (a), provided it does not refer to a side other than CE AND ‘their \(E\hat{C}G\) = 30(°)’. |
| = 8·5(…cm2) or \(\dfrac{25 + 25\sqrt{3}}{8}\) (cm2) | A1 |