Higher November 2018 Paper 6 Q11
11 A regular polygon has \(n\) sides.
The polygon’s interior angle is 5 times the size of its exterior angle.
Find \(n\). [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 12 nfww | 5 | B1 for 5x and x soi and M1 for 6x = 180 oe and A1 for x = 30 and M1 for [n =] \(\dfrac{360}{\textit{their}\ 30}\) Alternative M1 for xn = 360 oe and M1 for 5xn = 180(n – 2) oe and M1 for 5×360 = 180(n – 2) oe and M1 for 10 = n – 2 Alternative M2 for use of two of [exterior angle =] 360/n [interior angle =] 180(n – 2)/n interior + exterior = 180 or M1 for use of one of the above AND M1dep for checking interior = 5 × exterior A1 for interior = 150 and exterior = 30 identified | For first M1 allow exterior = 360/n but not just 360/n Eliminates x Can be implied from a seen calculation or a list showing results of at least two trials (see the table of trials below) Dependent on M2 For full marks allow 12 as final answer from trial and improvement, provided interior angle = 150 and exterior angle = 30 are identified in working |
Appendix: table of trials for Q11
| sides | interior | exterior |
|---|---|---|
| 5 | 108.0 | 72.0 |
| 6 | 120.0 | 60.0 |
| 7 | 128.6 | 51.4 |
| 8 | 135.0 | 45.0 |
| 9 | 140.0 | 40.0 |
| 10 | 144.0 | 36.0 |
| 11 | 147.3 | 32.7 |
| 12 | 150.0 | 30.0 |
| 13 | 152.3 | 27.7 |
| 14 | 154.3 | 25.7 |
| 15 | 156.0 | 24.0 |