4400 Higher November 2010 Paper 4 Q13
13 The size of each interior angle of a regular polygon is 11 times the size of each exterior angle.
Work out the number of sides the polygon has.
(4)
| Scheme | Marks |
|---|---|
\(11x + x = 180\) or \(12x = 180\) or for \(\dfrac{360}{n}\) or \(\dfrac{180(n - 2)}{n}\) | M1 |
(exterior angle =) 15 or \(\dfrac{360}{n} \times 11 = \dfrac{180(n - 2)}{n}\) oe or \(180 - \dfrac{360}{n} = 11 \times \dfrac{360}{n}\) | A1 |
\(\dfrac{360}{\text{“15”}}\) or simplified correct equation in which \(n\) appears only once eg \(360 \times 11 = 180(n - 2)\) or \(360 \times 11 = 180n - 360\) or \(12 \times \dfrac{360}{n} = 180\) | M1 |
| 24 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: May be implied by \(\dfrac{180}{12}\) or 15
A1: cao Award 4 marks for an answer of 24 unless clearly obtained by an incorrect method.