Higher June 2023 Paper 2 Q11
11 Here is a 9-sided regular polygon \(ABCDEFGHJ\), with centre \(O\)
Diagram NOT accurately drawn
\(ODK\) and \(FEK\) are straight lines.
Work out the value of \(x\)
(3)
| Scheme | Marks |
|---|---|
eg \(DEK = \dfrac{360}{9}\) or 40 or interior angle \(= \dfrac{(9 - 2) \times 180}{9}\) or 140 or \(OFK\) = 140 ÷ 2 (= 70) or \(FOK = \dfrac{2}{9} \times 360\) (= 80) or \(EDK\) = 180 – 0.5 × 140 (=110) Angles marked correctly (any exterior or interior angle) gains this mark | M1 |
| \(EDK\) = 110 and \(DEK\) = 40 or \(FOK\) = 80 and \(OFK\) = 70 or \(ODE\) = 70 and \(DEK\) = 40 or \(FED\) = 140 and \(EDK\) = 110 oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 30 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: method to find interior or exterior angle or correct interior or exterior angle stated or shown correctly on diagram or for using 70° for \(OFK\) or 80° for \(FOK\) or 110 for \(EDK\)
If a student has only found an interior or exterior angle and has clearly mixed up interior and exterior angles this mark cannot be awarded but can still award for any of the others angles stated
M1: For two correct angles that can lead directly to the answer in a single step (eg 180 – both angles or one angle minus the other)