Higher January 2022 Paper 2 Q11
11 The diagram shows a regular 10-sided polygon, \(ABCDEFGHIJ\)
Diagram NOT accurately drawn
Show that \(x = y\)
(4)
| Scheme | Marks |
|---|---|
\(\dfrac{360}{10}\) (= 36) ext angle or \(\dfrac{(10 - 2) \times 180}{10}\) (= 144) | M1 |
\(x = \text{``}{144}\text{''} - 90\;(= 54)\) or \(x = \dfrac{\text{``}{540}\text{''} - 3 \times \text{``}{144}\text{''}}{2}\;(= 54)\) or \(x = 90 - \text{``}{36}\text{''}\;(= 54)\) 54 on the diagram is insufficient – must see working | M1 |
| \(BAD = CDA = GDE = DGF = \dfrac{360 - 2 \times \text{``}{144}\text{''}}{2}\;(= 36)\) | M1 |
| There are other correct methods. Please check for correct working. Answer: \(x = 54\) \(y = 54\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: method to find interior or exterior angle.
(angles may be seen on diagram)
M1: method to find \(x\) (must show it is intended to be \(x\))
eg use of int angle − 90°
use of ext angle + \(x\) = 90°
use of pentagon \(GHIJA\)
All figures in “ ” must come from correct working
M1: A correct method to find an angle of 36° within the shape (not exterior angle)
or
36° shown in correct place in diagram
A1: dep on M3 to find each of \(x\) and \(y\) and the correct value of 54 for both from correct working
ALTERNATIVE
| Scheme | Marks |
|---|---|
| \(ADG = \text{``}{144}\text{''} - 2 \times \text{``}{36}\text{''}\;(= 72)\) | M1 |
| \(JA\) is parallel to \(GD\) | M1 |
| \(DGA = DAG\;(y)\) [isosceles triangle] | M1 |
| \(x = DGA = y\) There are other correct methods. Please check for correct working. Answer: shown | A1 |