Foundation January 2022 Paper 2 Q26
26 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 |
\(x\) = “144” – 90 (= 54) or \(x = \dfrac{\text{“}540\text{”} - 3 \times \text{“}144\text{”}}{2}\) (= 54) or \(x\) = 90 – “36” (= 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 |
| Working required 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
ALT
| Scheme | Marks |
|---|---|
| \(ADG\) = “144” – 2 × “36” (= 72) | M1 |
| \(JA\) is parallel to \(GD\) | M1 |
| \(DGA = DAG\) (\(y\)) [isosceles triangle] | M1 |
| Working required \(x = DGA = y\) There are other correct methods. Please check for correct working. Answer: shown | A1 |