Higher November 2020 Paper 1R Q22
22 The diagram shows a regular octagon \(ABCDEFGH\).
Diagram NOT accurately drawn
Each side of the octagon has length 10 cm.
Find the area of the shaded region \(ACDEH\).
Give your answer correct to the nearest cm²
(6)
| Scheme | Marks |
|---|---|
| Ext angle of octagon = 360 ÷ 8 (= 45) or Int angle of octagon (8 − 2) × 180 ÷ 8 oe (= 135) | M1 |
e.g. 10 + 2 × 10 × sin45 (= \(10 + 10\sqrt{2}\) or 24.1...) or e.g. \(\dfrac{10\sin 112.5}{\sin 22.5}\) (= 24.1…) | M1 |
e.g. 10 × (“\(10 + 10\sqrt{2}\)”) (= 100 + \(100\sqrt{2}\) or 241.4...) or 10 × “24.1…” (= 241.4…) | M1 |
e.g. 10 × sin45° (= \(5\sqrt{2}\) or 7.07...) or e.g. \(\sqrt{10^2 + 10^2 - 2 \times 10 \times 10 \times \cos\text{“}{135}\text{”}}\) (= 18.4…) or \(\dfrac{10\sin\text{“}{135}\text{”}}{\sin 22.5}\) (= 18.4…) | M1 |
| e.g. 0.5 × “24.1...” × “7.07…” (= 85.3...) or \(0.5 \times 10 \times \text{“}{18.4\ldots}\text{”} \times \sin 112.5\) (= 85.3…) | M1 |
| 327 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for method to find the size of one exterior or one interior angle of a regular octagon
M1: method to find \(HE\) or \(AD\)
22.5 comes from (180 – “135”) ÷ 2
112.5 comes from “135” – “22.5”
M1: area \(ADEH\)
M1: finds perpendicular height of triangle \(ACD\) (may be found before, but must realise this is also height of triangle) or finds the length of \(AC\)
22.5 comes from (180 – “135”) ÷ 2
M1: finds the area of triangle \(ACD\)
112.5 comes from “135” – “22.5”
A1: accept 326 – 327
Alternative (splitting octagon into triangles and subtracting trapezium and triangle)
| Scheme | Marks |
|---|---|
| Ext angle of octagon = 360 ÷ 8 (= 45) or Int angle of octagon (8 − 2) × 180 ÷ 8 oe (= 135) or one of 8 angles at centre = 360 ÷ 8 (= 45) | M1 |
e.g. 0.5 × 10 × 5 × tan67.5 (= 60.35...) or \(0.5 \times \left(\dfrac{10\sin 67.5}{\sin 45}\right)^2 \times \sin 45 (= 60.35\ldots)\) or Octagon = 8 × “60.35” (= 482.8...) | M1 |
| e.g. 10 + 2 × 10 × sin45° (= \(10 + 10\sqrt{2}\) = 24.14..) | M1 |
| 0.5 × (10 + \(10 + 10\sqrt{2}\)) × \(5\sqrt{2}\) (= 120.71...) | M1 |
| 0.5 × 10 × 10 × sin135° (= 35.35...) | M1 |
| 327 | A1 |
Notes
M1: for method to find the size of one exterior or one interior angle of a regular octagon or method to find one angle at centre of octagon when split into 8 equal triangles
M1: Area of one triangle (one-eighth of octagon) or octagon
M1: Method to find \(HE\)
M1: Method to find area of trapezium \(HEGF\)
M1: Method to find area of triangle \(ABC\)
A1: accept 326 – 327