Higher November 2018 Paper 2 Q20
20. Quadrilateral EFGH is shown below.
Diagram not drawn to scale
Calculate the area of EFGH. [6]
| Scheme | Marks | Comments |
|---|---|---|
| \(EG^2 = 24.4^2 \;+\;\)\(20.3^2\)\(\;-\; 2 \times 24.4 \times 20.3 \times \cos 36^\circ\) \((= 206.0054\ldots)\) | M1 | |
| \(EG = 14.35(2\ldots\text{cm})\) or 14.4 (cm) | A1 | |
| Area \(EFG = \tfrac12\times24.4\times20.3\times\sin 36^\circ\) | M1 | |
| Area \(EGH = \tfrac12\times19.6\times14.35\times\sin 49^\circ\) | M1 | FT ‘their 14.35...’ provided M1 awarded for the cosine rule |
| Area \(EFG\) = 145.5 (cm2) to 145.6 (cm2) AND Area \(EGH\) = 106.1 (cm2) to 106.6 (cm2) | A1 | Use of 14.4 gives 106.5(046...cm2) |
| Following correct working, answer in the range 251.6 (cm2) to 252.2 (cm2) | A1 | Depends on the award of all previous M1 marks |
| (6) |
Corrected by Westie's Workshop
- The board's mark scheme printed “\(20.3^3\)”. Changed to “\(20.3^2\)” because the cosine rule uses 20.3 squared, and the value 206.0054… printed underneath only comes from 20.3².
Original: see the board's mark scheme PDF (link above).