Higher November 2020 Paper 1 Q11
11 The diagram shows trapezium \(ABCD\) in which \(BC\) and \(AD\) are parallel.
Diagram NOT accurately drawn
The trapezium has exactly one line of symmetry.
\(BC = 8.4\) cm
\(AD = 17.6\) cm
The trapezium has area 179.4 cm2
Work out the size of angle \(ABC\).
Give your answer correct to 1 decimal place.
(6)
| Scheme | Marks |
|---|---|
| (\(AX\) =) (17.6 – 8.4) ÷ 2 (= 4.6) | M1 |
| 0.5 × (8.4 + 17.6) × \(h\) = 179.4 or 0.5 × ‘4.6’ × \(h\) + 0.5 × ‘4.6’ × \(h\) + 8.4 × \(h\) = 179.4 or 13 × \(h\) = 179.4 | M1 |
| (\(h\) =) 179.4 ÷ ‘13’ (=13.8) or (\(h\) =) 358.8 ÷ ‘26’ (=13.8) oe | M1 |
\(\tan ABX = \dfrac{\text{‘}4.6\text{’}}{\text{‘}13.8\text{’}}\) or \(\tan BAX = \dfrac{\text{‘}13.8\text{’}}{\text{‘}4.6\text{’}}\) | M1 |
\((ABX =)\; \tan^{-1}\left(\dfrac{\text{‘}4.6\text{’}}{\text{‘}13.8\text{’}}\right)\;(= 18.4)\) or \((BAX =)\; \tan^{-1}\left(\dfrac{\text{‘}13.8\text{’}}{\text{‘}4.6\text{’}}\right)\;(= 71.6)\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 108.4 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: where \(X\) is the foot of the perpendicular from \(B\) to \(AD\)
M1: ft their \(h\) dep on second M1
\((AB =)\; \sqrt{\text{‘}4.6\text{’}^2 + \text{‘}13.8\text{’}^2} = \sqrt{211.6}\) (= 14.546…) and one from
\(\sin ABX = \dfrac{\text{‘}4.6\text{’}}{\text{‘}\sqrt{211.6}\text{’}}\) or \(\sin BAX = \dfrac{\text{‘}13.8\text{’}}{\text{‘}\sqrt{211.6}\text{’}}\) or
\(\cos ABX = \dfrac{\text{‘}13.8\text{’}}{\text{‘}\sqrt{211.6}\text{’}}\) or \(\cos BAX = \dfrac{\text{‘}4.6\text{’}}{\text{‘}\sqrt{211.6}\text{’}}\) or
\(\sin ABX = \dfrac{\text{‘}4.6\text{’} \times \sin 90}{\text{‘}\sqrt{211.6}\text{’}}\) or
\(\cos ABX = \dfrac{\text{‘}\sqrt{211.6}\text{’}^2 + \text{‘}13.8\text{’}^2 - \text{‘}4.6\text{’}^2}{2 \times \text{‘}\sqrt{211.6}\text{’} \times \text{‘}13.8\text{’}}\)
A1: awrt 108.4