Higher June 2022 Paper 1R Q11
11 The diagram shows a quadrilateral \(ABCD\)
Diagram NOT accurately drawn
In the diagram, \(ABC\) and \(DAC\) are right-angled triangles.
\(BC = 6\) cm \(AC = 7.5\) cm
The area of quadrilateral \(ABCD\) is 31.5 cm2
Work out the length of \(AD\)
(6)
| Scheme | Marks |
|---|---|
(\(AB^2\) =) \(7.5^2 - 6^2\) (= 20.25) or eg (\(BAC\) =) \(\sin^{-1}\left(\dfrac{6}{7.5}\right)\) (= 53.1…) or \(\cos(BCA) = \dfrac{6}{7.5}\) (= 0.8) | M1 |
(\(AB\) =) \(\sqrt{7.5^2 - 6^2}\) (= 4.5) or (\(AB\) =) \(\dfrac{6}{\tan \text{``}{53.1}\text{''}}\) (= 4.5…) or (\(AB\) =) \(7.5\cos \text{``}{53.1}\text{''}\) (= 4.5…) or (\(BCA\) =) \(\cos^{-1}\left(\dfrac{6}{7.5}\right)\) (= 36.8…) | M1 |
(Area \(ABC\) =) \(\dfrac{1}{2} \times 6 \times \text{``}{4.5}\text{''}\) (= 13.5) or (Area \(ABC\) =) \(\dfrac{1}{2} \times 6 \times 7.5 \times \sin(\text{``}{36.8}\text{''})\) (= 13.47… or 13.5) | M1 |
| (Area \(DAC\) =) 31.5 – “13.5” (= 18) or \(\text{``}{13.5}\text{''} + 0.5 \times 7.5 \times AD = 31.5\) oe | M1 |
| (\(AD\) =) (“18” ÷ 7.5) ÷ 0.5 oe | M1 |
| 4.8 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for a correct first step to find \(AB\) or a complete method to find angle \(BAC\) or a correct first step to find angle \(BCA\)
M1: for a complete method to find \(AB\) or angle \(BCA\)
M1: ft [their labelled \(AB\)] or [their labelled \(BCA\)]
eg for \(\dfrac{1}{2} \times 6 \times\) [their labelled \(AB\)] or \(\dfrac{1}{2} \times 6 \times 7.5 \times \sin\)[their labelled \(BCA\)]
M1: ft (dep on previous M1)
allow 31.5 – [their area]
M1: for a complete method to find \(AD\), dependent on correct working
A1: accept 4.78 – 4.81