Higher January 2022 Paper 2R Q9
9 In the diagram, \(ABC\) is a right-angled triangle and \(DEF\) is a semicircular arc.
Diagram NOT accurately drawn
In triangle \(ABC\)
\(BC = 24\) cm angle \(ABC = 90^\circ\) angle \(BCA = 30^\circ\)
The points \(D\) and \(F\) lie on \(AC\) so that \(DF\) is the diameter of the semicircular arc \(DEF\)
The radius of the semicircular arc is 3 cm.
Work out the length of \(AFEDC\)
Give your answer correct to 2 significant figures.
(5)
| Scheme | Marks |
|---|---|
\(\cos 30 = \dfrac{24}{(AC)}\) or \(\sin \text{‘}{60}\text{’} = \dfrac{24}{(AC)}\) or \(\dfrac{\sin \text{‘}{60}\text{’}}{24} = \dfrac{\sin 90}{(AC)}\) oe | M1 |
\((AC =)\ \dfrac{24}{\cos 30}\ (= 16\sqrt{3} = 27.712\ldots)\) or \((AC =)\ \dfrac{24}{\sin \text{‘}{60}\text{’}}\ (= 16\sqrt{3} = 27.712\ldots)\) or \((AC =)\ \dfrac{24 \times \sin 90}{\sin \text{‘}{60}\text{’}}\) | M1 |
| \(\dfrac{1}{2} \times 2 \times \pi \times 3\ (= 3\pi = 9.424\ldots)\) | M1 |
| ‘27.712…’ + ‘9.424…’ – 2 × 3 | M1 |
| 31 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for correct trig ratio involving \(AC\)
M1: for a correct trig ratio for \(AC\)
M1: for using \(\pi \times 2 \times 3\) or \(2\pi \times 3\)
M1: for a complete method to find the length \(AFEDC\)
A1: accept answers in range from 31 to 31.15
M2 for use of tan and Pythagoras to obtain \(AC\)
\((AB =)\ 24\tan 30\ (= 13.856\ldots)\)
and
\(\sqrt{\text{‘}{13.856\ldots}\text{’}^2 + 24^2}\ (= 27.712\ldots)\)
If not M2, then M1 for use of tan and Pythagoras to obtain \(AC^2\)
\((AB =)\ 24\tan 30\ (= 13.856\ldots)\)
and
\(\text{‘}{13.856\ldots}\text{’}^2 + 24^2\ (= 768)\)