Foundation January 2022 Paper 2R Q26
26 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° angle \(BCA\) = 30°
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…) or \((AC =)\dfrac{24}{\sin \text{‘}60\text{’}}\) (= \(16\sqrt{3}\) = 27.712…) 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…) | 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 \(2\pi \times 3\) correctly to find the arc length of the semicircle, or circumference of a circle with radius 3.
M1: for a complete correct 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...) and \(\sqrt{\text{‘}13.856...\text{’}^2 + 24^2}\) (= 27.712...)
If not M2, then M1 for use of tan and Pythagoras to obtain \(AC^2\)
(\(AB\) =) 24 tan 30 (= 13.856...) and \(\text{‘}13.856...\text{’}^2 + 24^2\) (= 768)