4MA0 Higher June 2017 Paper 4R Q24
24 The diagram shows triangle \(KLM\).
Diagram NOT accurately drawn
\(KLP\) is a sector of a circle with centre \(L\) and radius 10.4 cm.
The region of the triangle outside the sector is shown shaded in the diagram.
Calculate the area of the shaded region.
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
Eg \(12.6^2 = 10.4^2 + 18^2 - 2 \times 10.4 \times 18 \times \cos L\) or \(158.76 = 108.16 + 324 - 374.4\cos L\) Note: \(\cos K = \dfrac{-57.08}{262.08}\) (= −0.217(796…)) and \(K\) = 102.(579…) \(\cos M = \dfrac{374.6}{453.6}\) (= 0.825(837…)) and \(M\) = 34.3(264…) | M1 |
Eg \(\cos L = \dfrac{10.4^2 + 18^2 - 12.6^2}{2 \times 10.4 \times 18}\) or \(\cos L = \dfrac{273.4}{374.4}\) or \(\cos L\) = 0.73(0235...) oe or \(L\) = 43.0(938...) | A1 |
| (Area of sector =) \(\dfrac{\text{“}43.0(938...)\text{”}}{360} \times \pi \times 10.4^2\) or 40.6(752...) | M1 |
| B1 | |
| 23.3 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: Correct substitution into cosine rule to find \(L\) or
For \(\cos K = \dfrac{10.4^2 + 12.6^2 - 18^2}{2 \times 10.4 \times 12.6}\) oe
AND
\(\sin L = \dfrac{12.6\sin\text{“}102.(579...)\text{”}}{18}\) or
For \(\cos M = \dfrac{18^2 + 12.6^2 - 10.4^2}{2 \times 18 \times 12.6}\) oe
AND
\(\sin L = \dfrac{12.6\sin\text{“}34.3(264...)\text{”}}{10.4}\)
A1: Rearranging cosine rule correctly.
Accept \(L = 43^\circ\)
Accept 43.0(938...) rounded or truncated to at least 3 SF
M1: Dep on at least M1
Accept 40.5 – 40.7
B1: For (area of triangle=) 63.9(471...)
Accept 63.8 – 64.0
A1: Accept 23.2 – 23.3