Higher November 2020 Unit 2 Q20
20. ACB is a sector of a circle with radius \(x\) cm and centre A, as shown below.
\(C\hat{E}A\) = 34°, \(A\hat{C}E\) = 100°, \(C\hat{A}E\) = 46° and CE = 12 cm.
Diagram not drawn to scale
Calculate the area of the shaded region BCE.
You must show all your working. [8]
| Scheme | Marks | Comments |
|---|---|---|
| \((x =)\ \dfrac{12}{\sin 46} \times \sin 34\) | M2 | A correct and complete method involving multiple trigonometric relationships leading to the correct answer may be seen at any stage to gain the method mark(s). M1 for \(\dfrac{x}{\sin 34} = \dfrac{12}{\sin 46}\) or equivalent. |
| \((x =)\) 9·3(28…cm) OR 9·32(cm) | A1 | |
| (Area of sector ACB=) \(\dfrac{46}{360} \times \pi \times 9{\cdot}3(28\ldots)^2\) | M1 | FT ‘their derived 9·3(28…)’. |
| = 34·9(3…cm2) | A1 | Answers in the range 34·7(cm2) to 35(cm2) or equivalent range on FT. |
| (Area of ACE=) ½×9·3(28…)×12×sin(100) | M1 | FT ‘their derived 9·3(28…cm) |
| = 55·1(2…cm2) | A1 | Answers in the range 54·95(cm2) to 55·13(cm2) or equivalent range on FT. |
| (Area of the shaded region BCE = 55·1… – 34·9… ) = 20·18(8…cm2) OR 20·2(cm2) | B1 | FT ‘their 34·9(…cm2)’ and ‘their 55·1(…cm2)’ provided previous M1, M1 (from area calculations) awarded AND ‘area of the shaded region’ > 0. This answer must be derived from the subtraction of ‘their areas’. |