C2 June 2014 Q5
5.
The shape \(ABCDEA\), as shown in Figure 2, consists of a right-angled triangle \(EAB\) and a triangle \(DBC\) joined to a sector \(BDE\) of a circle with radius 5 cm and centre \(B\).
The points \(A\), \(B\) and \(C\) lie on a straight line with \(BC = 7.5\) cm.
Angle \(EAB = \dfrac{\pi}{2}\) radians, angle \(EBD = 1.4\) radians and \(CD = 6.1\) cm.
| Scheme | Marks |
|---|---|
| Area \(BDE = \dfrac{1}{2}(5)^2(1.4)\) | M1 |
| \(= 17.5\ (\text{cm}^2)\) | A1 |
| (2) |
Notes
M1: Use of the correct formula or method for the area of the sector
A1: 17.5 oe
| Scheme | Marks |
|---|---|
| Parts (b) and (c) can be marked together | |
| \(6.1^2 = 5^2 + 7.5^2 - (2 \times 5 \times 7.5\cos DBC)\) or \(\cos DBC = \dfrac{5^2 + 7.5^2 - 6.1^2}{2 \times 5 \times 7.5}\) (or equivalent) | M1 |
| Angle \(DBC = 0.943201\ldots\) awrt 0.943 | A1 |
| (2) |
Notes
M1: A correct statement involving the angle \(DBC\)
Note that work for (b) may be seen on the diagram or in part (c)
| Scheme | Marks |
|---|---|
| Note that candidates may work in degrees in (c) (Angle \(DBC = 54.04\ldots.\text{degrees}\)) | |
| Area \(CBD = \dfrac{1}{2}5(7.5)\sin(0.943)\) | M1 |
| Angle \(EBA = \pi - 1.4 - \text{"}0.943\text{"}\) (Maybe seen on the diagram) | M1 |
| \(AB = 5\cos(\pi - 1.4 - \text{"}0.943\text{"})\) or \(AE = 5\sin(\pi - 1.4 - \text{"}0.943\text{"})\) | M1 |
| Area \(EAB = \tfrac{1}{2}5\cos(\pi - 1.4 - \text{"}0.943\text{"}) \times 5\sin(\pi - 1.4 - \text{"}0.943\text{"})\) | dM1 |
| Area \(ABCDE = 15.17\ldots + 17.5 + 6.24\ldots = 38.92\ldots\) awrt 38.9 | A1cso |
| (5) | |
| Total 9 |
Notes
1st M1: Area \(CBD = \tfrac{1}{2}5(7.5)\sin(\text{their } 0.943)\) or awrt 15.2. (Note area of \(CBD = 15.177\ldots\)) A correct method for the area of triangle \(CBD\) which can be implied by awrt 15.2
2nd M1: \(\pi - 1.4 - \text{"their } 0.943\text{"}\)
A value for angle \(EBA\) of awrt 0.8 (from 0.7985926536... or 0.7983916536...) or value for angle \(EBA\) of \((1.74159\ldots - \text{their angle } DBC)\) would imply this mark.
3rd M1: \(AB = 5\cos(\pi - 1.4 - \text{their } 0.943)\)
\(AB = 5\cos(0.79859\ldots) = 3.488577938\ldots\)
Allow M1 for \(AB =\) awrt 3.49
Or
\(AE = 5\sin(\pi - 1.4 - \text{their } 0.943)\)
\(AE = 5\sin(0.79859\ldots) = 3.581874365688\ldots\)
Allow M1 for \(AE =\) awrt 3.58
It must be clear that \(\pi - 1.4 - \text{"}0.943\text{"}\) is being used for angle EBA.
Note that some candidates use the sin rule here but it must be used correctly – do not allow mixing of degrees and radians.
dM1: This is dependent on the previous M1 and there must be no other errors in finding the area of triangle EAB
Allow M1 for area \(EAB =\) awrt 6.2
Note that a sign error in (b) can give the obtuse angle (2.198....) and could lead to the correct answer in (c) – this would lose the final mark in (c)