C2 June 2014 (R) Q5
5.
Figure 2 shows the shape \(ABCDEA\) which consists of a right-angled triangle \(BCD\) joined to a sector \(ABDEA\) of a circle with radius 7 cm and centre \(B\).
\(A\), \(B\) and \(C\) lie on a straight line with \(AB = 7\) cm.
Given that the size of angle \(ABD\) is exactly 2.1 radians,
| Scheme | Marks |
|---|---|
| Length \(DEA = 7(2.1) = 14.7\) | M1A1 |
| (2) |
Notes
M1: \(7 \times 2.1\) only
A1: 14.7
| Scheme | Marks |
|---|---|
| Angle \(CBD = \pi - 2.1\) | M1 |
| Both \(7\cos(\pi - 2.1)\) and \(7\sin(\pi - 2.1)\) or Both \(7\cos(\pi - 2.1)\) and \(\sqrt{7^2 - \left(7\cos(\pi - 2.1)\right)^2}\) or Both \(7\sin(\pi - 2.1)\) and \(\sqrt{7^2 - \left(7\sin(\pi - 2.1)\right)^2}\) Or equivalents to these | dM1 |
| Note that 2.1 radians is 120 degrees (to 3sf) which if used gives angle CBD as 60 degrees. If used this gives a correct perimeter of 31.3 and could score full marks. | |
| \(\mathrm{P} = 7\cos(\pi - 2.1) + 7\sin(\pi - 2.1) + 7 + 14.7\) | ddM1 |
| \(= 31.2764\ldots\) Awrt 31.3 | A1 |
| (4) | |
| Total 6 |
Notes
M1: May be seen on the diagram (allow awrt 1.0 and allow 180 – 120). Could score for sight of Angle CBD = awrt 60 degrees.
dM1: A correct attempt to find BC and BD. You can ignore how the candidate assigns \(BC\) and \(CD\). \(7\cos(\pi - 2.1)\) can be implied by awrt 3.5 and \(7\sin(\pi - 2.1)\) can be implied by awrt 6. Note if the sin rule is used, do not allow mixing of degrees and radians unless their answer implies a correct interpretation. Dependent on the previous method mark.
ddM1: their BC + their CD + 7 + their DEA Dependent on both previous method marks