C2 January 2012 Q7
7.
Figure 2 shows \(ABC\), a sector of a circle of radius 6 cm with centre \(A\). Given that the size of angle \(BAC\) is 0.95 radians, find
The point \(D\) lies on the line \(AC\) and is such that \(AD = BD\). The region \(R\), shown shaded in Figure 2, is bounded by the lines \(CD\), \(DB\) and the arc \(BC\).
Find
| Scheme | Marks |
|---|---|
| \(r\theta = 6 \times 0.95, = 5.7\) (cm) | M1, A1 |
| (2) |
Notes
M1: Needs \(\theta\) in radians for this formula. Could convert to degrees and use degrees formula.
A1: Does not need units
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}r^2\theta = \dfrac{1}{2} \times 6^2 \times 0.95, = 17.1\ \left(\text{cm}^2\right)\) | M1, A1 |
| (2) |
Notes
M1: Needs \(\theta\) in radians for this formula. Could convert to degrees and use degrees formula.
A1: Does not need units
| Scheme | Marks |
|---|---|
| Let \(AD = x\) then \(\dfrac{x}{\sin 0.95} = \dfrac{6}{\sin 1.24}\) so \(x = 5.16\) * OR \(x = 3/\cos 0.95\) OR so \(x = 3/\sin 0.62\) so \(x = 5.16\) * OR \(x^2 = 6^2 + x^2 - 12x\cos 0.95\) leading to \(x =\) , so \(x = 5.16\) * | M1 A1 |
| (2) |
Notes
M1: Needs complete correct trig method to achieve \(x =\)
May have worked in degrees, using 54.4 degrees and 71.1 degrees
Using angles of triangle sum to 360degrees is not correct method so is M0
A1: accept answers which round to 5.16 (NB This is given answer)
If the answer 5.16 is assumed and verified award M1A0 for correct work
| Scheme | Marks |
|---|---|
| Perimeter = ‘5.7’ + 5.16 + 6 – 5.16 = “11.7” or 6 + their 5.7 | M1A1 ft |
| (2) |
Notes
M1: Accept answer only as implying method, or just 6 + 5.7
A1 : can be scored even following wrong answer to part (c)
| Scheme | Marks |
|---|---|
| Area of triangle \(ABD = \tfrac{1}{2} \times 6 \times 5.16 \times \sin 0.95 = 12.6\) or \(\tfrac{1}{2} \times 6 \times 3 \times \tan 0.95 = 12.6\) (½ base x height) or \(\tfrac{1}{2} \times 5.16 \times 5.16 \times \sin 1.24 = 12.6\) | M1 A1 |
| So Area of \(R\) = ‘17.1’ – ‘12.6’ = 4.5 | M1 A1 |
| (4) | |
| 12 |
Notes
M1: needs complete method for area of triangle \(ABD\) not \(ABC\)
A1: Accept awrt 12.6 (If area of triangle is not evaluated or is given as 12.5 (truncated) this mark may be implied by 4.5 later)
M1: Uses area of \(R\) = area of sector – area of triangle \(ABD\) (not \(ABC\))
A1: Answers wrt 4.5
Alternative For part (e)
Finds area of segment and area of triangle \(BDC\) by correct methods M1
Obtains 2.4585 and 2.0498 – accept answers wrt 2.5, 2.1 A1
Uses area of segment + area of triangle \(BDC\), to obtain 4.5 (not 4.6) M1, A1
NB Just finding area of segment is M0