C2 June 2006 Q8
8.
Figure 2 shows the cross section \(ABCD\) of a small shed.
The straight line \(AB\) is vertical and has length 2.12 m.
The straight line \(AD\) is horizontal and has length 1.86 m.
The curve \(BC\) is an arc of a circle with centre \(A\), and \(CD\) is a straight line.
Given that the size of \(\angle BAC\) is 0.65 radians, find
| Scheme | Marks |
|---|---|
| \(r\theta = 2.12 \times 0.65 \qquad 1.38\ (\text{m})\) | M1 A1 |
| (2) |
Notes
M1: Use of \(r\theta\) with \(r = 2.12\) or 1.86, and \(\theta = 0.65\), or equiv. method for the angle changed to degrees (allow awrt 37°).
Angle changed to degrees wrongly and used throughout (a), (b) and (c):
Penalise ‘method’ only once, so could score M0A0, M1A0, M1A0.
Failure to round to 2 d.p.:
Penalise only once, on the first occurrence, then accept awrt.
If 0.65 is taken as degrees throughout: Only award marks in part (d).
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}r^2\theta = \dfrac{1}{2} \times 2.12^2 \times 0.65 \qquad 1.46\ (\text{m}^2)\) | M1 A1 |
| (2) |
Notes
M1: Use of \(\dfrac{1}{2}r^2\theta\) with \(r = 2.12\) or 1.86, and \(\theta = 0.65\), or equiv. method for the angle changed to degrees (allow awrt 37°).
| Scheme | Marks |
|---|---|
| \(\dfrac{\pi}{2} - 0.65 \qquad 0.92\) (radians) \(\quad (\alpha)\) | M1 A1 |
| (2) |
Notes
M1: Subtracting 0.65 from \(\dfrac{\pi}{2}\), or subtracting awrt 37 from 90 (degrees), (perhaps implied by awrt 53).
| Scheme | Marks |
|---|---|
| \(\Delta ACD:\ \dfrac{1}{2}(2.12)(1.86)\sin\alpha\) (With the value of \(\alpha\) from part (c)) | M1 |
| Area \(=\) “1.46” + “1.57”, \(\qquad 3.03\ (\text{m}^2)\) | M1 A1 |
| (3) | |
| (9 marks) |
Notes
First M1: Other area methods must be fully correct.
Second M1: Adding answer to (b) to their \(\Delta ACD\).