C2 January 2009 Q7
7.
The shape \(BCD\) shown in Figure 3 is a design for a logo.
The straight lines \(DB\) and \(DC\) are equal in length. The curve \(BC\) is an arc of a circle with centre \(A\) and radius 6 cm. The size of \(\angle BAC\) is 2.2 radians and \(AD = 4\) cm.
Find
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}r^2\theta = \dfrac{1}{2} \times 6^2 \times 2.2 = 39.6\quad\left(\text{cm}^2\right)\) | 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. Answer should be 39.6 exactly.
Answer with no working is M1 A1.
This M1A1 can only be awarded in part (a).
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{2\pi - 2.2}{2} =\right)\ \pi - 1.1 = 2.04\ \ (\text{rad})\) | M1 A1 |
| (2) |
Notes
M1: Needs full method to give angle in radians
A1: Allow answers which round to 2.04 (Just writes 2.04 – no working is 2/2)
| Scheme | Marks |
|---|---|
| \(\Delta DAC = \dfrac{1}{2} \times 6 \times 4\sin 2.04\qquad (\approx 10.7)\) | M1 A1ft |
| Total area = sector + 2 triangles = 61 \(\left(\text{cm}^2\right)\) | M1 A1 |
| (4) | |
| [8] |
Notes
M1: Use \(\dfrac{1}{2} \times 6 \times 4\sin A\) (if any other triangle formula e.g. \(\tfrac{1}{2}b \times h\) is used the method must be complete for this mark) (No value needed for \(A\), but should not be using 2.2)
A1: ft the value obtained in part (b) – need not be evaluated- could be in degrees
M1: Uses Total area = sector + 2 triangles or other complete method
A1: Allow answers which round to 61. (Do not need units)
Special case degrees: Could get M0A0, M0A0, M1A1M1A0
Special case: Use \(\Delta BDC - \Delta BAC\) Both areas needed for first M1
Total area = sector + area found is second M1
NB Just finding lengths BD, DC, and angle BDC then assuming area BDC is a sector to find area BDC is 0/4