C2 June 2015 Q4
4.
Figure 1 shows a sketch of a design for a scraper blade. The blade \(AOBCDA\) consists of an isosceles triangle \(COD\) joined along its equal sides to sectors \(OBC\) and \(ODA\) of a circle with centre \(O\) and radius 8 cm. Angles \(AOD\) and \(BOC\) are equal. \(AOB\) is a straight line and is parallel to the line \(DC\). \(DC\) has length 7 cm.
| Scheme | Marks |
|---|---|
| In triangle \(OCD\) complete method used to find angle \(COD\) so: Either \(\cos COD = \dfrac{8^2 + 8^2 - 7^2}{2 \times 8 \times 8}\) or uses \(\angle COD = 2 \times \arcsin\tfrac{3.5}{8}\) oe so \(\angle COD =\) | M1 |
| \((\angle COD = 0.9056(331894))\) \(= 0.906\) (3sf) * accept awrt 0.906 | A1 * |
| (2) |
Notes
M1: Either use correctly quoted cosine rule – may quote as \(7^2 = 8^2 + 8^2 - 2 \times 8 \times 8\cos\alpha \Rightarrow \alpha = \ldots..\)
Or split isosceles triangle into two right angled triangles and use arcsin or longer methods using Pythagoras and arcos (i.e. \(\pi - 2 \times \arccos\tfrac{3.5}{8}\)). There are many ways of showing this result.
Must conclude that \(\angle COD =\)
A1*: (NB this is a given answer) If any errors or over-approximation is seen this is A0. It needs correct work leading to stated answer of 0.906 or awrt 0.906 for A1. The cosine of \(COD\) is equal to 79/128 or awrt 0.617. Use of 0.62 (2sf) does not lead to printed answer. They may give 51.9 in degrees then convert to radians. This is fine.
The minimal solution \(7^2 = 8^2 + 8^2 - 2 \times 8 \times 8\cos\alpha \Rightarrow \alpha = \ldots..0.906\) (with no errors seen) can have M1A1 but errors rearranging result in M1A0
| Scheme | Marks |
|---|---|
| Uses \(s = 8\theta\) for any \(\theta\) in radians or \(\dfrac{\theta}{360} \times 2\pi \times 8\) for any \(\theta\) in degrees | M1 |
| \(\theta = \dfrac{\pi - \text{"}COD\text{"}}{2}\) \((= \textit{awrt}\ 1.12)\) or \(2\theta\ (= \textit{awrt}\ 2.24)\) and Perimeter = 23+( \(16 \times \theta\) ) | M1 |
| accept awrt 40.9 (cm) | A1 |
| (3) |
Notes
M1: Uses formula for arc length with \(r = 8\) and any angle i.e. \(s = 8\theta\) if working in rads or \(s = \dfrac{\theta}{360} \times 2\pi \times 8\) in degrees
(If the formula is quoted with \(r\) the 8 may be implied by the value of their \(r\theta\))
M1: Uses angles on straight line (or other geometry) to find angle \(BOC\) or \(AOD\) and uses Perimeter = 23 + arc lengths \(BC\) and \(AD\) (may make a slip – in calculation or miscopying)
A1: correct work leading to awrt 40.9 not 40.8 (do not need to see cm) This answer implies M1M1A1
| Scheme | Marks |
|---|---|
| Either Way 1: (Use of Area of two sectors + area of triangle) | |
| Area of triangle \(= \tfrac{1}{2} \times 8 \times 8 \times \sin 0.906\) (or 25.1781155 accept awrt 25.2)or \(\tfrac{1}{2} \times 8 \times 7 \times \sin 1.118\) or \(\tfrac{1}{2} \times 7 \times h\) after \(h\) calculated from correct Pythagoras or trig. | M1 |
| Area of sector \(= \tfrac{1}{2}8^2 \times \text{"}1.117979732\text{"}\) (or 35.77535142 accept awrt 35.8 ) | M1 |
| Total Area = Area of two sectors + area of triangle =awrt 96.7 or 96.8 or 96.9 (cm2) | A1 |
| (3) |
Notes
Or Way 2: (Use of area of semicircle – area of segment)
| Scheme | Marks |
|---|---|
| Area of semi-circle \(= \tfrac{1}{2} \times \pi \times 8 \times 8\) (or 100.5) | M1 |
| Area of segment \(= \tfrac{1}{2}8^2 \times (\text{"}0.906\text{"} - \sin\text{"}0.906\text{"})\) (or 3.807) | M1 |
| So area required = awrt 96.7 or 96.8 or 96.9 (cm2) | A1 |
| (3) | |
| [8] |
(c) Way 1: M1: Mark is given for correct statement of area of triangle \(\tfrac{1}{2} \times 8 \times 8 \times \sin 0.906\) (must use correct angle) or for correct answer (awrt 25.2) Accept alternative correct methods using Pythagoras and ½ base×height
M1: Mark is given for formula for area of sector \(\tfrac{1}{2}8^2 \times \text{"}1.117979732\text{"}\) with \(r = 8\) and their angle \(BOC\) or \(AOD\) or \((BOC + AOD)\) not \(COD\). May use \(A = \dfrac{\theta}{360} \times \pi \times 8^2\) if working in degrees
A1: Correct work leading to awrt 96.7, 96.8 or 96.9 (This answer implies M1M1A1)
NB. Solution may combine the two sectors for part (b) and (c) and so might use \(2 \times \angle BOC\) rather than \(\angle BOC\)
Way 2: M1: Mark is given for correct statement of area of semicircle \(\tfrac{1}{2} \times \pi \times 8 \times 8\) or for correct answer 100.5
M1: Mark is given for formula for area of segment \(\tfrac{1}{2}8^2 \times (\text{"}0.906\text{"} - \sin\text{"}0.906\text{"})\) with \(r = 8\) or 3.81 A1: As in Way 1