C2 June 2013 (R) Q8
8.

Figure 2 shows the design for a triangular garden \(ABC\) where \(AB = 7\) m, \(AC = 13\) m and \(BC = 10\) m.
Given that angle \(BAC = \theta\) radians,
The point \(D\) lies on \(AC\) such that \(BD\) is an arc of the circle centre \(A\), radius 7 m.
The shaded region \(S\) is bounded by the arc \(BD\) and the lines \(BC\) and \(DC\). The shaded region \(S\) will be sown with grass seed, to make a lawned area.
Given that 50 g of grass seed are needed for each square metre of lawn,
| Scheme | Marks |
|---|---|
| Way 1: \(10^2 = 7^2 + 13^2 - 2 \times 7 \times 13\cos\theta\) or \(\cos\theta = \dfrac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13}\) | M1 |
| \(\cos\theta = \dfrac{59}{91}\) or \(\cos\theta = \dfrac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13}\) or \(\cos\theta = 0.6483\) or \(0.8644\) | A1 o.e |
| \((\theta = 0.8653789549\ldots\ ) = 0.865\ *\) (to 3 dp) | A1* cso |
| (3) |
Way 2
| Scheme | Marks |
|---|---|
| Way 2: Uses \(\cos\theta = \dfrac{x}{7}\), where \(7^2 - x^2 = 10^2 - (13 - x)^2\) and finds \(x\) ( = 59/13) | M1 |
| \(\cos\theta = \dfrac{59}{91}\) and \((\theta = 0.8653789549\ldots\ ) = 0.865\ *\) (to 3 dp) – as in Way 1 | A1, A1 |
| (3) |
Notes
M1: use correct cosine formula in any form A1: give a value for \(\cos\theta\)
NB \(\cos\theta = \dfrac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13}\) earns M1A1
A1: deduce and state the printed answer \(\theta = 0.865\)
| Scheme | Marks |
|---|---|
| Area triangle \(ABC = \dfrac{1}{2} \times 13 \times 7\sin 0.865\) or \(\dfrac{1}{2} \times 13 \times 7\sin 49.6\) or \(20\sqrt{3}\) | M1 |
| Area sector \(ABD = \dfrac{1}{2} \times 7^2 \times 0.865\) or \(\dfrac{49.6}{360} \times \pi \times 7^2\) | M1 |
| =34.6 (triangle) or 21.2 (Sector) | A1 |
| Area of \(S = \dfrac{1}{2} \times 13 \times 7\sin 0.865 - \dfrac{1}{2} \times 7^2 \times 0.865 \quad (= 13.4)\) | M1 A1 |
| (Amount of seed = ) 13.4 x 50 = 670g or 680g (need one of these two answers) | M1 A1 |
| (7) | |
| Total 10 |
Notes
M1: Uses Correct method for area of the correct triangle i.e. \(ABC\)
M1: Uses Correct method for the area of the sector
A1: This is earned for one of the correct answers. May be implied if these answers are not calculated but the final answer is correct with no errors (or shaded area is 13.4 or13.5)
M1: Their area of Triangle \(ABC\)– Area of Sector (may have \(kr^2\theta\) but not \(k\theta\) )
A1: Correct expression or awrt 13.4 or 13.5 (may be implied by final answer)
M1: Multiply their previous answer by 50
A1: 670g or 680 g (There is an argument for rounding answer up to provide enough seed)
N.B. \(\left(\dfrac{1}{2} \times 13 \times 7\sin 0.865 - \dfrac{1}{2} \times 7^2 \times 0.865\right) \times 50 = 670\) or 680 earns full marks
\(\left(\dfrac{1}{2} \times 13 \times 7\sin 0.865 - \dfrac{1}{2} \times 7^2 \times 0.865\right) \times 50 =\) awrt 670 or 680 just loses last mark
\(\left(\dfrac{1}{2} \times 13 \times 7\sin 0.865 - \dfrac{1}{2} \times 7^2 \times 0.865\right) \times 50 =\) wrong answer M1M1A0M1A1M1A0