| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - area/perimeter with given values |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C2 topics: cosine rule to find an angle, sector area formula, and combining areas/arc lengths. Part (a) is given as 'show that', making it easier. The remaining parts require direct formula application with no novel problem-solving insight, making it slightly easier than average but not trivial due to the multi-step nature and need for careful calculation. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos BAC = \frac{81 + 25 - 100}{90} = \frac{1}{15}\), \(BAC = 1.504\) radians | M1 A1; A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}r^2\theta = \frac{1}{2} \times 9 \times 1.504 = 6.768\) cm² (6.77) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Area of triangle \(= \frac{1}{2} \times 45 \times \sin 1.504\) | M1 A1 | (= 22.450 cm²) |
| Answer | Marks | Guidance |
|---|---|---|
| Shaded area \(= 22.450 - 6.768 = 15.682\) cm² (15.68, 15.7) | A1 | (3 marks) |
| Arc length \(= r\theta = 3 \times 1.504 = 4.512\) cm | M1 A1 | |
| Perimeter \(= 10 + 6 + 2 + 4.512 = 22.512\) cm (22.51, 22.5) | M1 A1 ft | (4 marks) |
## (a)
$100 = 81 + 25 - (2 \times 9 \times 5 \cos BAC)$
$\cos BAC = \frac{81 + 25 - 100}{90} = \frac{1}{15}$, $BAC = 1.504$ radians | M1 A1; A1 | (3 marks)
## (b)
$\frac{1}{2}r^2\theta = \frac{1}{2} \times 9 \times 1.504 = 6.768$ cm² (6.77) | M1 A1 | (2 marks)
## (c)
Area of triangle $= \frac{1}{2} \times 45 \times \sin 1.504$ | M1 A1 | (= 22.450 cm²)
## (d)
Shaded area $= 22.450 - 6.768 = 15.682$ cm² (15.68, 15.7) | A1 | (3 marks)
Arc length $= r\theta = 3 \times 1.504 = 4.512$ cm | M1 A1 |
Perimeter $= 10 + 6 + 2 + 4.512 = 22.512$ cm (22.51, 22.5) | M1 A1 ft | (4 marks)
**Total: (12 marks)**
---\includegraphics{figure_2}
Triangle $ABC$ has $AB = 9$ cm, $BC = 10$ cm and $CA = 5$ cm. A circle, centre $A$ and radius 3 cm, intersects $AB$ and $AC$ at $P$ and $Q$ respectively, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Show that, to 3 decimal places, $\angle BAC = 1.504$ radians. [3]
\end{enumerate}
Calculate,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the area, in cm$^2$, of the sector $APQ$, [2]
\item the area, in cm$^2$, of the shaded region $BPQC$, [3]
\item the perimeter, in cm, of the shaded region $BPQC$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [12]}}