| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Simultaneous equations with arc/area |
| Difficulty | Moderate -0.3 This is a straightforward application of standard arc length and sector area formulas (s=rθ and A=½r²θ) leading to a simple simultaneous equation system. The division method to eliminate variables is routine, requiring only basic algebraic manipulation. While it involves two formulas and solving simultaneous equations, these are standard C2 techniques with no conceptual challenges or problem-solving insight needed. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r\theta = 6\) | B1 | Allow \(\frac{\theta}{2\pi}\times2\pi r=6\) but not \(\frac{\theta}{360}\times2\pi r=6\) |
| \(\frac{1}{2}r^2\theta = 20\) | B1 | Both equations required for second B1; allow \(\frac{\theta}{2\pi}\times\pi r^2=20\) but not \(\frac{\theta}{360}\times\pi r^2=20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(r\theta=6\) into \(\frac{1}{2}r^2\theta=20 \Rightarrow \frac{1}{2}\times6r=20\) | M1 | Combines two equations in \(r\) and \(\theta\) producing equation in one unknown |
| \(r = \frac{20}{3}\) | A1 | Or exact equivalent |
| Substitutes \(r=\frac{20}{3}\) in \(r\theta=6 \Rightarrow \theta=\frac{9}{10}\) | dM1, A1 | dM1: dependent on previous M1; correctly substituting to find second unknown; A1: both \(r=\frac{20}{3}\) and \(\theta=\frac{9}{10}\); condone \(6.\dot{6}\) for \(\frac{20}{3}\); do not allow \(6.67\) |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r\theta = 6$ | B1 | Allow $\frac{\theta}{2\pi}\times2\pi r=6$ but not $\frac{\theta}{360}\times2\pi r=6$ |
| $\frac{1}{2}r^2\theta = 20$ | B1 | Both equations required for second B1; allow $\frac{\theta}{2\pi}\times\pi r^2=20$ but not $\frac{\theta}{360}\times\pi r^2=20$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $r\theta=6$ into $\frac{1}{2}r^2\theta=20 \Rightarrow \frac{1}{2}\times6r=20$ | M1 | Combines two equations in $r$ and $\theta$ producing equation in one unknown |
| $r = \frac{20}{3}$ | A1 | Or exact equivalent |
| Substitutes $r=\frac{20}{3}$ in $r\theta=6 \Rightarrow \theta=\frac{9}{10}$ | dM1, A1 | dM1: dependent on previous M1; correctly substituting to find second unknown; A1: both $r=\frac{20}{3}$ and $\theta=\frac{9}{10}$; condone $6.\dot{6}$ for $\frac{20}{3}$; do not allow $6.67$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-18_387_397_255_794}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a circle with centre $O$ and radius $r \mathrm {~cm}$.
The points $A$ and $B$ lie on the circumference of this circle.
The minor arc $A B$ subtends an angle $\theta$ radians at $O$, as shown in Figure 3.\\
Given the length of minor $\operatorname { arc } A B$ is 6 cm and the area of minor sector $O A B$ is $20 \mathrm {~cm} ^ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item write down two different equations in $r$ and $\theta$.
\item Hence find the value of $r$ and the value of $\theta$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q8 [6]}}