| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Circular arc problems |
| Difficulty | Moderate -0.8 This is a straightforward application of standard arc length and sector area formulas (s=rθ, A=½r²θ) with clearly given values. Parts (c) and (d) require basic angle subtraction and area addition, but all steps are routine C2-level calculations with no 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 |
|---|---|---|
| (a) \(r\theta = 2.12 \times 0.65\) | M1, A1 | \(1.38\) (m) (2 marks) |
| (b) \(\frac{1}{2}r^2\theta = \frac{1}{2} \times 2.12^2 \times 0.65\) | M1, A1 | \(1.46\) (m²) (2 marks) |
| (c) \(\frac{\pi}{2} - 0.65\) | M1, A1 | \(0.92\) (radians) (\(\alpha\)) (2 marks) |
| (d) \(\triangle ACD: \frac{1}{2}(2.12)(1.86)\sin \alpha\) (With the value of \(\alpha\) from part (c)) | M1 | Area = "1.46" + "1.57" |
| (9 marks) | (a) M1: Use of \(r\theta\) with \(r = 2.12\) or \(1.86\), and \(\theta = 0.65\), or equiv. method for the angle changed to degrees (allow awrt 37°). (b) M1: Use of \(\frac{1}{2}r^2\theta\) with \(r = 2.12\) or \(1.86\), and \(\theta = 0.65\), or equiv. method for the angle changed to degrees (allow awrt 37°). (c) M1: Subtracting \(0.65\) from \(\frac{\pi}{2}\), or subtracting awrt 37 from 90 (degrees). (perhaps implied by awrt 53). Angle changed to degrees wrongly and used throughout (a), (b) and (c): Penalise 'method' only once, so could score M0A0, M1A0, M1A0. Failure to round to 2 d.p.: Penalise only once, on the first occurrence, then accept awrt. If 0.65 is taken as degrees throughout: Only award marks in part (d). |
**(a)** $r\theta = 2.12 \times 0.65$ | M1, A1 | $1.38$ (m) (2 marks) |
| --- | --- | --- |
| **(b)** $\frac{1}{2}r^2\theta = \frac{1}{2} \times 2.12^2 \times 0.65$ | M1, A1 | $1.46$ (m²) (2 marks) |
| **(c)** $\frac{\pi}{2} - 0.65$ | M1, A1 | $0.92$ (radians) ($\alpha$) (2 marks) |
| **(d)** $\triangle ACD: \frac{1}{2}(2.12)(1.86)\sin \alpha$ (With the value of $\alpha$ from part (c)) | M1 | Area = "1.46" + "1.57" | M1, A1 | $3.03$ (m²) (3 marks) |
| | (9 marks) | **(a)** M1: Use of $r\theta$ with $r = 2.12$ or $1.86$, and $\theta = 0.65$, or equiv. method for the angle changed to degrees (allow awrt 37°). **(b)** M1: Use of $\frac{1}{2}r^2\theta$ with $r = 2.12$ or $1.86$, and $\theta = 0.65$, or equiv. method for the angle changed to degrees (allow awrt 37°). **(c)** M1: Subtracting $0.65$ from $\frac{\pi}{2}$, or subtracting awrt 37 from 90 (degrees). (perhaps implied by awrt 53). **Angle changed to degrees wrongly and used throughout (a), (b) and (c):** Penalise 'method' only once, so could score M0A0, M1A0, M1A0. **Failure to round to 2 d.p.:** Penalise only once, on the first occurrence, then accept awrt. **If 0.65 is taken as degrees throughout:** Only award marks in part (d). |8.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{29c7baa1-6929-448a-a756-319ea75dffa7-10_620_636_301_660}
\end{center}
\end{figure}
Figure 2 shows the cross section $A B C D$ of a small shed. The straight line $A B$ is vertical and has length 2.12 m . The straight line $A D$ is horizontal and has length 1.86 m . The curve $B C$ is an arc of a circle with centre $A$, and $C D$ is a straight line. Given that the size of $\angle B A C$ is 0.65 radians, find
\begin{enumerate}[label=(\alph*)]
\item the length of the arc $B C$, in m , to 2 decimal places,
\item the area of the sector $B A C$, in $\mathrm { m } ^ { 2 }$, to 2 decimal places,
\item the size of $\angle C A D$, in radians, to 2 decimal places,
\item the area of the cross section $A B C D$ of the shed, in $\mathrm { m } ^ { 2 }$, to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2006 Q8 [9]}}