| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Optimization with sectors |
| Difficulty | Standard +0.3 This is a straightforward sector area problem requiring standard formulas (area = ½r²θ for sectors) and basic algebraic manipulation. Part (a) uses difference of two sector areas, (b) is simple substitution and solving, (c) applies arc length formula, and (d) uses cosine rule. All techniques are routine C2 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{1}{2}R^2\theta = \frac{49}{2}\theta\) or \(\frac{1}{2}r^2\theta = \frac{25}{2}\theta\) | B1 | |
| \(\frac{1}{2}R^2\theta - \frac{1}{2}r^2\theta = \frac{49}{2}\theta - \frac{25}{2}\theta = 12\theta\) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(12\theta = 15\), \(\theta = 1.25\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(R\theta = 7 \times 1.25\) (or \(r\theta = 5 \times 1.25\)) | B1 | |
| \(R\theta + r\theta + 4 = 8.75 + 6.25 + 4 = 19\) m | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\sin 0.625 = \dfrac{x}{5}\), \(AD = 2x\) \((= 5.851\) m\()\) | M1 | |
| \(6.25 - 5.85 = 0.399 \approx 40\) m | M1 A1 | (3 marks) |
## Question 6:
### Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{1}{2}R^2\theta = \frac{49}{2}\theta$ or $\frac{1}{2}r^2\theta = \frac{25}{2}\theta$ | B1 | |
| $\frac{1}{2}R^2\theta - \frac{1}{2}r^2\theta = \frac{49}{2}\theta - \frac{25}{2}\theta = 12\theta$ | M1 A1 | (3 marks) |
### Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $12\theta = 15$, $\theta = 1.25$ | M1 A1 | (2 marks) |
### Part (c):
| Working | Mark | Notes |
|---------|------|-------|
| $R\theta = 7 \times 1.25$ (or $r\theta = 5 \times 1.25$) | B1 | |
| $R\theta + r\theta + 4 = 8.75 + 6.25 + 4 = 19$ m | M1 A1 | (3 marks) |
### Part (d):
| Working | Mark | Notes |
|---------|------|-------|
| $\sin 0.625 = \dfrac{x}{5}$, $AD = 2x$ $(= 5.851$ m$)$ | M1 | |
| $6.25 - 5.85 = 0.399 \approx 40$ m | M1 A1 | (3 marks) |
---6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{1033051d-18bf-4734-a556-4c8e1c789992-3_842_963_280_392}
\end{center}
\end{figure}
Fig. 1 shows a gardener's design for the shape of a flower bed with perimeter $A B C D . A D$ is an arc of a circle with centre $O$ and radius $5 \mathrm {~m} . B C$ is an arc of a circle with centre $O$ and radius $7 \mathrm {~m} . O A B$ and $O D C$ are straight lines and the size of $\angle A O D$ is $\theta$ radians.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $\theta$, an expression for the area of the flower bed.
Given that the area of the flower bed is $15 \mathrm {~m} ^ { 2 }$,
\item show that $\theta = 1.25$,
\item calculate, in m , the perimeter of the flower bed.
The gardener now decides to replace arc $A D$ with the straight line $A D$.
\item Find, to the nearest cm , the reduction in the perimeter of the flower bed.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [10]}}