| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Moderate -0.3 This is a straightforward application of the arc length formula (s = rθ) to find the radius, followed by using the sector area formula. The question requires two standard formulas with minimal problem-solving, making it slightly easier than average but not trivial since it involves a compound shape and requires careful identification of which measurements apply to which sector. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(s = r\theta \Rightarrow 3 = r \times 0.4\) | M1 | Attempts to use correct formula with \(s=3\) and \(\theta=0.4\) |
| \(\Rightarrow OD = 7.5\) cm | A1 | Answer of 7.5 cm implies correct formula; scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses angle \(AOB = (\pi - 0.4)\) or uses radius \((12 - \text{'7.5'})\) cm | M1 | \(AOB = \pi - 0.4\) may be implied by \(AOB \approx 2.74\); or uses radius \(12 - \text{their '7.5'}\) |
| Uses area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2}\times(12-7.5)^2\times(\pi-0.4)\) | M1 | Follow through on radius \((12 - \text{their } OD)\) and their angle |
| \(= 27.8 \text{ cm}^2\) | A1ft | Allow awrt \(27.8\text{ cm}^2\); do not follow through on negative radius |
## Question 2:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $s = r\theta \Rightarrow 3 = r \times 0.4$ | M1 | Attempts to use correct formula with $s=3$ and $\theta=0.4$ |
| $\Rightarrow OD = 7.5$ cm | A1 | Answer of 7.5 cm implies correct formula; scores both marks |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses angle $AOB = (\pi - 0.4)$ or uses radius $(12 - \text{'7.5'})$ cm | M1 | $AOB = \pi - 0.4$ may be implied by $AOB \approx 2.74$; or uses radius $12 - \text{their '7.5'}$ |
| Uses area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2}\times(12-7.5)^2\times(\pi-0.4)$ | M1 | Follow through on radius $(12 - \text{their } OD)$ and their angle |
| $= 27.8 \text{ cm}^2$ | A1ft | Allow awrt $27.8\text{ cm}^2$; do not follow through on negative radius |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f7994129-07ee-4f6d-9531-08a15a38b794-04_350_639_210_712}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shape $A B C D O A$, as shown in Figure 1, consists of a sector $C O D$ of a circle centre $O$ joined to a sector $A O B$ of a different circle, also centre $O$.
Given that arc length $C D = 3 \mathrm {~cm} , \angle C O D = 0.4$ radians and $A O D$ is a straight line of length 12 cm ,
\begin{enumerate}[label=(\alph*)]
\item find the length of $O D$,
\item find the area of the shaded sector $A O B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 Q2 [5]}}