| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Standard +0.3 This is a straightforward compound shape problem requiring application of standard sector area and arc length formulas. Part (a) uses angle properties (angles on a straight line), part (b) involves adding three sector areas with given radii, and part (c) requires calculating arc lengths. All steps are routine 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 theta1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Angle \(AOB = \dfrac{\pi - \theta}{2}\) | B1 | Deduces correct expression for angle \(AOB\). Note \(\dfrac{180-\theta}{2}\) scores B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area \(= 2 \times \dfrac{1}{2}r^2\left(\dfrac{\pi-\theta}{2}\right) + \dfrac{1}{2}(2r)^2\theta\) | M1 | Fully correct strategy using angle from (a). Need \(2\times\dfrac{1}{2}r^2\alpha\) or \(r^2\alpha\) where \(\alpha\) is their angle from (a), \(+ \dfrac{1}{2}(2r)^2\theta\) |
| \(= \dfrac{1}{2}r^2\pi - \dfrac{1}{2}r^2\theta + 2r^2\theta = \dfrac{3}{2}r^2\theta + \dfrac{1}{2}r^2\pi = \dfrac{1}{2}r^2(3\theta+\pi)\)* | A1* | Correct proof. Brackets in \(\dfrac{1}{2}(2r)^2\theta\) may be condoned if recovered. First term must be expanded as \(\dfrac{1}{2}r^2\pi - \dfrac{1}{2}r^2\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Perimeter \(= 4r + 2r\left(\dfrac{\pi-\theta}{2}\right) + 2r\theta\) | M1 | Fully correct strategy using angle from (a). Need \(4r + 2r\alpha + 2r\theta\) where \(\alpha\) is their angle from (a) |
| \(= 4r + r\pi + r\theta\) or e.g. \(r(4+\pi+\theta)\) | A1 | Correct simplified expression |
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Angle $AOB = \dfrac{\pi - \theta}{2}$ | B1 | Deduces correct expression for angle $AOB$. Note $\dfrac{180-\theta}{2}$ scores B0 |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area $= 2 \times \dfrac{1}{2}r^2\left(\dfrac{\pi-\theta}{2}\right) + \dfrac{1}{2}(2r)^2\theta$ | M1 | Fully correct strategy using angle from (a). Need $2\times\dfrac{1}{2}r^2\alpha$ or $r^2\alpha$ where $\alpha$ is their angle from (a), $+ \dfrac{1}{2}(2r)^2\theta$ |
| $= \dfrac{1}{2}r^2\pi - \dfrac{1}{2}r^2\theta + 2r^2\theta = \dfrac{3}{2}r^2\theta + \dfrac{1}{2}r^2\pi = \dfrac{1}{2}r^2(3\theta+\pi)$* | A1* | Correct proof. Brackets in $\dfrac{1}{2}(2r)^2\theta$ may be condoned if recovered. First term must be expanded as $\dfrac{1}{2}r^2\pi - \dfrac{1}{2}r^2\theta$ |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Perimeter $= 4r + 2r\left(\dfrac{\pi-\theta}{2}\right) + 2r\theta$ | M1 | Fully correct strategy using angle from (a). Need $4r + 2r\alpha + 2r\theta$ where $\alpha$ is their angle from (a) |
| $= 4r + r\pi + r\theta$ or e.g. $r(4+\pi+\theta)$ | A1 | Correct simplified expression |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6c32000f-574f-473c-bd04-9cfe2c1bd715-12_487_784_292_644}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shape $O A B C D E F O$ shown in Figure 1 is a design for a logo.\\
In the design
\begin{itemize}
\item $O A B$ is a sector of a circle centre $O$ and radius $r$
\item sector $O F E$ is congruent to sector $O A B$
\item $O D C$ is a sector of a circle centre $O$ and radius $2 r$
\item $A O F$ is a straight line
\end{itemize}
Given that the size of angle $C O D$ is $\theta$ radians,
\begin{enumerate}[label=(\alph*)]
\item write down, in terms of $\theta$, the size of angle $A O B$
\item Show that the area of the logo is
$$\frac { 1 } { 2 } r ^ { 2 } ( 3 \theta + \pi )$$
\item Find the perimeter of the logo, giving your answer in simplest form in terms of $r , \theta$ and $\pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q6 [5]}}