| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.3 This is a straightforward application of standard formulas for arc length, chord length using cosine rule, and segment area. It requires multiple steps (3 parts) but each uses routine techniques from the radians/circle geometry topic with no novel problem-solving insight needed. Slightly easier than average due to clear setup and standard methods. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(XZ^2 = 3^2+3^2-2\times3\times3\cos1.3\), or \(\sin0.65=\frac{x}{3}\) so \(XZ=2x\) | M1 | Must use cosine rule correctly; or splits into right-angled triangles using \(\sin0.65\) and doubles |
| \(XZ = 3.63\) | A1 | awrt 3.63 |
| Total | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Arc length \(ZY = 3\times\theta = 3\times(\pi-1.3)\) \((= 5.52/5.53)\) | M1, A1 | Arc length formula \(r\theta\) with \(r=3\) and \(\theta=1.3\), \((\pi-1.3)\) or \((2\pi-1.3)\); accept 1.8 or 5.0 decimals; in degrees look for \(\frac{\theta}{360}\times2\pi r\) with \(\theta=74°\)–\(75°\) or \(105°\)–\(106°\) |
| Perimeter \(= 3+3+\text{arc }ZY+\text{chord }XZ = 15.2\) cm | dM1, A1 | dM1: complete method for perimeter; look for \(6+(a)+\text{arc length}\); A1: awrt 15.2 cm |
| Total | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of triangle \(OXZ = \frac{1}{2}\times3\times3\times\sin1.3\ (=4.34)\) | M1 | Uses area formula for triangle correctly |
| Area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2}\times3^2\times(\pi-1.3)\ (=8.28/8.29)\) | M1 | Uses \(\frac{1}{2}r^2\theta\) with correct angle; allow as minimum awrt 1.8 radians \((3.1-1.3)\) |
| Total area \(= \frac{1}{2}\times3^2\times(\pi-1.3)+\frac{1}{2}\times3\times3\times\sin1.3\) | dM1 | Adds two correct area formulae; both M's must have been awarded |
| \(= 12.6\ \text{cm}^2\) | A1 | Accept awrt 12.6 |
| Total | [4] |
# Question 10:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $XZ^2 = 3^2+3^2-2\times3\times3\cos1.3$, or $\sin0.65=\frac{x}{3}$ so $XZ=2x$ | M1 | Must use cosine rule correctly; or splits into right-angled triangles using $\sin0.65$ and doubles |
| $XZ = 3.63$ | A1 | awrt 3.63 |
| **Total** | **[2]** | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Arc length $ZY = 3\times\theta = 3\times(\pi-1.3)$ $(= 5.52/5.53)$ | M1, A1 | Arc length formula $r\theta$ with $r=3$ and $\theta=1.3$, $(\pi-1.3)$ or $(2\pi-1.3)$; accept 1.8 or 5.0 decimals; in degrees look for $\frac{\theta}{360}\times2\pi r$ with $\theta=74°$–$75°$ or $105°$–$106°$ |
| Perimeter $= 3+3+\text{arc }ZY+\text{chord }XZ = 15.2$ cm | dM1, A1 | dM1: complete method for perimeter; look for $6+(a)+\text{arc length}$; A1: awrt 15.2 cm |
| **Total** | **[4]** | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of triangle $OXZ = \frac{1}{2}\times3\times3\times\sin1.3\ (=4.34)$ | M1 | Uses area formula for triangle correctly |
| Area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2}\times3^2\times(\pi-1.3)\ (=8.28/8.29)$ | M1 | Uses $\frac{1}{2}r^2\theta$ with correct angle; allow as minimum awrt 1.8 radians $(3.1-1.3)$ |
| Total area $= \frac{1}{2}\times3^2\times(\pi-1.3)+\frac{1}{2}\times3\times3\times\sin1.3$ | dM1 | Adds two correct area formulae; both M's must have been awarded |
| $= 12.6\ \text{cm}^2$ | A1 | Accept awrt 12.6 |
| **Total** | **[4]** | |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ce06b37a-aa57-4256-bec8-7277c8a9fc65-24_348_593_221_534}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 1 shows a semicircle with centre $O$ and radius $3 \mathrm {~cm} . X Y$ is the diameter of this semicircle. The point Z is on the circumference such that angle $X O Z = 1.3$ radians. The shaded region enclosed by the chord $X Z$, the arc $Z Y$ and the diameter $X Y$ is a template for a badge.
Find, giving each answer to 3 significant figures,
\begin{enumerate}[label=(\alph*)]
\item the length of the chord $X Z$,
\item the perimeter of the template $X Z Y X$,
\item the area of the template.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q10 [10]}}