| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Moderate -0.8 This is a straightforward application of arc length formula (s = rθ) and perimeter calculation using Pythagoras. All values are given directly, requiring only substitution into standard formulas with minimal problem-solving. Easier than average C2 content. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Length \(DEA = 7(2.1) = 14.7\) | M1 | \(7 \times 2.1\) only |
| \(= 14.7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Angle \(CBD = \pi - 2.1\) | M1 | May be seen on diagram; allow awrt 1.0 and allow \(180 - 120\). Could score for sight of Angle \(CBD =\) awrt 60 degrees |
| Both \(7\cos(\pi - 2.1)\) and \(7\sin(\pi - 2.1)\), or Both \(7\cos(\pi-2.1)\) and \(\sqrt{7^2-(7\cos(\pi-2.1))^2}\), or Both \(7\sin(\pi-2.1)\) and \(\sqrt{7^2-(7\sin(\pi-2.1))^2}\) | dM1 | Correct attempt to find BC and BD. \(7\cos(\pi-2.1)\) implied by awrt 3.5 and \(7\sin(\pi-2.1)\) implied by awrt 6. If sin rule used, do not allow mixing degrees and radians unless interpretation is correct. Dependent on previous method mark |
| \(P = 7\cos(\pi-2.1) + 7\sin(\pi-2.1) + 7 + 14.7\) | ddM1 | their \(BC +\) their \(CD + 7 +\) their \(DEA\). Dependent on both previous method marks |
| \(= 31.2764...\) | A1 | Awrt 31.3 |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Length $DEA = 7(2.1) = 14.7$ | M1 | $7 \times 2.1$ only |
| $= 14.7$ | A1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Angle $CBD = \pi - 2.1$ | M1 | May be seen on diagram; allow awrt 1.0 and allow $180 - 120$. Could score for sight of Angle $CBD =$ awrt 60 degrees |
| **Both** $7\cos(\pi - 2.1)$ **and** $7\sin(\pi - 2.1)$, or **Both** $7\cos(\pi-2.1)$ **and** $\sqrt{7^2-(7\cos(\pi-2.1))^2}$, or **Both** $7\sin(\pi-2.1)$ **and** $\sqrt{7^2-(7\sin(\pi-2.1))^2}$ | dM1 | Correct attempt to find BC and BD. $7\cos(\pi-2.1)$ implied by awrt 3.5 and $7\sin(\pi-2.1)$ implied by awrt 6. If sin rule used, do not allow mixing degrees and radians unless interpretation is correct. Dependent on previous method mark |
| $P = 7\cos(\pi-2.1) + 7\sin(\pi-2.1) + 7 + 14.7$ | ddM1 | their $BC +$ their $CD + 7 +$ their $DEA$. Dependent on **both** previous method marks |
| $= 31.2764...$ | A1 | Awrt 31.3 |
> Note: 2.1 radians is 120 degrees (to 3sf) which if used gives angle CBD as 60 degrees. If used this gives a correct perimeter of 31.3 and could score full marks.
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f07cc9ed-a820-46c8-a3a3-3c780cf20fa7-08_566_725_127_614}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the shape $A B C D E A$ which consists of a right-angled triangle $B C D$ joined to a sector $A B D E A$ of a circle with radius 7 cm and centre $B$.\\
$A , B$ and $C$ lie on a straight line with $A B = 7 \mathrm {~cm}$.\\
Given that the size of angle $A B D$ is exactly 2.1 radians,
\begin{enumerate}[label=(\alph*)]
\item find, in cm, the length of the arc $D E A$,
\item find, in cm, the perimeter of the shape $A B C D E A$, giving your answer to 1 decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q5 [6]}}