| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| 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 standard C2 compound shape problem requiring sector area formula (½r²θ), cosine rule for finding an angle, and combining areas of standard shapes. All techniques are routine applications of formulae with no novel insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(BDE = \frac{1}{2}(5)^2(1.4)\) | M1 | Use of correct formula or method for area of sector |
| \(= 17.5 \text{ cm}^2\) | A1 | 17.5 or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6.1^2 = 5^2 + 7.5^2 - (2 \times 5 \times 7.5\cos DBC)\) or \(\cos DBC = \frac{5^2 + 7.5^2 - 6.1^2}{2 \times 5 \times 7.5}\) | M1 | A correct statement involving angle \(DBC\) |
| Angle \(DBC = 0.943201...\) | A1 | awrt 0.943 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(CBD = \frac{1}{2}5(7.5)\sin(0.943)\) | M1 | Correct method for area of triangle \(CBD\); can be implied by awrt 15.2 |
| Angle \(EBA = \pi - 1.4 - \text{"0.943"}\) | M1 | A value for angle \(EBA\) of awrt 0.8 (from 0.7985926536... or 0.7983916536...) or value of \((1.74159... - \text{their angle } DBC)\) would imply this mark |
| \(AB = 5\cos(\pi - 1.4 - \text{"0.943"})\) or \(AE = 5\sin(\pi - 1.4 - \text{"0.943"})\) | M1 | \(AB = 5\cos(0.79859...) = 3.488...\) awrt 3.49; or \(AE = 5\sin(0.79859...) = 3.581...\) awrt 3.58. Must be clear \(\pi - 1.4 - \text{"0.943"}\) is used for angle \(EBA\) |
| Area \(EAB = \frac{1}{2}5\cos(\pi-1.4-\text{"0.943"}) \times 5\sin(\pi-1.4-\text{"0.943"})\) | dM1 | Dependent on previous M1; no other errors in finding area of triangle \(EAB\); allow M1 for area \(EAB\) = awrt 6.2 |
| Area \(ABCDE = 15.17... + 17.5 + 6.24... = 38.92...\) | A1cso | awrt 38.9 |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $BDE = \frac{1}{2}(5)^2(1.4)$ | M1 | Use of correct formula or method for area of sector |
| $= 17.5 \text{ cm}^2$ | A1 | 17.5 or equivalent |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6.1^2 = 5^2 + 7.5^2 - (2 \times 5 \times 7.5\cos DBC)$ or $\cos DBC = \frac{5^2 + 7.5^2 - 6.1^2}{2 \times 5 \times 7.5}$ | M1 | A correct statement involving angle $DBC$ |
| Angle $DBC = 0.943201...$ | A1 | awrt 0.943 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $CBD = \frac{1}{2}5(7.5)\sin(0.943)$ | M1 | Correct method for area of triangle $CBD$; can be implied by awrt 15.2 |
| Angle $EBA = \pi - 1.4 - \text{"0.943"}$ | M1 | A value for angle $EBA$ of awrt 0.8 (from 0.7985926536... or 0.7983916536...) or value of $(1.74159... - \text{their angle } DBC)$ would imply this mark |
| $AB = 5\cos(\pi - 1.4 - \text{"0.943"})$ or $AE = 5\sin(\pi - 1.4 - \text{"0.943"})$ | M1 | $AB = 5\cos(0.79859...) = 3.488...$ awrt 3.49; or $AE = 5\sin(0.79859...) = 3.581...$ awrt 3.58. Must be clear $\pi - 1.4 - \text{"0.943"}$ is used for angle $EBA$ |
| Area $EAB = \frac{1}{2}5\cos(\pi-1.4-\text{"0.943"}) \times 5\sin(\pi-1.4-\text{"0.943"})$ | dM1 | Dependent on previous M1; no other errors in finding area of triangle $EAB$; allow M1 for area $EAB$ = awrt 6.2 |
| Area $ABCDE = 15.17... + 17.5 + 6.24... = 38.92...$ | A1cso | awrt 38.9 |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e6b490c0-80c4-4e15-b587-ac052ee27db7-07_531_1127_264_411}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The shape $A B C D E A$, as shown in Figure 2, consists of a right-angled triangle $E A B$ and a triangle $D B C$ joined to a sector $B D E$ of a circle with radius 5 cm and centre $B$.
The points $A , B$ and $C$ lie on a straight line with $B C = 7.5 \mathrm {~cm}$.\\
Angle $E A B = \frac { \pi } { 2 }$ radians, angle $E B D = 1.4$ radians and $C D = 6.1 \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Find, in $\mathrm { cm } ^ { 2 }$, the area of the sector $B D E$.
\item Find the size of the angle $D B C$, giving your answer in radians to 3 decimal places.
\item Find, in $\mathrm { cm } ^ { 2 }$, the area of the shape $A B C D E A$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q5 [9]}}