Edexcel C2 2017 June — Question 4 8 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2017
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeCompound shape area
DifficultyStandard +0.3 This is a straightforward application of standard arc length and sector area formulas (s=rθ, A=½r²θ) with basic triangle area calculation. All necessary values are given directly, requiring only substitution into formulas and arithmetic. Slightly easier than average due to minimal problem-solving required.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{752efc6c-8d0e-46a6-b75d-5125956969d8-10_508_960_212_477} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Diagram not drawn to scale Figure 1 is a sketch representing the cross-section of a large tent \(A B C D E F\). \(A B\) and \(D E\) are line segments of equal length.
Angle \(F A B\) and angle \(D E F\) are equal. \(F\) is the midpoint of the straight line \(A E\) and \(F C\) is perpendicular to \(A E\). \(B C D\) is an arc of a circle of radius 3.5 m with centre at \(F\).
It is given that $$\begin{aligned} A F & = F E = 3.7 \mathrm {~m} \\ B F & = F D = 3.5 \mathrm {~m} \\ \text { angle } B F D & = 1.77 \text { radians } \end{aligned}$$ Find
  1. the length of the arc \(B C D\) in metres to 2 decimal places,
  2. the area of the sector \(F B C D\) in \(\mathrm { m } ^ { 2 }\) to 2 decimal places,
  3. the total area of the cross-section of the tent in \(\mathrm { m } ^ { 2 }\) to 2 decimal places.
AnswerMarks
(a) Usually answered in radians: Uses \(BCD=3.5\times(\text{angle})=3.5\times 1.77=6.195\) (m) (accept awrt 6.20)M1 A1
(b) Area \(=\frac{1}{2}(3.5)^{2}\times 1.77=10.84\) (m²)M1 A1
(c) Area of triangle \(=\frac{1}{2}\times 3.7\times 3.5\times\sin(\text{angle})=\frac{1}{2}\times 3.7\times 3.5\times\sin\left(\frac{\pi}{2}-\frac{1.77}{2}\right)\) (=awrt 4.1)M1, A1
Total area = "\(10.84"+2\times"4.101"\) = 19.04M1 A1cao
Notes:
- (a) M1: uses \(s=3.5\times\theta\) with \(\theta\) in radians or completely correct work in degrees; A1: awrt 6.20 or just 6.2 (do not need to see units). Correct answer can imply the method
- (b) M1: for attempt to use \(A=\frac{1}{2}\times 3.5^{2}\times\theta\) (Accept \(\theta\) in degrees); A1: for awrt 10.84 (do not need to see units). isw if correct answer is followed by 10.8. Correct answer can imply the method
- (c) M1: Uses area of triangle \(\frac{1}{2}\times 3.7\times 3.5\times\sin(\text{angle})\). Must be correct method for area of triangle but may be less direct; A1: Correct expression using correct angle i.e. \(\frac{\pi}{2}-\frac{1.77}{2}\) or awrt 0.69 or awrt 39 degrees (need at least 2 sf if no other work seen, but may be implied by correct final answer). If correct expression is given then isw (so e.g. isw an answer of 0.775 implying angle set to degrees on calculator); M1: Adds twice their second calculated area (even if rectangle or segment; errors in one or both formulae – such as missing \(\frac{1}{2}\) or mixture of degrees and radians or weak attempt at triangle area) so M0A0M1A0 is a possible mark distribution; A1cao: Common answer through insufficient accuracy is 19.08 which loses final mark
- Special Case: Mark profile M1A0M1A0M1A0 can be given if angle is misunderstood as \(1.77\pi\) or as \(AFB\) for example. But use of \(3.5\times 3.7\sin(\text{angle})\) and later doubled and added to "\(10.84"\) is 1st M0, 2nd M1
| (a) **Usually answered in radians:** Uses $BCD=3.5\times(\text{angle})=3.5\times 1.77=6.195$ (m) (accept awrt 6.20) | M1 A1 | |
|---|---|---|
| (b) Area $=\frac{1}{2}(3.5)^{2}\times 1.77=10.84$ (m²) | M1 A1 | |
| (c) Area of triangle $=\frac{1}{2}\times 3.7\times 3.5\times\sin(\text{angle})=\frac{1}{2}\times 3.7\times 3.5\times\sin\left(\frac{\pi}{2}-\frac{1.77}{2}\right)$ (=awrt 4.1) | M1, A1 | |
| Total area = "$10.84"+2\times"4.101"$ = 19.04 | M1 A1cao | |

**Notes:**
- (a) M1: uses $s=3.5\times\theta$ with $\theta$ in radians or completely correct work in degrees; A1: awrt 6.20 or just 6.2 (do not need to see units). Correct answer can imply the method
- (b) M1: for attempt to use $A=\frac{1}{2}\times 3.5^{2}\times\theta$ (Accept $\theta$ in degrees); A1: for awrt 10.84 (do not need to see units). isw if correct answer is followed by 10.8. Correct answer can imply the method
- (c) M1: Uses area of triangle $\frac{1}{2}\times 3.7\times 3.5\times\sin(\text{angle})$. Must be correct method for area of triangle but may be less direct; A1: Correct expression using correct angle i.e. $\frac{\pi}{2}-\frac{1.77}{2}$ or awrt 0.69 or awrt 39 degrees (need at least 2 sf if no other work seen, but may be implied by correct final answer). If correct expression is given then isw (so e.g. isw an answer of 0.775 implying angle set to degrees on calculator); M1: Adds twice their second calculated area (even if rectangle or segment; errors in one or both formulae – such as missing $\frac{1}{2}$ or mixture of degrees and radians or weak attempt at triangle area) so M0A0M1A0 is a possible mark distribution; A1cao: Common answer through insufficient accuracy is 19.08 which loses final mark
- **Special Case:** Mark profile M1A0M1A0M1A0 can be given if angle is misunderstood as $1.77\pi$ or as $AFB$ for example. But use of $3.5\times 3.7\sin(\text{angle})$ and later doubled and added to "$10.84"$ is 1st M0, 2nd M1
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{752efc6c-8d0e-46a6-b75d-5125956969d8-10_508_960_212_477}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Diagram not drawn to scale

Figure 1 is a sketch representing the cross-section of a large tent $A B C D E F$. $A B$ and $D E$ are line segments of equal length.\\
Angle $F A B$ and angle $D E F$ are equal.\\
$F$ is the midpoint of the straight line $A E$ and $F C$ is perpendicular to $A E$.\\
$B C D$ is an arc of a circle of radius 3.5 m with centre at $F$.\\
It is given that

$$\begin{aligned}
A F & = F E = 3.7 \mathrm {~m} \\
B F & = F D = 3.5 \mathrm {~m} \\
\text { angle } B F D & = 1.77 \text { radians }
\end{aligned}$$

Find
\begin{enumerate}[label=(\alph*)]
\item the length of the arc $B C D$ in metres to 2 decimal places,
\item the area of the sector $F B C D$ in $\mathrm { m } ^ { 2 }$ to 2 decimal places,
\item the total area of the cross-section of the tent in $\mathrm { m } ^ { 2 }$ to 2 decimal places.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2 2017 Q4 [8]}}
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