| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Circular arc problems |
| Difficulty | Standard +0.3 This is a standard circular segment problem requiring basic geometry (perpendicular from center bisects chord), Pythagoras' theorem, inverse trigonometry, and sector/triangle area formulas. All steps are routine applications of well-practiced techniques with clear scaffolding through parts (a) and (b), making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| States \(r^2=1.2^2+(r-0.4)^2\) | M1 | Attempts Pythagoras with \(r\), \((r-0.4)\) and \(1.2\) in correct positions |
| \(0.8r=1.60 \Rightarrow r=2\) | A1* | Proceeds to \(r=2\) with no errors; if alternative method used must state "hence \(r=2\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \('\frac{1}{2}\theta'=\arcsin\left(\frac{1.2}{2}\right) \Rightarrow '\frac{1}{2}\theta'=\) awrt \(37°\) or awrt \(0.64\) rads | M1 | Attempts to find angle or half-angle; accept \(\arctan\left(\frac{1.2}{1.6}\right)\), \(\arcsin\left(\frac{1.2}{2}\right)\), \(\arccos\left(\frac{1.6}{2}\right)\) |
| cso \(AOB=1.2870\) | A1* | Given answer; allow from value where \(\frac{1}{2}\) angle lacks 4 d.p. accuracy |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area sector \(=\frac{'73.74'}{360}\times\pi\times2^2\) or \(\frac{1}{2}\times2^2\times'1.2870'\) | M1 | Correct method for area of sector, radius 2, angle 1.2870 |
| \(\text{Area}=\frac{1}{2}\times2\times2\times\sin'73.74°'\) or \(\frac{1}{2}\times2\times2\times\sin'1.2870'\) | M1 | Correct method for area of isosceles triangle; also accept \(\frac{1}{2}bh=\frac{1}{2}\times2.4\times1.6\) |
| \(=\frac{1}{2}\times2^2\times'1.29'-\frac{1}{2}\times2^2\times\sin'1.29'=0.654\,(m^2)\) | dM1, A1 | dM1: correct combination sector \(-\) triangle, dependent on both M's. A1: awrt \(0.654\,(m^2)\) |
## Question 11:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| States $r^2=1.2^2+(r-0.4)^2$ | M1 | Attempts Pythagoras with $r$, $(r-0.4)$ and $1.2$ in correct positions |
| $0.8r=1.60 \Rightarrow r=2$ | A1* | Proceeds to $r=2$ with no errors; if alternative method used must state "hence $r=2$" |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $'\frac{1}{2}\theta'=\arcsin\left(\frac{1.2}{2}\right) \Rightarrow '\frac{1}{2}\theta'=$ awrt $37°$ or awrt $0.64$ rads | M1 | Attempts to find angle or half-angle; accept $\arctan\left(\frac{1.2}{1.6}\right)$, $\arcsin\left(\frac{1.2}{2}\right)$, $\arccos\left(\frac{1.6}{2}\right)$ |
| cso $AOB=1.2870$ | A1* | Given answer; allow from value where $\frac{1}{2}$ angle lacks 4 d.p. accuracy |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area sector $=\frac{'73.74'}{360}\times\pi\times2^2$ or $\frac{1}{2}\times2^2\times'1.2870'$ | M1 | Correct method for area of sector, radius 2, angle 1.2870 |
| $\text{Area}=\frac{1}{2}\times2\times2\times\sin'73.74°'$ or $\frac{1}{2}\times2\times2\times\sin'1.2870'$ | M1 | Correct method for area of isosceles triangle; also accept $\frac{1}{2}bh=\frac{1}{2}\times2.4\times1.6$ |
| $=\frac{1}{2}\times2^2\times'1.29'-\frac{1}{2}\times2^2\times\sin'1.29'=0.654\,(m^2)$ | dM1, A1 | dM1: correct combination sector $-$ triangle, dependent on both M's. A1: awrt $0.654\,(m^2)$ |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-16_892_825_228_548}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 2 shows the design for a sail $A P B C A$.
The curved edge $A P B$ of the sail is an arc of a circle centre $O$ and radius $r \mathrm {~m}$.
The straight edge $A C B$ is a chord of the circle.
The height $A B$ of the sail is 2.4 m .
The maximum width $C P$ of the sail is 0.4 m .
\begin{enumerate}[label=(\alph*)]
\item Show that $r = 2$
\item Show, to 4 decimal places, that angle $A O B = 1.2870$ radians.
\item Hence calculate the area of the sail, giving your answer, in $\mathrm { m } ^ { 2 }$, to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q11 [8]}}