| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Logo and design problems |
| Difficulty | Standard +0.3 This is a standard C2 sector/triangle geometry problem requiring basic trigonometry (cosine rule) to find an angle, then straightforward application of arc length and sector area formulas. The multi-part structure and need to work with radians adds slight complexity, but all techniques are routine for this level with no novel problem-solving required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.03h Parametric equations: in modelling contexts1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\cos C\hat{O}D=\dfrac{8^2+8^2-7^2}{2\times8\times8}\) or uses \(\angle COD=2\times\arcsin\frac{3.5}{8}\) | M1 | Either correctly quoted cosine rule, or split isosceles triangle using Pythagoras/arcsin |
| \(\angle COD=0.906\) (3sf) | A1* | Accept awrt 0.906; needs correct work leading to stated answer [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Uses \(s=8\theta\) for any \(\theta\) in radians or \(\dfrac{\theta}{360}\times2\pi\times8\) for any \(\theta\) in degrees | M1 | Formula for arc length with \(r=8\) |
| \(\theta=\dfrac{\pi-\text{"COD"}}{2}\) (\(=\) awrt 1.12) or \(2\theta\) (\(=\) awrt 2.24) | M1 | Uses angles on straight line (or other geometry) to find angle \(BOC\) or \(AOD\) |
| Perimeter \(=23+(16\times\theta)\), accept awrt 40.9 (cm) | A1 | Correct work leading to awrt 40.9, not 40.8 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area of triangle \(=\frac{1}{2}\times8\times8\times\sin0.906\) (or 25.1781..., accept awrt 25.2) | M1 | Correct statement for area of triangle using correct angle; or \(\frac{1}{2}\times8\times7\times\sin1.118\) or \(\frac{1}{2}\times7\times h\) after \(h\) from correct Pythagoras |
| Area of sector \(=\frac{1}{2}(8)^2\times\text{"1.117979732"}\) (or 35.775..., accept awrt 35.8) | M1 | Formula for area of sector \(\frac{1}{2}r^2\theta\) with \(r=8\) and angle \(BOC\) or \(AOD\) or \((BOC+AOD)\) not \(COD\) |
| Total Area \(=\) Area of two sectors \(+\) area of triangle \(=\) awrt 96.7 or 96.8 or 96.9 (cm\(^2\)) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area of semi-circle \(=\frac{1}{2}\times\pi\times8\times8\) (or 100.5) | M1 | Correct statement for area of semicircle |
| Area of segment \(=\frac{1}{2}(8)^2\times(\text{"0.906"}-\sin\text{"0.906"})\) (or 3.807) | M1 | Formula for area of segment with \(r=8\) |
| Area required \(=\) awrt 96.7 or 96.8 or 96.9 (cm\(^2\)) | A1 | [3] |
# Question 4:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\cos C\hat{O}D=\dfrac{8^2+8^2-7^2}{2\times8\times8}$ or uses $\angle COD=2\times\arcsin\frac{3.5}{8}$ | M1 | Either correctly quoted cosine rule, or split isosceles triangle using Pythagoras/arcsin |
| $\angle COD=0.906$ (3sf) | A1* | Accept awrt 0.906; needs correct work leading to stated answer **[2]** |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Uses $s=8\theta$ for any $\theta$ in radians or $\dfrac{\theta}{360}\times2\pi\times8$ for any $\theta$ in degrees | M1 | Formula for arc length with $r=8$ |
| $\theta=\dfrac{\pi-\text{"COD"}}{2}$ ($=$ awrt 1.12) or $2\theta$ ($=$ awrt 2.24) | M1 | Uses angles on straight line (or other geometry) to find angle $BOC$ or $AOD$ |
| Perimeter $=23+(16\times\theta)$, accept awrt 40.9 (cm) | A1 | Correct work leading to awrt 40.9, not 40.8 **[3]** |
## Part (c) — Way 1
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area of triangle $=\frac{1}{2}\times8\times8\times\sin0.906$ (or 25.1781..., accept awrt 25.2) | M1 | Correct statement for area of triangle using correct angle; or $\frac{1}{2}\times8\times7\times\sin1.118$ or $\frac{1}{2}\times7\times h$ after $h$ from correct Pythagoras |
| Area of sector $=\frac{1}{2}(8)^2\times\text{"1.117979732"}$ (or 35.775..., accept awrt 35.8) | M1 | Formula for area of sector $\frac{1}{2}r^2\theta$ with $r=8$ and angle $BOC$ or $AOD$ or $(BOC+AOD)$ **not** $COD$ |
| Total Area $=$ Area of two sectors $+$ area of triangle $=$ awrt 96.7 or 96.8 or 96.9 (cm$^2$) | A1 | **[3]** |
**Way 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area of semi-circle $=\frac{1}{2}\times\pi\times8\times8$ (or 100.5) | M1 | Correct statement for area of semicircle |
| Area of segment $=\frac{1}{2}(8)^2\times(\text{"0.906"}-\sin\text{"0.906"})$ (or 3.807) | M1 | Formula for area of segment with $r=8$ |
| Area required $=$ awrt 96.7 or 96.8 or 96.9 (cm$^2$) | A1 | **[3]** |
**Total: [8]**4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a7593c3-4f0b-4351-afae-7bd98cfc351d-06_513_775_269_589}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a design for a scraper blade. The blade $A O B C D A$ consists of an isosceles triangle $C O D$ joined along its equal sides to sectors $O B C$ and $O D A$ of a circle with centre $O$ and radius 8 cm . Angles $A O D$ and $B O C$ are equal. $A O B$ is a straight line and is parallel to the line $D C . D C$ has length 7 cm .
\begin{enumerate}[label=(\alph*)]
\item Show that the angle $C O D$ is 0.906 radians, correct to 3 significant figures.
\item Find the perimeter of $A O B C D A$, giving your answer to 3 significant figures.
\item Find the area of $A O B C D A$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2015 Q4 [8]}}