| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.3 This is a standard C2 sector/segment question requiring routine application of formulas. Part (a) uses cosine rule to find an angle, parts (b-d) apply standard area formulas (sector, triangle, segment), and part (e) calculates perimeter using arc length. While multi-part with several steps, each component is straightforward application of learned techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05c 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 | Marks | Guidance |
| \(\cos PQR = \frac{6^2+6^2-(6\sqrt{3})^2}{2\times6\times6} \left\{=-\frac{1}{2}\right\}\) | M1, A1 | Use of cosine rule for \(\cos PQR\); allow \(A\), \(\theta\) or other symbol. Apply usual rules: formula not stated must be correct; correct formula stated, allow one sign slip. Correct expression \(\frac{6^2+6^2-(6\sqrt{3})^2}{2\times6\times6}\) e.g. \(-\frac{36}{72}\) or \(-\frac{1}{2}\). Also allow invisible brackets \([6\sqrt{3}^2]\) |
| \(PQR = \frac{2\pi}{3}\) | A1 | \(\frac{2\pi}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin\theta = \frac{a\sqrt{3}}{6}\) where \(\theta\) is any symbol and \(a<6\) | M1 | |
| \(\sin\theta = \frac{3\sqrt{3}}{6}\) where \(\theta\) is any symbol | A1 | |
| \(\frac{2\pi}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area \(= \frac{1}{2}\times6^2\times\frac{2\pi}{3}\) m² | M1 | Use of \(\frac{1}{2}r^2\theta\) with \(r=6\) and \(\theta=\) their (a); for M mark \(\theta\) does not have to be exact. M0 if using degrees |
| \(= 12\pi\) m² | A1cso | \(12\pi\) c.s.o. (answer given in question) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of \(\Delta = \frac{1}{2}\times6\times6\times\sin\frac{2\pi}{3}\) m² | M1 | Use of \(\frac{1}{2}r^2\sin\theta\) with \(r=6\) and their (a); or attempt to find \(h\) using trig/Pythagoras and use in \(\frac{1}{2}bh\) |
| \(= 9\sqrt{3}\) m² | A1cso | \(9\sqrt{3}\) c.s.o. Must be exact; correct approx. followed by \(9\sqrt{3}\) is okay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of segment \(= 12\pi - 9\sqrt{3}\) m² | M1 | Use of area of sector \(-\) area of \(\Delta\), or use of \(\frac{1}{2}r^2(\theta-\sin\theta)\) |
| \(= 22.1\) m² | A1 | Any value to 1 d.p. or more which rounds to 22.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Perimeter \(= 6+6+\left[6\times\frac{2\pi}{3}\right]\) m | M1 | \(6+6+[6\times\text{their (a)}]\) |
| \(= 24.6\) m | A1ft | Correct for their (a) to 1 d.p. or more |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos PQR = \frac{6^2+6^2-(6\sqrt{3})^2}{2\times6\times6} \left\{=-\frac{1}{2}\right\}$ | M1, A1 | Use of cosine rule for $\cos PQR$; allow $A$, $\theta$ or other symbol. Apply usual rules: formula not stated must be correct; correct formula stated, allow one sign slip. Correct expression $\frac{6^2+6^2-(6\sqrt{3})^2}{2\times6\times6}$ e.g. $-\frac{36}{72}$ or $-\frac{1}{2}$. Also allow invisible brackets $[6\sqrt{3}^2]$ |
| $PQR = \frac{2\pi}{3}$ | A1 | $\frac{2\pi}{3}$ |
**Alternative for (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta = \frac{a\sqrt{3}}{6}$ where $\theta$ is any symbol and $a<6$ | M1 | |
| $\sin\theta = \frac{3\sqrt{3}}{6}$ where $\theta$ is any symbol | A1 | |
| $\frac{2\pi}{3}$ | A1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\times6^2\times\frac{2\pi}{3}$ m² | M1 | Use of $\frac{1}{2}r^2\theta$ with $r=6$ and $\theta=$ their (a); for M mark $\theta$ does not have to be exact. M0 if using degrees |
| $= 12\pi$ m² | A1cso | $12\pi$ c.s.o. (answer given in question) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of $\Delta = \frac{1}{2}\times6\times6\times\sin\frac{2\pi}{3}$ m² | M1 | Use of $\frac{1}{2}r^2\sin\theta$ with $r=6$ and their (a); or attempt to find $h$ using trig/Pythagoras and use in $\frac{1}{2}bh$ |
| $= 9\sqrt{3}$ m² | A1cso | $9\sqrt{3}$ c.s.o. Must be exact; correct approx. followed by $9\sqrt{3}$ is okay |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of segment $= 12\pi - 9\sqrt{3}$ m² | M1 | Use of area of sector $-$ area of $\Delta$, or use of $\frac{1}{2}r^2(\theta-\sin\theta)$ |
| $= 22.1$ m² | A1 | Any value to 1 d.p. or more which rounds to 22.1 |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perimeter $= 6+6+\left[6\times\frac{2\pi}{3}\right]$ m | M1 | $6+6+[6\times\text{their (a)}]$ |
| $= 24.6$ m | A1ft | Correct for their (a) to 1 d.p. or more |9.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{872356ab-68d3-43ee-8b76-650a2697d80e-11_627_965_338_502}
\end{center}
\end{figure}
Figure 2 shows a plan of a patio. The patio $P Q R S$ is in the shape of a sector of a circle with centre $Q$ and radius 6 m .
Given that the length of the straight line $P R$ is $6 \sqrt { } 3 \mathrm {~m}$,
\begin{enumerate}[label=(\alph*)]
\item find the exact size of angle $P Q R$ in radians.
\item Show that the area of the patio $P Q R S$ is $12 \pi \mathrm {~m} ^ { 2 }$.
\item Find the exact area of the triangle $P Q R$.
\item Find, in $\mathrm { m } ^ { 2 }$ to 1 decimal place, the area of the segment $P R S$.
\item Find, in $m$ to 1 decimal place, the perimeter of the patio $P Q R S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2007 Q9 [11]}}