| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of standard sector formulas (arc length, sector area) with basic trigonometry. Part (a) requires recognizing CD as a chord (2r sin θ), parts (b) and (c) involve direct substitution into memorized formulas with θ = π/6 giving exact values. Slightly easier than average due to the routine nature and helpful exact angle. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin\theta = \frac{r}{OC} \rightarrow OC = \frac{r}{\sin\theta}\) | M1 A1 | |
| \(CD = r + \frac{r}{\sin\theta}\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Radius of arc \(AB = 4 + \frac{4}{\sin\frac{\pi}{6}} = 4 + 8 = 12\) | B1 | SOI |
| Arc \(AB =\) their \(12 \times \frac{2\pi}{6}\) or \(\left(\frac{1}{2}AB =\right)\) their \(12 \times \frac{\pi}{6}\) | M1 | Expect \(4\pi\), must use their \(CD\), not 4 |
| Perimeter \(= 24 + 4\pi\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(FOC = \frac{1}{2} \times 4 \times\) their \(OC \times \sin\frac{\pi}{3}\) | M1 | |
| \(8\sqrt{3}\) | A1 | |
| Area sector \(FOE = \frac{1}{2} \times \frac{2\pi}{3} \times 4^2 = \frac{16\pi}{3}\) | B1 | |
| Shaded area \(= 16\sqrt{3} - \frac{16\pi}{3}\) | A1 | |
| Alternative method: | ||
| \(FC = \sqrt{(\text{their } OC)^2 - 4^2}\) | M1 | \(\sqrt{48}\) or \(4\sqrt{3}\) |
| Area \(FOC = \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3}\) | A1 | |
| Area of half sector \(FOE = \frac{1}{2} \times \frac{\pi}{3} \times 4^2 = \frac{8\pi}{3}\) | B1 | |
| Shaded area \(= 16\sqrt{3} - \frac{16\pi}{3}\) | A1 | |
| Total | 4 |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin\theta = \frac{r}{OC} \rightarrow OC = \frac{r}{\sin\theta}$ | M1 A1 | |
| $CD = r + \frac{r}{\sin\theta}$ | A1 | |
| **Total** | **3** | |
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Radius of arc $AB = 4 + \frac{4}{\sin\frac{\pi}{6}} = 4 + 8 = 12$ | B1 | SOI |
| Arc $AB =$ their $12 \times \frac{2\pi}{6}$ or $\left(\frac{1}{2}AB =\right)$ their $12 \times \frac{\pi}{6}$ | M1 | Expect $4\pi$, must use their $CD$, not 4 |
| Perimeter $= 24 + 4\pi$ | A1 | |
| **Total** | **3** | |
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $FOC = \frac{1}{2} \times 4 \times$ their $OC \times \sin\frac{\pi}{3}$ | M1 | |
| $8\sqrt{3}$ | A1 | |
| Area sector $FOE = \frac{1}{2} \times \frac{2\pi}{3} \times 4^2 = \frac{16\pi}{3}$ | B1 | |
| Shaded area $= 16\sqrt{3} - \frac{16\pi}{3}$ | A1 | |
| **Alternative method:** | | |
| $FC = \sqrt{(\text{their } OC)^2 - 4^2}$ | M1 | $\sqrt{48}$ or $4\sqrt{3}$ |
| Area $FOC = \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3}$ | A1 | |
| Area of half sector $FOE = \frac{1}{2} \times \frac{\pi}{3} \times 4^2 = \frac{8\pi}{3}$ | B1 | |
| Shaded area $= 16\sqrt{3} - \frac{16\pi}{3}$ | A1 | |
| **Total** | **4** | |\begin{tikzpicture}[>=stealth, scale=0.65]
% --- Geometry ---
\def\R{7}
\coordinate (O) at (0,4);
\coordinate (C) at (0,9);
\coordinate (F) at (-1.833,4.8);
\coordinate (E) at ( 1.833,4.8);
\coordinate (A) at ({0+\R*cos(246.4)},{9+\R*sin(246.4)});
\coordinate (B) at ({0+\R*cos(293.6)},{9+\R*sin(293.6)});
\coordinate (D) at (0,2);
% --- Shaded region: between tangent lines CF, CE and arc FE over top ---
\fill[gray!60] (C) -- (E) arc(23.6:156.4:2) -- cycle;
% --- Small circle ---
\draw (O) circle (2);
% --- Tangent lines from C to A and B ---
\draw (A) -- (C) -- (B);
% --- Arc ADB centred at C (curving downward) ---
\draw (A) arc(246.4:293.6:\R);
% --- Dashed vertical line C--O--D ---
\draw[dashed] (C) -- (D);
% --- Dashed radius O--F with label ---
\draw[dashed] (O) -- (F);
\node[below, font=\small] at ($(O)!0.55!(F)$) {$r$};
% --- Angle arc at C (theta rad) ---
\pic[draw, angle radius=8mm, angle eccentricity=1.35]
{angle = F--C--E};
\node[font=\small] (thetalabel) at (-1.8,7.8) {$\theta\,\mathrm{rad}$};
% --- Point labels ---
\node[above] at (C) {$C$};
\node[below left] at (A) {$A$};
\node[below right] at (B) {$B$};
\node[below right] at (D) {$D$};
\node[left] at (F) {$F$};
\node[right] at (E) {$E$};
\node[below right, inner sep=1pt] at (O) {$O$};
\end{tikzpicture}
The diagram shows a sector $CAB$ which is part of a circle with centre $C$. A circle with centre $O$ and radius $r$ lies within the sector and touches it at $D$, $E$, and $F$, where $COD$ is a straight line and angle $ACD$ is $\theta$ radians.
\begin{enumerate}[label=(\alph*)]
\item Find $C D$ in terms of $r$ and $\sin \theta$.\\
It is now given that $r = 4$ and $\theta = \frac { 1 } { 6 } \pi$.
\item Find the perimeter of sector $C A B$ in terms of $\pi$.
\item Find the area of the shaded region in terms of $\pi$ and $\sqrt { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q10 [10]}}