| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.8 This question requires geometric insight to prove angle relationships using circle theorems (part i), then combines sector and triangle areas with algebraic manipulation (part ii), before applying these results to a novel equilateral triangle configuration (part b). While individual components are standard P1 topics, the multi-step reasoning, proof element, and application to a non-routine context elevate it above average difficulty. |
| 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 theta1.05q Trig in context: vectors, kinematics, forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BAO = OBA = \dfrac{\pi}{2} - \alpha\) | Allow use of 90° or 180° | |
| \(AOB = \pi - \left(\dfrac{\pi}{2} - \alpha\right) - \left(\dfrac{\pi}{2} - \alpha\right) = 2\alpha\) AG | M1A1 [2] | Or other valid reasoning |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{2}r^2(2\alpha) - \dfrac{1}{2}r^2\sin 2\alpha\) oe | B2,1,0 [2] | SCB1 for reversed subtraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(\alpha = \dfrac{\pi}{6}\), \(r = 4\) | B1B1 | |
| 1 segment \(S = \left(\dfrac{1}{2}\right)4^2\left(\dfrac{\pi}{3}\right) - \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3}\) | ||
| \(= \left(\dfrac{8\pi}{3} - 4\sqrt{3}\right)\) | M1 | Ft *their* (ii), \(\alpha, r\) |
| Area \(ABC\): \(T = \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3}\) \(\left(= 4\sqrt{3}\right)\) | B1 | Or \(AXB = \dfrac{T}{3} = 4\tan\dfrac{\pi}{6}\) or \(\dfrac{1}{2}\left(\dfrac{4}{\sqrt{3}}\right)^2\sin\dfrac{2\pi}{3}\left(= \dfrac{4\sqrt{3}}{3}\right)\) |
| \(T - 3S = \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3} - 3\left[\left(\dfrac{1}{2}\right)4^2\left(\dfrac{\pi}{3}\right) - \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3}\right]\) | M1 | Or \(3\left[\dfrac{T}{3} - S\right] = 3\left[\dfrac{4\sqrt{3}}{3} - \left(\dfrac{8\pi}{3} - 4\sqrt{3}\right)\right]\) |
| \(16\sqrt{3} - 8\pi\) cao | A1 [6] |
## Question 9:
### Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BAO = OBA = \dfrac{\pi}{2} - \alpha$ | | Allow use of 90° or 180° |
| $AOB = \pi - \left(\dfrac{\pi}{2} - \alpha\right) - \left(\dfrac{\pi}{2} - \alpha\right) = 2\alpha$ **AG** | M1A1 [2] | Or other valid reasoning |
### Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{2}r^2(2\alpha) - \dfrac{1}{2}r^2\sin 2\alpha$ oe | B2,1,0 [2] | SCB1 for reversed subtraction |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $\alpha = \dfrac{\pi}{6}$, $r = 4$ | B1B1 | |
| 1 segment $S = \left(\dfrac{1}{2}\right)4^2\left(\dfrac{\pi}{3}\right) - \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3}$ | | |
| $= \left(\dfrac{8\pi}{3} - 4\sqrt{3}\right)$ | M1 | Ft *their* (ii), $\alpha, r$ |
| Area $ABC$: $T = \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3}$ $\left(= 4\sqrt{3}\right)$ | B1 | Or $AXB = \dfrac{T}{3} = 4\tan\dfrac{\pi}{6}$ or $\dfrac{1}{2}\left(\dfrac{4}{\sqrt{3}}\right)^2\sin\dfrac{2\pi}{3}\left(= \dfrac{4\sqrt{3}}{3}\right)$ |
| $T - 3S = \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3} - 3\left[\left(\dfrac{1}{2}\right)4^2\left(\dfrac{\pi}{3}\right) - \left(\dfrac{1}{2}\right)4^2\sin\dfrac{\pi}{3}\right]$ | M1 | Or $3\left[\dfrac{T}{3} - S\right] = 3\left[\dfrac{4\sqrt{3}}{3} - \left(\dfrac{8\pi}{3} - 4\sqrt{3}\right)\right]$ |
| $16\sqrt{3} - 8\pi$ cao | A1 [6] | |
---9
\begin{enumerate}[label=(\alph*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0f58de6c-aba7-4a79-a962-c23be3ee0aa9-4_433_476_264_872}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
In Fig. 1, $O A B$ is a sector of a circle with centre $O$ and radius $r$. $A X$ is the tangent at $A$ to the arc $A B$ and angle $B A X = \alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that angle $A O B = 2 \alpha$.
\item Find the area of the shaded segment in terms of $r$ and $\alpha$.
\end{enumerate}\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0f58de6c-aba7-4a79-a962-c23be3ee0aa9-4_451_503_1162_861}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
In Fig. 2, $A B C$ is an equilateral triangle of side 4 cm . The lines $A X , B X$ and $C X$ are tangents to the equal circular $\operatorname { arcs } A B , B C$ and $C A$. Use the results in part (a) to find the area of the shaded region, giving your answer in terms of $\pi$ and $\sqrt { } 3$.\\[0pt]
[6]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2016 Q9 [10]}}