| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| 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 straightforward multi-part question combining basic trigonometry with sector geometry. Part (a) requires simple arc length formula manipulation (θ = s/r = 2/10), and part (b) involves finding a triangle area using standard formulas (½r²sinθ) then subtracting from a sector area. All techniques are standard P1 content with no novel insight required, 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 | Mark | Guidance |
| \(A\hat{O}B = \frac{2}{10}\) | B1 | OE. Sight of \(0.2\) from \(s=r\theta\) but \(10\theta=2\) is not enough. ISW if \(\frac{2}{10}=\frac{\pi}{5}\) |
| \(B\hat{O}C = \frac{5\pi+6}{30}\) or \(\frac{1}{6}\pi+0.2\) | B1 | OE e.g. \(0.724^c\) AWRT or \(41.5\) degrees AWRT. But not \(\frac{2+\frac{5\pi}{3}}{10}\) – fraction within a fraction. ISW incorrect simplifications |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Arc \(AC = \frac{10\pi}{6}\) or Arc \(BC = \frac{10\pi}{6}+2\) or \(7.2\) | B1 | AWRT. Sight of \(\frac{10\pi}{6}\) or \(5.2\) or \(7.2\) |
| \(B\hat{O}C = \frac{5\pi+6}{30}\) or \(\frac{1}{6}\pi+0.2\) | B1 | OE e.g. \(0.724^c\) AWRT or \(41.5\) degrees AWRT. But not \(\frac{2+\frac{5\pi}{3}}{10}\). ISW incorrect simplifications |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([BP]=10\sin\left(\frac{5\pi+6}{30}\right)\) and \([OP]=10\cos\left(\frac{5\pi+6}{30}\right)\) \([=6.6208...]\) and \([=7.494...]\) OR \([BP]=10\sin\left(\frac{5\pi+6}{30}\right)\) and \([O\hat{B}P]=\left(\frac{5\pi-3}{15}\right)\) \([=6.6208...]\) and \([=0.84719...]\) | M1 | OE. Any correct method for both lengths, for their angle BOC (which may have been incorrectly 'simplified' but not 0.2) or length BP and \(O\hat{B}P\). May be seen as part of \(\frac{1}{2}ab\sin C\). Sight of correct method enough. Can be implied by next A1 |
| Area of \(\triangle OBP = \frac{1}{2}\times10\sin\left(\frac{5\pi+6}{30}\right)\times10\cos\left(\frac{5\pi+6}{30}\right)\) or \(\frac{1}{2}\times10\times10\sin\left(\frac{5\pi+6}{30}\right)\times\sin\left(\frac{5\pi-3}{15}\right)\) \([=24.809]\) | A1 | OE. Can be implied by any answer in range \((24.7, 24.9)\) or a final answer in range \((11.3, 11.5)\) WWW |
| \([\text{Sector }BOC]=\frac{1}{2}\times10^2\times\text{their}\left(\frac{5\pi+6}{30}\right)\) \(\left[=50\left(\frac{5\pi+6}{30}\right)=36.1799...\right]\) | M1 | Use of \(\frac{1}{2}r^2\theta\) with their angle BOC (may have been incorrectly 'simplified' but not \(0.2\)) |
| Area of region \(BPC = 11.4\) | A1 | CAO |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A\hat{O}B = \frac{2}{10}$ | B1 | OE. Sight of $0.2$ from $s=r\theta$ but $10\theta=2$ is not enough. ISW if $\frac{2}{10}=\frac{\pi}{5}$ |
| $B\hat{O}C = \frac{5\pi+6}{30}$ or $\frac{1}{6}\pi+0.2$ | B1 | OE e.g. $0.724^c$ AWRT or $41.5$ degrees AWRT. But not $\frac{2+\frac{5\pi}{3}}{10}$ – fraction within a fraction. ISW incorrect simplifications |
**Alternative method for 7(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Arc $AC = \frac{10\pi}{6}$ or Arc $BC = \frac{10\pi}{6}+2$ or $7.2$ | B1 | AWRT. Sight of $\frac{10\pi}{6}$ or $5.2$ or $7.2$ |
| $B\hat{O}C = \frac{5\pi+6}{30}$ or $\frac{1}{6}\pi+0.2$ | B1 | OE e.g. $0.724^c$ AWRT or $41.5$ degrees AWRT. But not $\frac{2+\frac{5\pi}{3}}{10}$. ISW incorrect simplifications |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[BP]=10\sin\left(\frac{5\pi+6}{30}\right)$ and $[OP]=10\cos\left(\frac{5\pi+6}{30}\right)$ $[=6.6208...]$ and $[=7.494...]$ **OR** $[BP]=10\sin\left(\frac{5\pi+6}{30}\right)$ and $[O\hat{B}P]=\left(\frac{5\pi-3}{15}\right)$ $[=6.6208...]$ and $[=0.84719...]$ | M1 | OE. Any correct method for **both** lengths, for their angle BOC (which may have been incorrectly 'simplified' but not 0.2) or length BP and $O\hat{B}P$. May be seen as part of $\frac{1}{2}ab\sin C$. Sight of correct method enough. Can be implied by next A1 |
| Area of $\triangle OBP = \frac{1}{2}\times10\sin\left(\frac{5\pi+6}{30}\right)\times10\cos\left(\frac{5\pi+6}{30}\right)$ or $\frac{1}{2}\times10\times10\sin\left(\frac{5\pi+6}{30}\right)\times\sin\left(\frac{5\pi-3}{15}\right)$ $[=24.809]$ | A1 | OE. Can be implied by any answer in range $(24.7, 24.9)$ or a final answer in range $(11.3, 11.5)$ WWW |
| $[\text{Sector }BOC]=\frac{1}{2}\times10^2\times\text{their}\left(\frac{5\pi+6}{30}\right)$ $\left[=50\left(\frac{5\pi+6}{30}\right)=36.1799...\right]$ | M1 | Use of $\frac{1}{2}r^2\theta$ with their angle BOC (may have been incorrectly 'simplified' but not $0.2$) |
| Area of region $BPC = 11.4$ | A1 | CAO |
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\includegraphics[max width=\textwidth, alt={}, center]{bb7595c9-93ae-49e8-9cc5-9ecc802e6060-10_593_841_260_651}
The diagram shows a sector $O B A C$ of a circle with centre $O$ and radius 10 cm . The point $P$ lies on $O C$ and $B P$ is perpendicular to $O C$. Angle $A O C = \frac { 1 } { 6 } \pi$ and the length of the $\operatorname { arc } A B$ is 2 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the angle $B O C$.
\item Hence find the area of the shaded region $B P C$ giving your answer correct to 3 significant figures. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q7 [6]}}