| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| 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 application of trigonometry and sector geometry. Part (a) requires finding a triangle area using sin formula with a given angle. Part (b) involves using the perpendicular height to find the angle, then calculating arc length and using Pythagoras. All techniques are standard for this topic with no novel insight required, making it slightly easier than average. |
| Spec | 1.05c 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 | Marks | Guidance |
| Sector area \(= \frac{1}{2}r^2\left(\frac{\pi}{6}\right)\left[=\frac{\pi}{12}r^2\right]\) | B1 | Using \(\frac{1}{2}r^2\theta\) with \(\theta\) in radians SOI. B0 if using a value for \(r\) |
| \(BD = \sin\frac{\pi}{6}r\left[=\frac{1}{2}r\right]\) and \(AD = \cos\frac{\pi}{6}r\left[=\frac{\sqrt{3}}{2}r\right]\), so triangle area \(= \frac{1}{2}\left(\sin\frac{\pi}{6}r\right)\left(\cos\frac{\pi}{6}r\right)\left[=\frac{1}{2}\times\frac{1}{2}r\times\frac{\sqrt{3}}{2}r\right]\) | B1 | SOI finding triangle area. Decimals B0 unless exact values seen in working |
| Area of \(BCD = \frac{1}{12}\pi r^2 - \frac{\sqrt{3}}{8}r^2\) | B1 | OE e.g. \(\frac{r^2}{4}\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)\) with \(\cos\frac{\pi}{6}\) and \(\sin\frac{\pi}{6}\) evaluated. Must be exact, in terms of \(r^2\). ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Angle \(BAC = \sin^{-1}\left(\frac{\frac{\sqrt{3}}{2}r}{r}\right)\left[=\frac{\pi}{3}\right]\) | B1 | SOI by length of \(AD\), \(CD\) or arc, or by perimeter |
| Length \(AD = \cos\frac{\pi}{3}r\left[=\frac{1}{2}r\right]\) [so length \(CD = \frac{1}{2}r\)] | M1 | SOI finding length by Pythagoras, or by trigonometry with their angle \(BAC\), provided \(BAC \neq \frac{\pi}{6}\) |
| Length of arc \(BC = r \times \frac{\pi}{3}\) | M1 | SOI using \(r\theta\) with \(\theta\) in radians. Condone \(\theta = \frac{\pi}{6}\) |
| Perimeter of \(BCD = \frac{\sqrt{3}}{2}r + \frac{1}{2}r + \frac{\pi}{3}r\) | A1 | OE e.g. \(r\left(\frac{\sqrt{3}+1}{2} + \frac{\pi}{3}\right)\) with e.g. \(\cos\frac{\pi}{3}\) evaluated. Must be exact, in terms of \(r\). ISW |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sector area $= \frac{1}{2}r^2\left(\frac{\pi}{6}\right)\left[=\frac{\pi}{12}r^2\right]$ | B1 | Using $\frac{1}{2}r^2\theta$ with $\theta$ in radians SOI. B0 if using a value for $r$ |
| $BD = \sin\frac{\pi}{6}r\left[=\frac{1}{2}r\right]$ and $AD = \cos\frac{\pi}{6}r\left[=\frac{\sqrt{3}}{2}r\right]$, so triangle area $= \frac{1}{2}\left(\sin\frac{\pi}{6}r\right)\left(\cos\frac{\pi}{6}r\right)\left[=\frac{1}{2}\times\frac{1}{2}r\times\frac{\sqrt{3}}{2}r\right]$ | B1 | SOI finding triangle area. Decimals B0 unless exact values seen in working |
| Area of $BCD = \frac{1}{12}\pi r^2 - \frac{\sqrt{3}}{8}r^2$ | B1 | OE e.g. $\frac{r^2}{4}\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)$ with $\cos\frac{\pi}{6}$ and $\sin\frac{\pi}{6}$ evaluated. Must be exact, in terms of $r^2$. ISW |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Angle $BAC = \sin^{-1}\left(\frac{\frac{\sqrt{3}}{2}r}{r}\right)\left[=\frac{\pi}{3}\right]$ | B1 | SOI by length of $AD$, $CD$ or arc, or by perimeter |
| Length $AD = \cos\frac{\pi}{3}r\left[=\frac{1}{2}r\right]$ [so length $CD = \frac{1}{2}r$] | M1 | SOI finding length by Pythagoras, or by trigonometry with their angle $BAC$, provided $BAC \neq \frac{\pi}{6}$ |
| Length of arc $BC = r \times \frac{\pi}{3}$ | M1 | SOI using $r\theta$ with $\theta$ in radians. Condone $\theta = \frac{\pi}{6}$ |
| Perimeter of $BCD = \frac{\sqrt{3}}{2}r + \frac{1}{2}r + \frac{\pi}{3}r$ | A1 | OE e.g. $r\left(\frac{\sqrt{3}+1}{2} + \frac{\pi}{3}\right)$ with e.g. $\cos\frac{\pi}{3}$ evaluated. Must be exact, in terms of $r$. ISW |5\\
\includegraphics[max width=\textwidth, alt={}, center]{574b96b2-62f2-41b3-a178-8e68e16429ff-08_509_654_264_751}
The diagram shows a sector $A B C$ of a circle with centre $A$ and radius $r$. The line $B D$ is perpendicular to $A C$. Angle $C A B$ is $\theta$ radians.
\begin{enumerate}[label=(\alph*)]
\item Given that $\theta = \frac { 1 } { 6 } \pi$, find the exact area of $B C D$ in terms of $r$.
\item Given instead that the length of $B D$ is $\frac { \sqrt { 3 } } { 2 } r$, find the exact perimeter of $B C D$ in terms of $r$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q5 [7]}}