| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Standard +0.3 This is a straightforward sector problem requiring standard formulas (perimeter = 2r + rθ, area = ½r²θ) to set up simultaneous equations, then basic triangle area calculation. While it involves multiple steps and solving simultaneous equations, the techniques are routine for A-level and require no novel insight—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 | Marks | Guidance |
| State \(2r + r\theta = 65\) and \(\frac{1}{2}r^2\theta = 225\) | B1 | |
| Form a 3-term quadratic or cubic in \(r\) or \(\theta\) or \(r\theta\) from correct arc and sector formula | \*M1 | Condone sign errors. |
| Solve their 3 term quadratic or cubic to obtain values of \(r\) or \(\theta\) | DM1 | Expect \(2r^2 - 65r + 450 = (2r-45)(r-10)\) or \(18\theta^2 - 97\theta + 72 = (9\theta - 8)(2\theta - 9)\) |
| \(r = 10\) and \(\theta = 4.5\); ignore \(r = 22.5\) and \(\theta = \frac{8}{9}\); do not ignore \(r = 0\) | A1 | B1 SC if no quadratic or cubic solution. If \(r = 0\) included A0 or B0 SC. |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct formula for area of triangle with clear use of angle being \(2\pi - \) their \(\theta\) | M1 | Expect 1.783 or 102.2°, their \(\theta\) must be reflex. |
| 48.9 | A1 | AWRT, WWW or a second answer. Or greater accuracy; condone absence of units. |
| 2 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $2r + r\theta = 65$ and $\frac{1}{2}r^2\theta = 225$ | B1 | |
| Form a 3-term quadratic or cubic in $r$ or $\theta$ or $r\theta$ from correct arc and sector formula | \*M1 | Condone sign errors. |
| Solve their 3 term quadratic or cubic to obtain values of $r$ or $\theta$ | DM1 | Expect $2r^2 - 65r + 450 = (2r-45)(r-10)$ or $18\theta^2 - 97\theta + 72 = (9\theta - 8)(2\theta - 9)$ |
| $r = 10$ and $\theta = 4.5$; ignore $r = 22.5$ and $\theta = \frac{8}{9}$; do not ignore $r = 0$ | A1 | **B1 SC** if no quadratic or cubic solution. If $r = 0$ included A0 or B0 SC. |
| | **4** | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct formula for area of triangle with clear use of angle being $2\pi - $ their $\theta$ | M1 | Expect 1.783 or 102.2°, their $\theta$ must be reflex. |
| 48.9 | A1 | AWRT, WWW or a second answer. Or greater accuracy; condone absence of units. |
| | **2** | |3\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-05_483_561_287_753}
The diagram shows a sector of a circle with centre $C$. The radii $C A$ and $C B$ each have length $r \mathrm {~cm}$ and the size of the reflex angle $A C B$ is $\theta$ radians. The sector, shaded in the diagram, has a perimeter of 65 cm and an area of $225 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $r$ and $\theta$.
\item Find the area of triangle $A C B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q3 [6]}}