| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Standard +0.3 This is a straightforward sector/circle area problem requiring students to set up an equation (area of large sector minus area of small sector equals area of circle) and solve for the angle, then calculate perimeter using standard arc length formulas. The algebra is simple and all techniques are standard P1 material, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2} \cdot 3^2\pi = \frac{1}{2} \cdot 9^2\theta - \frac{1}{2} \cdot 3^2\theta\) | M1 A1 | M1 needs \(\frac{1}{2}r^2\theta\) once. A1 all correct |
| \(\rightarrow \theta = \frac{1}{4}\pi\) | A1 [3] | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P = 6+6+3\times\frac{1}{4}\pi + 9\times\frac{1}{4}\pi = 21.4\) cm, or \(12 + 3\pi\) | M1 A1 [2] | M1 is for use of \(s = r\theta\) once |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2} \cdot 3^2\pi = \frac{1}{2} \cdot 9^2\theta - \frac{1}{2} \cdot 3^2\theta$ | M1 A1 | M1 needs $\frac{1}{2}r^2\theta$ once. A1 all correct |
| $\rightarrow \theta = \frac{1}{4}\pi$ | A1 [3] | Answer given |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P = 6+6+3\times\frac{1}{4}\pi + 9\times\frac{1}{4}\pi = 21.4$ cm, or $12 + 3\pi$ | M1 A1 [2] | M1 is for use of $s = r\theta$ once |
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\includegraphics[max width=\textwidth, alt={}, center]{13cfb59a-7781-4786-a625-919b01a2a4f0-2_501_641_461_753}
The diagram shows a circle $C$ with centre $O$ and radius 3 cm . The radii $O P$ and $O Q$ are extended to $S$ and $R$ respectively so that $O R S$ is a sector of a circle with centre $O$. Given that $P S = 6 \mathrm {~cm}$ and that the area of the shaded region is equal to the area of circle $C$,\\
(i) show that angle $P O Q = \frac { 1 } { 4 } \pi$ radians,\\
(ii) find the perimeter of the shaded region.
\hfill \mbox{\textit{CAIE P1 2013 Q2 [5]}}