| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector area calculation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on sector areas and arc lengths using standard formulas. Part (i) is direct substitution, part (ii) requires setting up and solving a simple equation involving perimeters, and part (iii) involves basic trigonometry to find chord length AB. All parts use routine techniques with no novel problem-solving 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 |
|---|---|---|
| (i) \(\theta = 1\) angle \(BOC = \pi - \theta\); Area = \(1/2r^2\theta = 68.5\) or \(32(\pi-1)\) (or 1/2 circle-sector) | B1, M1, A1 [3] | For \(\pi - \theta\) or for \(1/2mr^2\) – sector. Use of \(1/2r^2\theta\). Co. NB. 32 gets M1 only |
| (ii) \(8 + 8 + 8\theta = 1/2(8 + 8 + 8(\pi-\theta))\); Solution of this eqn | M1, M1 [2] | Relevant use of \(s = r\theta\) twice. Needs \(\theta\) – collected – needs perimeters. Co. |
| \(\rightarrow 0.381\) or \(1/3(\pi-2)\) | A1 [3] | |
| (iii) \(\theta = \pi/3\); \(AB = 8\) cm; \(BC = 2 \times 8\sin\pi/3 = 8\sqrt{3}\) | B1, M1, A1 [3] | Co. Valid method for BC – cos rule, Pyth allow decimals here. Everything OK. Answer given. NB. Decimal check loses this mark |
| Perimeter = \(24 + 8\sqrt{3}\) | [3] |
**(i)** $\theta = 1$ angle $BOC = \pi - \theta$; Area = $1/2r^2\theta = 68.5$ or $32(\pi-1)$ (or 1/2 circle-sector) | B1, M1, A1 [3] | For $\pi - \theta$ or for $1/2mr^2$ – sector. Use of $1/2r^2\theta$. Co. NB. 32 gets M1 only
**(ii)** $8 + 8 + 8\theta = 1/2(8 + 8 + 8(\pi-\theta))$; Solution of this eqn | M1, M1 [2] | Relevant use of $s = r\theta$ twice. Needs $\theta$ – collected – needs perimeters. Co.
$\rightarrow 0.381$ or $1/3(\pi-2)$ | A1 [3]
**(iii)** $\theta = \pi/3$; $AB = 8$ cm; $BC = 2 \times 8\sin\pi/3 = 8\sqrt{3}$ | B1, M1, A1 [3] | Co. Valid method for BC – cos rule, Pyth allow decimals here. Everything OK. Answer given. NB. Decimal check loses this mark
Perimeter = $24 + 8\sqrt{3}$ | [3]9\\
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The diagram shows a semicircle $A B C$ with centre $O$ and radius 8 cm . Angle $A O B = \theta$ radians.\\
(i) In the case where $\theta = 1$, calculate the area of the sector BOC.\\
(ii) Find the value of $\theta$ for which the perimeter of sector $A O B$ is one half of the perimeter of sector BOC.\\
(iii) In the case where $\theta = \frac { 1 } { 3 } \pi$, show that the exact length of the perimeter of triangle $A B C$ is $( 24 + 8 \sqrt { } 3 ) \mathrm { cm }$.
\hfill \mbox{\textit{CAIE P1 2003 Q9 [9]}}