| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Standard +0.3 This is a standard sector/segment problem requiring identification of geometric regions, application of sector area formula, and use of trigonometry to find lengths. While it involves multiple steps and careful geometric reasoning, the techniques are routine for A-level: sector area (½r²θ), triangle area (½r²sin2θ), and basic trigonometry. The perimeter calculation in part (ii) is straightforward once the geometry is understood. This is slightly easier than average as it follows a well-practiced template for this topic. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(AS = r \tan \theta\) | M1 | Or (AB) = 2r tan θ or (AO) = \(\frac{r}{\cos \theta}\) |
| Area \(OAB = r^2 \tan \theta\) or (OAS) = ½r² tan θ | A1 | Or \(OAB = \frac{1}{2}\cos 2\theta \sin 2\theta\) |
| Area of sector = ½ r² × 2θ (= r²θ) | B1 | |
| Shaded area = r² (tan θ − θ) OE | A1 [4] | Or area sector (OPS) = ½r²θ Allow e.g. r² tan θ − ½r² 2θ |
| (ii) \(\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12\) | M1 | |
| \(AP = 6\) | A1 | |
| \(AS = 6 \tan \frac{\pi}{3} (\Rightarrow AB = 12\sqrt{3})\) | B1 | |
| Arc (PST) = \(12 \frac{\pi}{3}\) | B1 | Or arc (PS) = \(6 \frac{\pi}{3}\) or arc (ST) = \(6 \frac{\pi}{3}\) |
| Perimeter = \(12 + 12\sqrt{3} + 4\pi\) | A1 [5] | Allow unsimplified 4π |
**(i)** $AS = r \tan \theta$ | M1 | Or (AB) = 2r tan θ or (AO) = $\frac{r}{\cos \theta}$
Area $OAB = r^2 \tan \theta$ or (OAS) = ½r² tan θ | A1 | Or $OAB = \frac{1}{2}\cos 2\theta \sin 2\theta$
Area of sector = ½ r² × 2θ (= r²θ) | B1 |
Shaded area = r² (tan θ − θ) OE | A1 [4] | Or area sector (OPS) = ½r²θ Allow e.g. r² tan θ − ½r² 2θ
**(ii)** $\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12$ | M1 |
$AP = 6$ | A1 |
$AS = 6 \tan \frac{\pi}{3} (\Rightarrow AB = 12\sqrt{3})$ | B1 |
Arc (PST) = $12 \frac{\pi}{3}$ | B1 | Or arc (PS) = $6 \frac{\pi}{3}$ or arc (ST) = $6 \frac{\pi}{3}$
Perimeter = $12 + 12\sqrt{3} + 4\pi$ | A1 [5] | Allow unsimplified 4π
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\includegraphics[max width=\textwidth, alt={}, center]{53839c8c-07ea-4545-9c00-a6884aa2afc3-3_387_1175_1781_486}
In the diagram, $O A B$ is an isosceles triangle with $O A = O B$ and angle $A O B = 2 \theta$ radians. Arc $P S T$ has centre $O$ and radius $r$, and the line $A S B$ is a tangent to the $\operatorname { arc } P S T$ at $S$.\\
(i) Find the total area of the shaded regions in terms of $r$ and $\theta$.\\
(ii) In the case where $\theta = \frac { 1 } { 3 } \pi$ and $r = 6$, find the total perimeter of the shaded regions, leaving your answer in terms of $\sqrt { } 3$ and $\pi$.
\hfill \mbox{\textit{CAIE P1 2011 Q9 [9]}}