| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Challenging +1.2 This is a multi-part geometry problem requiring circle properties, sector formulas, and tangent relationships. Part (i) involves setting up equations using tangent properties (moderate setup), parts (ii-iii) require standard sector area and arc length formulas with careful subtraction. More challenging than routine integration but less demanding than proof-based questions requiring novel insight. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OQ = x + OC = 20\) | B1 | Used somewhere – needs "20" |
| \(\sin 0.6 = \frac{x}{OC} \rightarrow OC = \frac{x}{\sin 0.6}\) | M1 | Use of trig in 90° triangle |
| \(x + \frac{x}{\sin 0.6} = 20 \rightarrow x = 7.218\) | M1 A1 [4] | Soln of linear equation. (answer given, ensure there is a correct method) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(= \frac{1}{2} \cdot 20^2 \times 1.2 - \pi \times 7.218^2 = 76.3\) | M1 A1 [2] | Use of \(\frac{1}{2}r^2\theta\) - needs \(r=20\) and \(\theta = 1.2\); co |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Angle \(PCR = \pi - 1.2\) | B1 | Use of \(s = r\theta\) with \(r = 7.218\) - any \(\theta\) - even \(\frac{2\pi}{3}\) |
| Arc \(PR = 7.218 \times (\pi - 1.2) = (14.01)\) | M1 | |
| \(OP = OR = \frac{x}{\tan 0.6}\) | M1 A1 | Correct use of trig or Pythagoras; co |
| \(\rightarrow\) Perimeter of \(35.1\) cm | [4] |
# Question 11:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OQ = x + OC = 20$ | B1 | Used somewhere – needs "20" |
| $\sin 0.6 = \frac{x}{OC} \rightarrow OC = \frac{x}{\sin 0.6}$ | M1 | Use of trig in 90° triangle |
| $x + \frac{x}{\sin 0.6} = 20 \rightarrow x = 7.218$ | M1 A1 [4] | Soln of linear equation. (answer given, ensure there is a correct method) |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= \frac{1}{2} \cdot 20^2 \times 1.2 - \pi \times 7.218^2 = 76.3$ | M1 A1 [2] | Use of $\frac{1}{2}r^2\theta$ - needs $r=20$ and $\theta = 1.2$; co |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Angle $PCR = \pi - 1.2$ | B1 | Use of $s = r\theta$ with $r = 7.218$ - any $\theta$ - even $\frac{2\pi}{3}$ |
| Arc $PR = 7.218 \times (\pi - 1.2) = (14.01)$ | M1 | |
| $OP = OR = \frac{x}{\tan 0.6}$ | M1 A1 | Correct use of trig or Pythagoras; co |
| $\rightarrow$ Perimeter of $35.1$ cm | [4] | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{11bfe5bd-604c-43e5-81e7-4c1f5676bcbb-4_611_668_1699_737}
The diagram shows a sector of a circle with centre $O$ and radius 20 cm . A circle with centre $C$ and radius $x \mathrm {~cm}$ lies within the sector and touches it at $P , Q$ and $R$. Angle $P O R = 1.2$ radians.\\
(i) Show that $x = 7.218$, correct to 3 decimal places.\\
(ii) Find the total area of the three parts of the sector lying outside the circle with centre $C$.\\
(iii) Find the perimeter of the region $O P S R$ bounded by the $\operatorname { arc } P S R$ and the lines $O P$ and $O R$.
\hfill \mbox{\textit{CAIE P1 2012 Q11 [10]}}