| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Standard +0.3 This is a standard circle geometry problem requiring arc length formula, area of sector, and area of segment calculations. While it involves multiple steps and careful geometric reasoning about the regions, the techniques are all routine P1 content with no novel insights required. Slightly above average due to the multi-part nature and need to identify the correct regions, but well within typical A-level expectations. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| Pythagoras \(\rightarrow r = \sqrt{72}\) OE, or \(\cos 45 = \frac{6}{r} \rightarrow r = \frac{6}{\cos 45} = 6\sqrt{2}\) | M1 | Correct method leading to \(r =\) |
| Arc \(DC = \sqrt{72} \times \frac{1}{4}\pi = \frac{3\sqrt{2}}{2}\pi\), 2.12\(\pi\), 6.66 | M1 A1 | Use of \(s=r\theta\) with their \(r\) (NOT 6) and \(\frac{1}{4}\pi\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area of sector-\(BDC\) is \(\frac{1}{2} \times 72 \times \frac{1}{4}\pi\) (\(= 9\pi\) or 28.274…) | *M1 | Use of \(\frac{1}{2}r^2\theta\) with their \(r\) (NOT 6) and \(\frac{1}{4}\pi\) |
| Area \(Q = 9\pi - 18\) (10.274…) | DM1 | Subtracts their \(\frac{1}{2} \times 6 \times 6\) from their \(\frac{1}{2}r^2\theta\) |
| Area \(P\) is (\(\frac{1}{4}\pi 6^2 -\) area \(Q\)) \(= 18\) | M1 | Uses \(\{\frac{1}{4}\pi 6^2 -\) (their area \(Q\) using \(\sqrt{72}\))\(\}\) |
| Ratio is \(\frac{18}{9\pi - 18}\) \(\left(\frac{18}{10.274}\right) \rightarrow 1.75\) | A1 | |
| Total: 4 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Pythagoras $\rightarrow r = \sqrt{72}$ OE, or $\cos 45 = \frac{6}{r} \rightarrow r = \frac{6}{\cos 45} = 6\sqrt{2}$ | M1 | Correct method leading to $r =$ |
| Arc $DC = \sqrt{72} \times \frac{1}{4}\pi = \frac{3\sqrt{2}}{2}\pi$, 2.12$\pi$, 6.66 | M1 A1 | Use of $s=r\theta$ with their $r$ (NOT 6) and $\frac{1}{4}\pi$ |
| **Total: 3** | | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of sector-$BDC$ is $\frac{1}{2} \times 72 \times \frac{1}{4}\pi$ ($= 9\pi$ or 28.274…) | *M1 | Use of $\frac{1}{2}r^2\theta$ with their $r$ (NOT 6) and $\frac{1}{4}\pi$ |
| Area $Q = 9\pi - 18$ (10.274…) | DM1 | Subtracts their $\frac{1}{2} \times 6 \times 6$ from their $\frac{1}{2}r^2\theta$ |
| Area $P$ is ($\frac{1}{4}\pi 6^2 -$ area $Q$) $= 18$ | M1 | Uses $\{\frac{1}{4}\pi 6^2 -$ (their area $Q$ using $\sqrt{72}$)$\}$ |
| Ratio is $\frac{18}{9\pi - 18}$ $\left(\frac{18}{10.274}\right) \rightarrow 1.75$ | A1 | |
| **Total: 4** | | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{518bb805-5b14-4b41-94fd-38a31a90c218-06_401_698_255_721}
The diagram shows a semicircle with centre $O$ and radius 6 cm . The radius $O C$ is perpendicular to the diameter $A B$. The point $D$ lies on $A B$, and $D C$ is an arc of a circle with centre $B$.\\
(i) Calculate the length of the $\operatorname { arc } D C$.\\
(ii) Find the value of\\
$\frac { \text { area of region } P } { \text { area of region } Q }$,\\
giving your answer correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P1 2017 Q4 [7]}}