| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Standard +0.3 This is a straightforward compound area problem requiring basic trigonometry (arctan to find angles), sector area formula, and simple subtraction. All steps are routine applications of standard formulas with no novel insight needed. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin^{-1}\left(\frac{3}{5}\right) = 0.6435\) AG | M1 | OR \((PBC =)\cos^{-1}\left(\frac{3}{5}\right) = 0.9273 \Rightarrow (ABP =)\frac{\pi}{2} - 0.9273 = 0.6435\). Or other valid method. Check working and diagram for evidence of incorrect method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use (once) of sector area \(= \frac{1}{2}r^2\theta\) | M1 | |
| Area sector \(BAP = \frac{1}{2} \times 5^2 \times 0.6435 = 8.04\) | A1 | |
| Area sector \(DAQ = \frac{1}{2} \times \frac{1}{2}\pi \times 3^2 = 7.07\), Allow \(\frac{9\pi}{4}\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: Region \(=\) sect \(+\) sect \(-\) (rect \(- \Delta\)) or sect \(-\) [rect \(-\) (sect \(+ \Delta\))] | (M1) | Use of correct strategy |
| (Area \(\triangle BPC =) \frac{1}{2} \times 3 \times 4 = 6\) Seen | A1 | |
| \(8.04 + 7.07 - (15 - 6) = 6.11\) | A1) | |
| OR1: Region \(=\) sector \(ADQ -\) (trap \(ABPD -\) sector \(ABP\)) | (M1) | Use of correct strategy |
| (Area trap \(ABPD =) \frac{1}{2}(5+1) \times 3 = 9\) Seen | A1 | |
| \(7.07 - (9 - 8.04) = 7.07 - 0.96 = 6.11\) | A1) | |
| OR2: Area segment \(AP = 2.5686\), Area segment \(AQ = 0.5438\), Region \(=\) segment \(AP +\) segment \(AQ + \triangle APQ\) | (M1) | Use of correct strategy |
| (Area \(\triangle APQ =) \frac{1}{2} \times 2 \times 3 = 3\) Seen | A1 | |
| \(2.57 + 0.54 + 3 = 6.11\) | A1) | |
| Total | 3 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin^{-1}\left(\frac{3}{5}\right) = 0.6435$ AG | M1 | OR $(PBC =)\cos^{-1}\left(\frac{3}{5}\right) = 0.9273 \Rightarrow (ABP =)\frac{\pi}{2} - 0.9273 = 0.6435$. Or other valid method. Check working and diagram for evidence of incorrect method |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use (once) of sector area $= \frac{1}{2}r^2\theta$ | M1 | |
| Area sector $BAP = \frac{1}{2} \times 5^2 \times 0.6435 = 8.04$ | A1 | |
| Area sector $DAQ = \frac{1}{2} \times \frac{1}{2}\pi \times 3^2 = 7.07$, Allow $\frac{9\pi}{4}$ | A1 | |
| **Total** | **3** | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** Region $=$ sect $+$ sect $-$ (rect $- \Delta$) or sect $-$ [rect $-$ (sect $+ \Delta$)] | (M1) | Use of correct strategy |
| (Area $\triangle BPC =) \frac{1}{2} \times 3 \times 4 = 6$ Seen | A1 | |
| $8.04 + 7.07 - (15 - 6) = 6.11$ | A1) | |
| **OR1:** Region $=$ sector $ADQ -$ (trap $ABPD -$ sector $ABP$) | (M1) | Use of correct strategy |
| (Area trap $ABPD =) \frac{1}{2}(5+1) \times 3 = 9$ Seen | A1 | |
| $7.07 - (9 - 8.04) = 7.07 - 0.96 = 6.11$ | A1) | |
| **OR2:** Area segment $AP = 2.5686$, Area segment $AQ = 0.5438$, Region $=$ segment $AP +$ segment $AQ + \triangle APQ$ | (M1) | Use of correct strategy |
| (Area $\triangle APQ =) \frac{1}{2} \times 2 \times 3 = 3$ Seen | A1 | |
| $2.57 + 0.54 + 3 = 6.11$ | A1) | |
| **Total** | **3** | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{17ca6dd2-271b-4b06-8433-354493feaf06-10_401_561_260_790}
The diagram shows a rectangle $A B C D$ in which $A B = 5$ units and $B C = 3$ units. Point $P$ lies on $D C$ and $A P$ is an arc of a circle with centre $B$. Point $Q$ lies on $D C$ and $A Q$ is an arc of a circle with centre $D$.\\
(i) Show that angle $A B P = 0.6435$ radians, correct to 4 decimal places.\\
(ii) Calculate the areas of the sectors $B A P$ and $D A Q$.\\
(iii) Calculate the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2017 Q7 [7]}}