| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Moderate -0.3 This is a straightforward composite shape problem requiring basic trigonometry (finding OB using tan), perimeter calculation (sum of sides plus arc length), and area calculation (triangle plus sector). All techniques are standard P1/mechanics content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Length of \(OB = \frac{6}{\cos 0.6} = 7.270\) | M1 [1] | ag Any valid method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AB = 6\tan 0.6\) or \(4.1\) | B1 | Sight of in (ii) |
| Arc length \(= 7.27 \times (\frac{1}{2}\pi - 0.6) = (7.06)\) | M1 | Use of \(s = r\theta\) with sector angle |
| Perimeter \(= 6 + 7.27 + 7.06 + 6\tan 0.6 = 24.4\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of \(AOB = \frac{1}{2} \times 6 \times 7.27 \times \sin 0.6\) | M1 | Use of any correct area method |
| Area of \(OBC = \frac{1}{2} \times 7.27^2 \times (\frac{1}{2}\pi - 0.6)\) | M1 | Use of \(\frac{1}{2}r^2\theta\) |
| \(\rightarrow\) area \(= 12.31 + 25.65 = 38.0\) | A1 [3] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Length of $OB = \frac{6}{\cos 0.6} = 7.270$ | **M1** [1] | ag Any valid method |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB = 6\tan 0.6$ or $4.1$ | **B1** | Sight of in (ii) |
| Arc length $= 7.27 \times (\frac{1}{2}\pi - 0.6) = (7.06)$ | **M1** | Use of $s = r\theta$ with sector angle |
| Perimeter $= 6 + 7.27 + 7.06 + 6\tan 0.6 = 24.4$ | **A1** [3] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of $AOB = \frac{1}{2} \times 6 \times 7.27 \times \sin 0.6$ | **M1** | Use of any correct area method |
| Area of $OBC = \frac{1}{2} \times 7.27^2 \times (\frac{1}{2}\pi - 0.6)$ | **M1** | Use of $\frac{1}{2}r^2\theta$ |
| $\rightarrow$ area $= 12.31 + 25.65 = 38.0$ | **A1** [3] | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{9cdb00a6-1e86-4185-bb73-ed3ecab981ba-3_560_506_258_822}
The diagram shows a metal plate $O A B C$, consisting of a right-angled triangle $O A B$ and a sector $O B C$ of a circle with centre $O$. Angle $A O B = 0.6$ radians, $O A = 6 \mathrm {~cm}$ and $O A$ is perpendicular to $O C$.\\
(i) Show that the length of $O B$ is 7.270 cm , correct to 3 decimal places.\\
(ii) Find the perimeter of the metal plate.\\
(iii) Find the area of the metal plate.
\hfill \mbox{\textit{CAIE P1 2015 Q5 [7]}}