| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| 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 geometry problem requiring basic trigonometry (finding CD and OD using sin/cos), arc length formula, and area calculation by subtraction (sector minus triangle). While it involves multiple steps, all techniques are routine for A-level and the problem structure is familiar from textbook exercises. Slightly easier than average due to straightforward setup and clear diagram. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(CD = r\cos\theta\), \(BD = r - r\sin\theta\) oe | B1 B1 | Allow degrees but not for last B1 |
| Arc \(CB = r\left(\frac{1}{2}\pi - \theta\right)\) oe | B1 | |
| \(\rightarrow P = r\cos\theta + r - r\sin\theta + r\left(\frac{1}{2}\pi - \theta\right)\) oe | B1\(\checkmark\) | \(\checkmark\) sum – assuming trig used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sector \(= \frac{1}{2} \cdot 5^2 \cdot \left(\frac{1}{2}\pi - 0.6\right)\) (12.135) | M1 | Uses \(\frac{1}{2}r^2\theta\) |
| Triangle \(= \frac{1}{2} \cdot 5\cos 0.6 \cdot 5\sin 0.6\) (5.825) | M1 | Uses \(\frac{1}{2}bh\) with some use of trig |
| \(\rightarrow \text{Area} = 6.31\) (or \(\frac{1}{4}\) circle − triangle − sector) | A1 |
## Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $CD = r\cos\theta$, $BD = r - r\sin\theta$ oe | **B1 B1** | Allow degrees but not for last B1 |
| Arc $CB = r\left(\frac{1}{2}\pi - \theta\right)$ oe | **B1** | |
| $\rightarrow P = r\cos\theta + r - r\sin\theta + r\left(\frac{1}{2}\pi - \theta\right)$ oe | **B1$\checkmark$** | $\checkmark$ sum – assuming trig used |
## Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sector $= \frac{1}{2} \cdot 5^2 \cdot \left(\frac{1}{2}\pi - 0.6\right)$ (12.135) | **M1** | Uses $\frac{1}{2}r^2\theta$ |
| Triangle $= \frac{1}{2} \cdot 5\cos 0.6 \cdot 5\sin 0.6$ (5.825) | **M1** | Uses $\frac{1}{2}bh$ with some use of trig |
| $\rightarrow \text{Area} = 6.31$ (or $\frac{1}{4}$ circle − triangle − sector) | **A1** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{b6ae63ce-a8a8-45ef-9c75-2fab30de8ad9-3_408_451_721_845}
In the diagram, $A O B$ is a quarter circle with centre $O$ and radius $r$. The point $C$ lies on the arc $A B$ and the point $D$ lies on $O B$. The line $C D$ is parallel to $A O$ and angle $A O C = \theta$ radians.\\
(i) Express the perimeter of the shaded region in terms of $r , \theta$ and $\pi$.\\
(ii) For the case where $r = 5 \mathrm {~cm}$ and $\theta = 0.6$, find the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2016 Q7 [7]}}