| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard arc length and sector area formulas with given values. Part (i) requires writing perimeter = 2r + rθ, and part (ii) involves direct substitution into area = ½r²θ minus triangle area. Both parts are routine calculations with no problem-solving or geometric insight required, making this easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((AB) = 2r\sin\theta\) (or \(r\sqrt{2-2\cos 2\theta}\) or \(\frac{r\sin 2\theta}{\sin\left(\frac{\pi}{2}-\theta\right)}\)) | B1 | Allow unsimplified throughout e.g. \(r + r\), \(\frac{2\theta}{2}\) etc |
| \((\text{Arc } AB) = 2r\theta\) | B1 | |
| \((P =)\ 2r + 2r\theta + 2r\sin\theta\) (or \(r\sqrt{2-2\cos 2\theta}\) or \(\frac{r\sin 2\theta}{\sin\left(\frac{\pi}{2}-\theta\right)}\)) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area sector \(AOB = \left(\frac{1}{2}r^2 \cdot 2\theta\right)\frac{25\pi}{6}\) or \(13.1\) | B1 | Use of segment formula gives 2.26 B1B1 |
| Area triangle \(AOB = \frac{1}{2}\times 2r\sin\theta \times r\cos\theta\) or \(\frac{1}{2}\times r^2\sin 2\theta = \frac{25\sqrt{3}}{4}\) or \(10.8\) | B1 | |
| Area rectangle \(ABCD = (r \times 2r\sin\theta)\ 25\) | B1 | |
| (Area \(=\)) Either \(25 - \left(\frac{25\pi}{6} - \frac{25\sqrt{3}}{4}\right)\) or \(22.7\) | B1 | Correct final answer gets B4. |
## Question 4:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(AB) = 2r\sin\theta$ (or $r\sqrt{2-2\cos 2\theta}$ or $\frac{r\sin 2\theta}{\sin\left(\frac{\pi}{2}-\theta\right)}$) | **B1** | Allow unsimplified throughout e.g. $r + r$, $\frac{2\theta}{2}$ etc |
| $(\text{Arc } AB) = 2r\theta$ | **B1** | |
| $(P =)\ 2r + 2r\theta + 2r\sin\theta$ (or $r\sqrt{2-2\cos 2\theta}$ or $\frac{r\sin 2\theta}{\sin\left(\frac{\pi}{2}-\theta\right)}$) | **B1** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area sector $AOB = \left(\frac{1}{2}r^2 \cdot 2\theta\right)\frac{25\pi}{6}$ or $13.1$ | **B1** | Use of segment formula gives 2.26 **B1B1** |
| Area triangle $AOB = \frac{1}{2}\times 2r\sin\theta \times r\cos\theta$ or $\frac{1}{2}\times r^2\sin 2\theta = \frac{25\sqrt{3}}{4}$ or $10.8$ | **B1** | |
| Area rectangle $ABCD = (r \times 2r\sin\theta)\ 25$ | **B1** | |
| (Area $=$) Either $25 - \left(\frac{25\pi}{6} - \frac{25\sqrt{3}}{4}\right)$ or $22.7$ | **B1** | Correct final answer gets **B4**. |
---(i) Express the perimeter of the shaded region in terms of $r$ and $\theta$.\\
(ii) In the case where $r = 5$ and $\theta = \frac { 1 } { 6 } \pi$, find the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2017 Q4 [7]}}