| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - single tangent line |
| Difficulty | Standard +0.3 This is a straightforward application of standard circle geometry formulas. Part (i) requires identifying that the perimeter consists of an arc (rα) and two straight lines (tangent PT and radius QT), using the tangent-radius relationship to find PT = r tan α. Part (ii) involves substituting values and calculating area as (sector - triangle), both standard formulas. While it requires multiple steps and careful geometry, the techniques are routine for A-level students who have covered radians and sectors. |
| 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 |
| \(PT = r\tan\alpha\) | B1 | |
| \(QT = OT - OQ = \frac{r}{\cos\alpha} - r\) or \(\sqrt{r^2 + r^2\tan^2\alpha} - r\) | B1 | |
| Perimeter = sum of the 3 parts including \(r\alpha\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of triangle \(= \frac{1}{2} \times 10 \times 10\tan\frac{\pi}{3}\) | M1 | Correct formula used, \(50\sqrt{3}\), 86.6 |
| Area of sector \(= \frac{1}{2} \times 10^2 \times \frac{1}{3}\pi\) | M1 | Correct formula used, \(\frac{50\pi}{3}\), 52.36 |
| Shaded region has area 34 (2sf) | A1 |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $PT = r\tan\alpha$ | B1 | |
| $QT = OT - OQ = \frac{r}{\cos\alpha} - r$ or $\sqrt{r^2 + r^2\tan^2\alpha} - r$ | B1 | |
| Perimeter = sum of the 3 parts including $r\alpha$ | B1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of triangle $= \frac{1}{2} \times 10 \times 10\tan\frac{\pi}{3}$ | M1 | Correct formula used, $50\sqrt{3}$, 86.6 |
| Area of sector $= \frac{1}{2} \times 10^2 \times \frac{1}{3}\pi$ | M1 | Correct formula used, $\frac{50\pi}{3}$, 52.36 |
| Shaded region has area 34 (2sf) | A1 | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{616a6177-0d5c-49f7-b0c1-9138a13c1963-3_552_734_255_703}
The diagram shows a circle with radius $r \mathrm {~cm}$ and centre $O$. The line $P T$ is the tangent to the circle at $P$ and angle $P O T = \alpha$ radians. The line $O T$ meets the circle at $Q$.\\
(i) Express the perimeter of the shaded region $P Q T$ in terms of $r$ and $\alpha$.\\
(ii) In the case where $\alpha = \frac { 1 } { 3 } \pi$ and $r = 10$, find the area of the shaded region correct to 2 significant figures.
\hfill \mbox{\textit{CAIE P1 2016 Q6 [6]}}