| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - two tangents from external point |
| Difficulty | Standard +0.3 This is a standard sector-tangent problem requiring routine application of arc length, sector area, and tangent properties. Part (i) involves straightforward geometric reasoning (triangle area minus sector area), and part (ii) is direct substitution into formulas. Slightly above average due to the algebraic manipulation in part (i), but still a textbook exercise with no novel insight required. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(QR = r\tan\theta\). Area shaded = \(\frac{1}{2}r^2\tan\theta - \frac{1}{2}r^2\theta\) | B1, B1 | Correct somewhere – in (ii) ok; All correct – answer given, beware fortuitous |
| (ii) Arc \(PQ = 15 \times 0.8 = 12\) OR \(r = r\cos\theta\) (21.53) | B1, M1 | Anywhere (could be implied); Must be correct with \(r\) and \(\theta\) or Pythagoras |
| Perimeter = \(r\tan\theta +\) arc \(PQ + (r - r\cos\theta)\) = 34.0 (33.9 ok) | M1, A1 | Putting 4 things together – even if algebraic; Correct only |
**(i)** $QR = r\tan\theta$. Area shaded = $\frac{1}{2}r^2\tan\theta - \frac{1}{2}r^2\theta$ | B1, B1 | Correct somewhere – in (ii) ok; All correct – answer given, beware fortuitous
**(ii)** Arc $PQ = 15 \times 0.8 = 12$ OR $r = r\cos\theta$ (21.53) | B1, M1 | Anywhere (could be implied); Must be correct with $r$ and $\theta$ or Pythagoras
Perimeter = $r\tan\theta +$ arc $PQ + (r - r\cos\theta)$ = 34.0 (33.9 ok) | M1, A1 | Putting 4 things together – even if algebraic; Correct only3\\
\includegraphics[max width=\textwidth, alt={}, center]{a10ad459-6f86-4845-ba28-e4e394bf3d1e-2_330_634_753_758}
In the diagram, $O P Q$ is a sector of a circle, centre $O$ and radius $r \mathrm {~cm}$. Angle $Q O P = \theta$ radians. The tangent to the circle at $Q$ meets $O P$ extended at $R$.\\
(i) Show that the area, $A \mathrm {~cm} ^ { 2 }$, of the shaded region is given by $A = \frac { 1 } { 2 } r ^ { 2 } ( \tan \theta - \theta )$.\\
(ii) In the case where $\theta = 0.8$ and $r = 15$, evaluate the length of the perimeter of the shaded region.
\hfill \mbox{\textit{CAIE P1 2002 Q3 [6]}}