| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - algebraic/general expressions |
| Difficulty | Standard +0.3 This is a straightforward application of standard sector and triangle area formulas. Students must recognize the shaded region as triangle ABC minus sector DAE, then apply basic formulas (area = ½r²θ for sector, ½bh for triangle) and perimeter calculation. The algebra is simple and the setup is clear, making this slightly easier than average but requiring correct formula application. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) area \(\triangle = \frac{1}{2} \times 4 \times 4\tan\alpha\) oe soi | B1 | \(4\tan\alpha = \sqrt{16/\cos^2\alpha - 16}\) (Can also score in answer) Accept \(\theta\) throughout |
| Area sector \(= \frac{1}{2} \times 2^2\alpha\) oe soi | B1 | |
| Shaded area \(= 8\tan\alpha - 2\alpha\) cao | B1 [3] | Little/no working – accept terms in answer |
| (ii) \(DC = \frac{4}{\cos\alpha} - 2\) oe soi | B1 | |
| Arc \(DE = 2\alpha\) soi anywhere provided clear | B1 | \(\frac{4}{\cos\alpha} = \sqrt{16 + 16\tan^2\alpha}\) Can score in answer |
| Perimeter \(= \frac{4}{\cos\alpha} + 4\tan\alpha + 2\alpha\) cao | B1 [3] | Little/no working – accept terms in answer |
(i) area $\triangle = \frac{1}{2} \times 4 \times 4\tan\alpha$ oe soi | B1 | $4\tan\alpha = \sqrt{16/\cos^2\alpha - 16}$ (Can also score in answer) Accept $\theta$ throughout
Area sector $= \frac{1}{2} \times 2^2\alpha$ oe soi | B1 |
Shaded area $= 8\tan\alpha - 2\alpha$ cao | B1 [3] | Little/no working – accept terms in answer
(ii) $DC = \frac{4}{\cos\alpha} - 2$ oe soi | B1 |
Arc $DE = 2\alpha$ soi anywhere provided clear | B1 | $\frac{4}{\cos\alpha} = \sqrt{16 + 16\tan^2\alpha}$ Can score in answer
Perimeter $= \frac{4}{\cos\alpha} + 4\tan\alpha + 2\alpha$ cao | B1 [3] | Little/no working – accept terms in answer6\\
\includegraphics[max width=\textwidth, alt={}, center]{62f7f1e2-a8e7-4574-a432-8e9b20b54d7a-3_625_897_260_623}
The diagram shows triangle $A B C$ in which $A B$ is perpendicular to $B C$. The length of $A B$ is 4 cm and angle $C A B$ is $\alpha$ radians. The arc $D E$ with centre $A$ and radius 2 cm meets $A C$ at $D$ and $A B$ at $E$. Find, in terms of $\alpha$,\\
(i) the area of the shaded region,\\
(ii) the perimeter of the shaded region.
\hfill \mbox{\textit{CAIE P1 2014 Q6 [6]}}