| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2005 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Moderate -0.5 This is a straightforward application of standard arc length and sector area formulas with two concentric circles. Part (i) requires direct substitution into the sector area formula (difference of two sectors), while part (ii) involves setting up and solving a simple linear equation for α using arc length formulas. Both parts are routine calculations with no conceptual challenges beyond recalling the basic formulas s=rθ and A=½r²θ. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Area \(= \frac{1}{2}.16^2.0.8 - \frac{1}{2}.10^2.0.8 \rightarrow 62.4 \text{ cm}^2\) | M1 A1 | Use of \(\frac{1}{2}r^2\theta\) once.. Co. |
| Answer | Marks | Guidance |
|---|---|---|
| \(12 + 10a + 16a = 28.9 \rightarrow a = 0.65\) | M1 DM1 A1 | Use of \(s=r\theta\) once. Forming an eqn, including the 6 + 6. co |
$a = 0.8$, radii 10cm and 16 cm
(i) Area $= \frac{1}{2}.16^2.0.8 - \frac{1}{2}.10^2.0.8 \rightarrow 62.4 \text{ cm}^2$ | M1 A1 | Use of $\frac{1}{2}r^2\theta$ once.. Co. | [2]
(ii) Arcs are $10a$ and $16a$
$12 + 10a + 16a = 28.9 \rightarrow a = 0.65$ | M1 DM1 A1 | Use of $s=r\theta$ once. Forming an eqn, including the 6 + 6. co | [3]2\\
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In the diagram, $O A B$ and $O C D$ are radii of a circle, centre $O$ and radius 16 cm . Angle $A O C = \alpha$ radians. $A C$ and $B D$ are arcs of circles, centre $O$ and radii 10 cm and 16 cm respectively.\\
(i) In the case where $\alpha = 0.8$, find the area of the shaded region.\\
(ii) Find the value of $\alpha$ for which the perimeter of the shaded region is 28.9 cm .
\hfill \mbox{\textit{CAIE P1 2005 Q2 [5]}}