| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Shaded region between arcs |
| Difficulty | Moderate -0.3 This is a straightforward application of arc length and sector area formulas with some basic geometry. Students need to recognize that BD has radius AB = 6√2 (from right triangle AOB), convert 90° to π/2 radians, then apply standard formulas. The shaded area requires subtracting a sector from a triangle. While it involves multiple steps, each is routine and the geometric setup is clearly given, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
B1: $r = 6$ cm
(i) B1: $AB = \sqrt{6^2 + 6^2} = \sqrt{72}$ — Use of Pythagoras or trig (8.5 ok)
B1: Angle $BAD = \frac{1}{4}\pi$ or $45°$ — In degrees or radians
M1: Arc length = $\sqrt{72} \times \frac{1}{4}\pi = 6.67$ (or better) — Use of $s=r\theta$ with $\theta$ in rads only or correct with degrees. Use of $r = 6$ M0.
A1
[4]
(ii) M1: Sector area = $\frac{1}{2}r^2\theta = \frac{1}{2} \times 72 \times \frac{1}{4}\pi$ — Use of $\frac{1}{2}r^2\theta$ with $\theta$ in rad, and $r \neq 6$
B1: Area of triangle = $\frac{1}{2} \times 6 \times 6$
A1: Shaded area = $10.3$ or $9\pi − 18$
[3]5\\
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The diagram shows a semicircle $A B C$ with centre $O$ and radius 6 cm . The point $B$ is such that angle $B O A$ is $90 ^ { \circ }$ and $B D$ is an arc of a circle with centre $A$. Find\\
(i) the length of the $\operatorname { arc } B D$,\\
(ii) the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2009 Q5 [7]}}