| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Standard +0.3 This is a standard A-level geometry problem combining circle sectors and segments with straightforward application of arc length and area formulas. While it requires multiple steps and careful organization (finding chord length via cosine rule, then using it as diameter for semicircle, then computing perimeter and area), all techniques are routine P1 content with no novel insight required. Slightly easier than average due to the guided structure and standard formulas. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{r}{10} = \sin 0.6\) or \(\frac{r}{10} = \cos 0.97\) or \(BD = \sqrt{200 - 200\cos 1.2} (= 11.3)\) | M1 | Or other valid alternative |
| \(r = 10 \times 0.5646\), \(r = 10 \times \sin 0.6\), \(r = 10 \times \cos 0.971\) or \(r = \frac{1}{2}BD \rightarrow r = 5.646\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Major arc \(= 10(\theta) (= 50.832)\), \(\theta = 2\pi - 1.2 (= 5.083)\) | M1 | \(\theta = 2\pi - 1.2\) or \(\pi - 1.2\), implied by 5.1 |
| or \(C = 2\pi \times 10\), Minor arc \(= 1.2 \times 10\) | B1 | |
| Semicircle \(= 5.646\pi (= 17.737)\) | ||
| Major arc + semicircle \(= 68.6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of major sector \(= \frac{1}{2}10^2(\theta) (= 254.159)\) | M1 | \(\theta = 2\pi - 1.2\) or \(\pi - 1.2\) |
| Area of triangle \(OBD = \frac{1}{2}10^2\sin 1.2 (= 46.602)\) | M1 | Use of \(\frac{1}{2}ab\sin C\) or other complete method |
| Area = semicircle + sector + triangle \((= 50.1 + 254.2 + 46.6) = 351\) | A1 |
## Question 6(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{r}{10} = \sin 0.6$ or $\frac{r}{10} = \cos 0.97$ or $BD = \sqrt{200 - 200\cos 1.2} (= 11.3)$ | M1 | Or other valid alternative |
| $r = 10 \times 0.5646$, $r = 10 \times \sin 0.6$, $r = 10 \times \cos 0.971$ or $r = \frac{1}{2}BD \rightarrow r = 5.646$ | A1 | AG |
## Question 6(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Major arc $= 10(\theta) (= 50.832)$, $\theta = 2\pi - 1.2 (= 5.083)$ | M1 | $\theta = 2\pi - 1.2$ or $\pi - 1.2$, implied by 5.1 |
| or $C = 2\pi \times 10$, Minor arc $= 1.2 \times 10$ | B1 | |
| Semicircle $= 5.646\pi (= 17.737)$ | | |
| Major arc + semicircle $= 68.6$ | A1 | |
## Question 6(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of major sector $= \frac{1}{2}10^2(\theta) (= 254.159)$ | M1 | $\theta = 2\pi - 1.2$ or $\pi - 1.2$ |
| Area of triangle $OBD = \frac{1}{2}10^2\sin 1.2 (= 46.602)$ | M1 | Use of $\frac{1}{2}ab\sin C$ or other complete method |
| Area = semicircle + sector + triangle $(= 50.1 + 254.2 + 46.6) = 351$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{3a631b88-5ba5-49e7-a312-dfd8a6d8a24e-2_615_809_1535_667}
The diagram shows a metal plate $A B C D$ made from two parts. The part $B C D$ is a semicircle. The part $D A B$ is a segment of a circle with centre $O$ and radius 10 cm . Angle $B O D$ is 1.2 radians.\\
(i) Show that the radius of the semicircle is 5.646 cm , correct to 3 decimal places.\\
(ii) Find the perimeter of the metal plate.\\
(iii) Find the area of the metal plate.
\hfill \mbox{\textit{CAIE P1 2016 Q6 [8]}}