| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | November |
| Marks | 7 |
| 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 sector/segment problem requiring basic circle geometry formulas (sector area minus triangle area) and straightforward trigonometry. Part (i) is a guided 'show that' requiring sector area - triangle area calculation with simple algebraic manipulation. Part (ii) involves substituting values and computing arc length + two straight sides using basic trig. Slightly easier than average due to the structured guidance and routine application of standard formulas. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) area of sector \(= \frac{1}{4}r^2\theta\) used. \(AX = r\sin\theta\), \(OX = r\cos\theta\); Area of \(\triangle\frac{1}{2}bh\) used → \(A = \frac{r^2}{2}(\theta - \sin\theta\cos\theta)\) | M1, M1, A1 [3] | Used correctly for the sector. Realises the need to use trig in \(\triangle OAX\). (beware \(AX, OX\) wrong way round) ag- be careful of above error that scores 2/3 |
| (ii) \(AX = 12\sin\frac{1}{6}\pi = 6\); \(OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}\); \(BX = 12 - 6\sqrt{3}\); Arc \(AB = 12x\frac{1}{6}\pi = 2\pi\) → \(P = 18 - 6\sqrt{3} + 2\pi\) (17.0) | B1, B1, M1, A1 [4] | co – anywhere even if in an area. co – anywhere (for \(6\sqrt{3}\) or \(\frac{1}{2}×12\sqrt{3}\)) Use of \(s=r\theta\) co. allow \(6 + 12\) instead of 18. |
**(i)** area of sector $= \frac{1}{4}r^2\theta$ used. $AX = r\sin\theta$, $OX = r\cos\theta$; Area of $\triangle\frac{1}{2}bh$ used → $A = \frac{r^2}{2}(\theta - \sin\theta\cos\theta)$ | M1, M1, A1 [3] | Used correctly for the sector. Realises the need to use trig in $\triangle OAX$. (beware $AX, OX$ wrong way round) ag- be careful of above error that scores 2/3
**(ii)** $AX = 12\sin\frac{1}{6}\pi = 6$; $OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}$; $BX = 12 - 6\sqrt{3}$; Arc $AB = 12x\frac{1}{6}\pi = 2\pi$ → $P = 18 - 6\sqrt{3} + 2\pi$ (17.0) | B1, B1, M1, A1 [4] | co – anywhere even if in an area. co – anywhere (for $6\sqrt{3}$ or $\frac{1}{2}×12\sqrt{3}$) Use of $s=r\theta$ co. allow $6 + 12$ instead of 18.7\\
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In the diagram, $A B$ is an arc of a circle, centre $O$ and radius $r \mathrm {~cm}$, and angle $A O B = \theta$ radians. The point $X$ lies on $O B$ and $A X$ is perpendicular to $O B$.\\
(i) Show that the area, $A \mathrm {~cm} ^ { 2 }$, of the shaded region $A X B$ is given by
$$A = \frac { 1 } { 2 } r ^ { 2 } ( \theta - \sin \theta \cos \theta ) .$$
(ii) In the case where $r = 12$ and $\theta = \frac { 1 } { 6 } \pi$, find the perimeter of the shaded region $A X B$, leaving your answer in terms of $\sqrt { } 3$ and $\pi$.
\hfill \mbox{\textit{CAIE P1 2007 Q7 [7]}}