| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.8 This is a multi-step geometry problem requiring trigonometry to find angles in two different triangles, then applying arc length and sector area formulas. The setup with two different circle centers and finding angle AXB (not the central angle AOB) requires careful geometric reasoning beyond routine sector problems. However, the steps are guided and use standard techniques, placing it moderately above average difficulty. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OM = 8\) (Pythagoras) \(\rightarrow XM = 2\) | B1 | could find \(\sqrt{40}\) and use \(\sin^{-1}\) or \(\cos^{-1}\) |
| \(\tan AXM = \frac{6}{2}\), \(AXB = 2\tan^{-1}3 = 2.498\) | M1 A1 | AG Needs \(\times 2\) and correct trig for M1 |
| Alt 1: \(\sin AOM = \frac{6}{10}\), \(AOM = 0.6435\), \(AXB = \pi - 0.6435\) | Alt 1: Use of isosceles triangles, B1 for AOM, M1,A1 for completion. Alt 2: Use of circle theorem, B1 for AOB, M1,A1 for completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AX = \sqrt{6^2 + 2^2} = \sqrt{40}\) | B1 | CAO, could be gained in part (i) or part (iii) |
| Arc \(AYB = r\theta = \sqrt{40} \times 2.498\) | M1 | Allow for incorrect \(\sqrt{40}\) (not \(r = 6\) or \(12\) or \(10\)) |
| Perimeter \(= 12 + \text{arc} = 27.8\) cm | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area of sector \(AXBY = \frac{1}{2} \times (\sqrt{40})^2 \times 2.498\) | M1 | Use of \(\frac{1}{2}r^2\theta\) with their \(r\), (not \(r = 6\) or \(r = 10\)) |
| Area of triangle \(AXB = \frac{1}{2} \times 12 \times 2\), subtract these \(\rightarrow 38.0\) cm² | M1 A1 | Use of \(\frac{1}{2}bh\) and subtraction. Could gain M1 with \(r = 10\) |
## Question 8:
**Part 8(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OM = 8$ (Pythagoras) $\rightarrow XM = 2$ | B1 | could find $\sqrt{40}$ and use $\sin^{-1}$ or $\cos^{-1}$ |
| $\tan AXM = \frac{6}{2}$, $AXB = 2\tan^{-1}3 = 2.498$ | M1 A1 | AG Needs $\times 2$ and correct trig for M1 |
| Alt 1: $\sin AOM = \frac{6}{10}$, $AOM = 0.6435$, $AXB = \pi - 0.6435$ | | Alt 1: Use of isosceles triangles, B1 for AOM, M1,A1 for completion. Alt 2: Use of circle theorem, B1 for AOB, M1,A1 for completion |
**Part 8(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AX = \sqrt{6^2 + 2^2} = \sqrt{40}$ | B1 | CAO, could be gained in part (i) or part (iii) |
| Arc $AYB = r\theta = \sqrt{40} \times 2.498$ | M1 | Allow for incorrect $\sqrt{40}$ (not $r = 6$ or $12$ or $10$) |
| Perimeter $= 12 + \text{arc} = 27.8$ cm | A1 | |
**Part 8(iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of sector $AXBY = \frac{1}{2} \times (\sqrt{40})^2 \times 2.498$ | M1 | Use of $\frac{1}{2}r^2\theta$ with their $r$, (not $r = 6$ or $r = 10$) |
| Area of triangle $AXB = \frac{1}{2} \times 12 \times 2$, subtract these $\rightarrow 38.0$ cm² | M1 A1 | Use of $\frac{1}{2}bh$ and subtraction. Could gain M1 with $r = 10$ |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{028c7979-6b24-42d0-9857-c616a169b2b2-14_590_691_260_726}
In the diagram, $O A X B$ is a sector of a circle with centre $O$ and radius 10 cm . The length of the chord $A B$ is 12 cm . The line $O X$ passes through $M$, the mid-point of $A B$, and $O X$ is perpendicular to $A B$. The shaded region is bounded by the chord $A B$ and by the arc of a circle with centre $X$ and radius $X A$.\\
(i) Show that angle $A X B$ is 2.498 radians, correct to 3 decimal places.\\
(ii) Find the perimeter of the shaded region.\\
(iii) Find the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2017 Q8 [9]}}