| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Standard +0.3 This is a straightforward geometry problem requiring Pythagoras' theorem to find the larger radius (which is essentially given away in part (i)), then calculating areas using standard circle formulas and basic subtraction. The perpendicular diameters setup is clear, and the problem requires only routine application of well-practiced techniques with no novel insight needed. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BC^2 = r^2 + r^2 = 2r^2 \rightarrow BC = r\sqrt{2}\) | B1 | AG |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area sector \(BCFD = \frac{1}{4}\pi(r\sqrt{2})^2\) soi | M1 | Expect \(\frac{1}{2}\pi r^2\) |
| Area \(\triangle BCAD = \frac{1}{2}(2r)r\) | M1* | Expect \(r^2\) (could be embedded) |
| Area segment \(CFDA = \frac{1}{2}\pi r^2 - r^2\) oe | A1 | |
| Area semi-circle \(CADE = \frac{1}{2}\pi r^2\) | ||
| Shaded area \(\frac{1}{2}\pi r^2 - \left(\frac{1}{2}\pi r^2 - r^2\right)\) | B1 | |
| or \(\pi r^2 - \left(\frac{1}{2}\pi r^2 + \left(\frac{1}{2}\pi r^2 - r^2\right)\right)\) | DM1 | Depends on the area \(\triangle BCD\) |
| \(= r^2\) | A1 | |
| Total: 6 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BC^2 = r^2 + r^2 = 2r^2 \rightarrow BC = r\sqrt{2}$ | B1 | AG |
| **Total: 1** | | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area sector $BCFD = \frac{1}{4}\pi(r\sqrt{2})^2$ soi | M1 | Expect $\frac{1}{2}\pi r^2$ |
| Area $\triangle BCAD = \frac{1}{2}(2r)r$ | M1* | Expect $r^2$ (could be embedded) |
| Area segment $CFDA = \frac{1}{2}\pi r^2 - r^2$ oe | A1 | |
| Area semi-circle $CADE = \frac{1}{2}\pi r^2$ | | |
| Shaded area $\frac{1}{2}\pi r^2 - \left(\frac{1}{2}\pi r^2 - r^2\right)$ | B1 | |
| or $\pi r^2 - \left(\frac{1}{2}\pi r^2 + \left(\frac{1}{2}\pi r^2 - r^2\right)\right)$ | DM1 | Depends on the area $\triangle BCD$ |
| $= r^2$ | A1 | |
| **Total: 6** | | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{097c5d00-9f92-4c3e-8056-7de09347fbb6-10_716_899_258_621}
The diagram shows a circle with centre $A$ and radius $r$. Diameters $C A D$ and $B A E$ are perpendicular to each other. A larger circle has centre $B$ and passes through $C$ and $D$.\\
(i) Show that the radius of the larger circle is $r \sqrt { } 2$.\\
(ii) Find the area of the shaded region in terms of $r$.\\
\hfill \mbox{\textit{CAIE P1 Q7 [7]}}