| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| 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 circle's radius, then calculating areas using standard circle formulas (πr² and sector areas). The perpendicular diameters setup is clear, and the multi-step calculation is routine for A-level, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| \(BC^2 = r^2 + r^2 = 2r^2 \rightarrow BC = r\sqrt{2}\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Area sector \(BCFD = \frac{1}{4}\pi(r\sqrt{2})^2\) seen or implied | M1 | Expect \(\frac{1}{2}\pi r^2\). (\(F\) is intersection of large circle with \(AE\)) |
| Area \(\triangle ABCD = \frac{1}{2}(2r)r\) | M1* | Expect \(r^2\) (could be embedded) |
| Area segment \(CFDA = \frac{1}{2}\pi r^2 - r^2\) | A1 | OE |
| Area semi circle \(CADE = \frac{1}{2}\pi r^2\) | B1 | |
| Shaded area \(= \frac{1}{2}\pi r^2 - (\frac{1}{2}\pi r^2 - r^2)\) OR \(\pi r^2 - [\frac{1}{2}\pi r^2 + (\frac{1}{2}\pi r^2 - r^2)]\) | DM1 | Depends on the area \(\triangle BCD\) |
| \(= r^2\) | A1 |
# Question 9:
## Part 9(a):
| $BC^2 = r^2 + r^2 = 2r^2 \rightarrow BC = r\sqrt{2}$ | B1 | AG |
## Part 9(b):
| Area sector $BCFD = \frac{1}{4}\pi(r\sqrt{2})^2$ seen or implied | M1 | Expect $\frac{1}{2}\pi r^2$. ($F$ is intersection of large circle with $AE$) |
| Area $\triangle ABCD = \frac{1}{2}(2r)r$ | M1* | Expect $r^2$ (could be embedded) |
| Area segment $CFDA = \frac{1}{2}\pi r^2 - r^2$ | A1 | OE |
| Area semi circle $CADE = \frac{1}{2}\pi r^2$ | B1 | |
| Shaded area $= \frac{1}{2}\pi r^2 - (\frac{1}{2}\pi r^2 - r^2)$ **OR** $\pi r^2 - [\frac{1}{2}\pi r^2 + (\frac{1}{2}\pi r^2 - r^2)]$ | DM1 | Depends on the area $\triangle BCD$ |
| $= r^2$ | A1 | |
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{9803d51b-215e-4d03-884f-a67fb7ed6442-14_713_912_258_573}
The diagram shows a circle with centre $A$ and radius $r$. Diameters CAD and BAE are perpendicular to each other. A larger circle has centre $B$ and passes through $C$ and $D$.\\
(a) Show that the radius of the larger circle is $r \sqrt { 2 }$.\\
(b) Find the area of the shaded region in terms of $r$.\\
\hfill \mbox{\textit{CAIE P1 2020 Q9 [7]}}