| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| 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 (given as 'show that'), then calculating areas of circular sectors and triangles. The perpendicular diameters and symmetry make the setup clear, and the methods are standard P1 techniques with no novel insight required. Slightly easier than average due to the guided 'show that' part and routine area calculations. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(BC^2 = r^2 + r^2 = 2r^2 \rightarrow BC = r\sqrt{2}\) | B1 [1] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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 \(CADE = \frac{1}{2}\pi r^2\) | B1 | |
| Shaded area \(= \frac{1}{2}\pi r^2 - \left(\frac{1}{2}\pi r^2 - r^2\right)\) | ||
| 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 [6] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $BC^2 = r^2 + r^2 = 2r^2 \rightarrow BC = r\sqrt{2}$ | B1 [1] | AG |
### Part (ii):
| Answer/Working | 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 $CADE = \frac{1}{2}\pi r^2$ | B1 | |
| Shaded area $= \frac{1}{2}\pi r^2 - \left(\frac{1}{2}\pi r^2 - r^2\right)$ | | |
| 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 [6] | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{a9e04003-1e43-40c4-991a-36aa3a93654b-3_718_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 2015 Q7 [7]}}