| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Moderate -0.5 This is a straightforward geometry problem requiring Pythagoras to find AC, then arc length and sector area formulas. All steps are routine applications of standard techniques with no conceptual challenges, making it slightly easier than average for A-level. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c 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 |
|---|---|---|
| \(\tan BDC = \frac{4}{3}\) or \(\sin BDC = \frac{4}{5}\) or \(\cos BDC = \frac{3}{5}\) used to find \(ADC\) | M1 | May use cosine rule or \(CAD = \tan^{-1}\frac{4}{8}\) |
| \(BDC = 0.927[3]\ \rightarrow ADC = \pi - 0.927[3]\ [= 2.214 \text{ to } 2.215]\) | A1 | Allow degrees \(126.87\), and \(0.7048\pi\) or \(0.705\pi\) |
| \(\text{Arc } AC = 5\times\text{their } 2.214\) | M1 | Use of \(r\theta\) or \(\frac{\theta}{360}\cdot 2\pi r\). Expect \(11.07\) |
| \(AC = \sqrt{8^2+4^2}\) or \(2\times5\times\sin 1.107\) | M1 | Expect \(8.94\) |
| \([\text{Perimeter} = 11.07 + 8.94 =]\ 20.0\) | A1 | Accept AWRT \([20.01, 20.02]\) |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Sector \(ACD = \frac{1}{2}\times5^2\times\text{their } 2.214\) | M1 | See use of \(\frac{1}{2}r^2\theta\) or \(\frac{\theta}{360}\cdot\pi r^2\). Expect \(27.7\) |
| Subtracting the area of \(\triangle ADC = \frac{1}{2}\times5\times4\) or \(\frac{1}{2}5^2\sin(\text{their } 2.214)\) or \(\frac{1}{2}\times8\times4 - \frac{1}{2}\times3\times4\) | M1 | Subtracting the area of \(\triangle ADC\), expect \(-10\) |
| Shaded area \(= 27.7 - 10 = 17.7\) | A1 | Accept AWRT \([17.67, 17.68]\). Correct answer cannot come from an angle of \(2.215\) |
| 3 |
## Question 8(a):
| $\tan BDC = \frac{4}{3}$ or $\sin BDC = \frac{4}{5}$ or $\cos BDC = \frac{3}{5}$ **used** to find $ADC$ | M1 | May use cosine rule or $CAD = \tan^{-1}\frac{4}{8}$ |
|---|---|---|
| $BDC = 0.927[3]\ \rightarrow ADC = \pi - 0.927[3]\ [= 2.214 \text{ to } 2.215]$ | A1 | Allow degrees $126.87$, and $0.7048\pi$ or $0.705\pi$ |
| $\text{Arc } AC = 5\times\text{their } 2.214$ | M1 | Use of $r\theta$ or $\frac{\theta}{360}\cdot 2\pi r$. Expect $11.07$ |
| $AC = \sqrt{8^2+4^2}$ or $2\times5\times\sin 1.107$ | M1 | Expect $8.94$ |
| $[\text{Perimeter} = 11.07 + 8.94 =]\ 20.0$ | A1 | Accept AWRT $[20.01, 20.02]$ |
| | **5** | |
## Question 8(b):
| Sector $ACD = \frac{1}{2}\times5^2\times\text{their } 2.214$ | M1 | See use of $\frac{1}{2}r^2\theta$ or $\frac{\theta}{360}\cdot\pi r^2$. Expect $27.7$ |
|---|---|---|
| Subtracting the area of $\triangle ADC = \frac{1}{2}\times5\times4$ or $\frac{1}{2}5^2\sin(\text{their } 2.214)$ or $\frac{1}{2}\times8\times4 - \frac{1}{2}\times3\times4$ | M1 | Subtracting the area of $\triangle ADC$, expect $-10$ |
| Shaded area $= 27.7 - 10 = 17.7$ | A1 | Accept AWRT $[17.67, 17.68]$. Correct answer cannot come from an angle of $2.215$ |
| | **3** | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{3bad1d9f-5b9e-4895-aa4e-3e6d9f6c072e-10_454_744_255_703}
The diagram shows triangle $A B C$ in which angle $B$ is a right angle. The length of $A B$ is 8 cm and the length of $B C$ is 4 cm . The point $D$ on $A B$ is such that $A D = 5 \mathrm {~cm}$. The sector $D A C$ is part of a circle with centre $D$.
\begin{enumerate}[label=(\alph*)]
\item Find the perimeter of the shaded region.
\item Find the area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q8 [8]}}