| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - area/perimeter with given values |
| Difficulty | Standard +0.3 This is a straightforward application of sector area and arc length formulas with minimal problem-solving required. Part (a) uses the sector area formula A = ½r²θ to find the angle (one-step rearrangement), and part (b) requires calculating arc length and using Pythagoras for the triangle side. While it combines geometric shapes, the techniques are standard and the question guides students through each step clearly, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 4^2 \times \text{angle } BAD = 10\) | M1 | Use of sector area formula |
| Angle \(BAD = 1.25\) | A1 | OE. Accept \(0.398\pi\), \(71.6°\) for SC B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Arc \(BD = 4 \times \textit{their } 1.25\) | M1 | Use of arc length formula. Expect 5. |
| \(BC = 4\tan(\textit{their } 1.25)\) | M1 | Expect 12.0(4). May use \(ACB = 0.321\) or \(18.4°\) |
| \(CD = \dfrac{4}{\cos(\textit{their } 1.25)} - 4\) or \(\sqrt{4^2 + (\textit{their } BC)^2} - 4\) | M1 | Expect \(12.69 - 4 = 8.69\). May use \(ACB\). |
| Perimeter \(= 5 + 12.0(4) + 8.69 = 25.7\) (cm) | A1 | AWRT |
## Question 5:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 4^2 \times \text{angle } BAD = 10$ | M1 | Use of sector area formula |
| Angle $BAD = 1.25$ | A1 | OE. Accept $0.398\pi$, $71.6°$ for **SC B1** only |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Arc $BD = 4 \times \textit{their } 1.25$ | M1 | Use of arc length formula. Expect 5. |
| $BC = 4\tan(\textit{their } 1.25)$ | M1 | Expect 12.0(4). May use $ACB = 0.321$ or $18.4°$ |
| $CD = \dfrac{4}{\cos(\textit{their } 1.25)} - 4$ or $\sqrt{4^2 + (\textit{their } BC)^2} - 4$ | M1 | Expect $12.69 - 4 = 8.69$. May use $ACB$. |
| Perimeter $= 5 + 12.0(4) + 8.69 = 25.7$ (cm) | A1 | AWRT |
---5
The diagram shows a triangle $A B C$, in which angle $A B C = 90 ^ { \circ }$ and $A B = 4 \mathrm {~cm}$. The sector $A B D$ is part of a circle with centre $A$. The area of the sector is $10 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find angle $B A D$ in radians.
\item Find the perimeter of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q5 [6]}}