| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.3 This is a straightforward application of standard sector and segment formulas with clear geometric setup. Part (a) uses symmetry of the semicircle (angle BOD = π - 2×0.7), part (b) requires sector area minus triangle area, and part (c) needs arc length plus chord length. All steps are routine for P1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\angle BOD = \pi - 2 \times 0.7 = 1.742\) | B1* | Correct working to achieve 1.742 or better; must state \(\angle BOD = 1.742\) as minimal conclusion; may work in degrees: \(180 - 2 \times \frac{0.7}{\pi} \times 180 \approx 99.8°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area of \(BOD = \frac{1}{2} \times 3^2 \sin 1.742\ (= \text{awrt } 4.43)\) | M1 | Correct strategy for area of triangle \(BOD\) using \(\angle BOD = 1.742\); may work in degrees |
| Area of \(R\): \(\frac{1}{2} \times 3^2 \times 1.742 - \frac{1}{2} \times 3^2 \sin 1.742\) or \(\frac{1}{2}\pi \times 3^2 - \frac{1}{2} \times 3^2 \sin 1.742 - 2 \times \frac{1}{2} \times 3^2 \times 0.7\) | dM1 | Applies correct method for area of \(R\); dependent on previous M mark |
| \(= \text{awrt } 3.4\ \text{m}^2\) | A1 | Do not award if other areas added/subtracted incorrectly |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(BD = \sqrt{3^2 + 3^2 - 2\times3\times3\cos 1.742}\ (= \text{awrt } 4.59)\) or \(BD = 2 \times 3\sin\!\left(\frac{1.742}{2}\right)\) or \(BD = 2 \times 3\cos 0.7\) or arc \(BCD = 3 \times 1.742\ (= 5.226)\) | M1 | Correct method for length \(BD\) (implied by awrt 4.59) or correct method for arc \(BCD\) (= 5.226); may work in degrees |
| Perimeter of \(R\): \(3 \times 1.742 + BD\) | dM1 | Fully correct method adding arc \(BCD\) to their \(BD\); both methods must be correct; dependent on previous M mark |
| \(= \text{awrt } 9.8\ \text{m}\) | A1 | Do not award if incorrect working seen |
## Question 5:
**Part (a):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\angle BOD = \pi - 2 \times 0.7 = 1.742$ | B1* | Correct working to achieve 1.742 or better; must state $\angle BOD = 1.742$ as minimal conclusion; may work in degrees: $180 - 2 \times \frac{0.7}{\pi} \times 180 \approx 99.8°$ |
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| Area of $BOD = \frac{1}{2} \times 3^2 \sin 1.742\ (= \text{awrt } 4.43)$ | M1 | Correct strategy for area of triangle $BOD$ using $\angle BOD = 1.742$; may work in degrees |
| Area of $R$: $\frac{1}{2} \times 3^2 \times 1.742 - \frac{1}{2} \times 3^2 \sin 1.742$ or $\frac{1}{2}\pi \times 3^2 - \frac{1}{2} \times 3^2 \sin 1.742 - 2 \times \frac{1}{2} \times 3^2 \times 0.7$ | dM1 | Applies correct method for area of $R$; dependent on previous M mark |
| $= \text{awrt } 3.4\ \text{m}^2$ | A1 | Do not award if other areas added/subtracted incorrectly |
**Part (c):**
| Working | Mark | Guidance |
|---------|------|----------|
| $BD = \sqrt{3^2 + 3^2 - 2\times3\times3\cos 1.742}\ (= \text{awrt } 4.59)$ or $BD = 2 \times 3\sin\!\left(\frac{1.742}{2}\right)$ or $BD = 2 \times 3\cos 0.7$ or arc $BCD = 3 \times 1.742\ (= 5.226)$ | M1 | Correct method for length $BD$ (implied by awrt 4.59) or correct method for arc $BCD$ (= 5.226); may work in degrees |
| Perimeter of $R$: $3 \times 1.742 + BD$ | dM1 | Fully correct method adding arc $BCD$ to their $BD$; both methods must be correct; dependent on previous M mark |
| $= \text{awrt } 9.8\ \text{m}$ | A1 | Do not award if incorrect working seen |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6c320b71-8793-461a-a078-e4f64c144a3a-12_401_677_219_635}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a plan view of a semicircular garden $A B C D E O A$
The semicircle has
\begin{itemize}
\item centre $O$
\item diameter $A O E$
\item radius 3 m
\end{itemize}
The straight line $B D$ is parallel to $A E$ and angle $B O A$ is 0.7 radians.
\begin{enumerate}[label=(\alph*)]
\item Show that, to 4 significant figures, angle $B O D$ is 1.742 radians.
The flowerbed $R$, shown shaded in Figure 2, is bounded by $B D$ and the arc $B C D$.
\item Find the area of the flowerbed, giving your answer in square metres to one decimal place.
\item Find the perimeter of the flowerbed, giving your answer in metres to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q5 [7]}}