| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Standard +0.3 This is a standard geometric area problem requiring identification of a sector and triangle, then subtraction to find a segment area. The geometry is straightforward (equilateral triangle with 60° angles), and the calculation involves routine formulas for sector area and triangle area. While it requires careful setup and exact form manipulation, it's a typical textbook exercise slightly easier than average A-level questions. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies angle \(BAO = \frac{\pi}{3}\) or angle \(BOC = \frac{2\pi}{3}\) | B1 | May be seen on diagram or embedded in formula. Implied by e.g. \(\frac{1}{6}\times\pi\times 5^2\) for minor sector area |
| Area(segment) \(= \frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3} \left\{= \frac{25\pi}{6} - \frac{25\sqrt{3}}{4}\right\}\) or Area\((AOB) = 2\times\frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3}\) | M1 | Uses correct process with angle \(\frac{\pi}{3}\) radians or 60°. Use of 60° must be in correct formula in degrees. \(0.5\times 5^2\times 60\) scores M0 |
| Area of \(R = \frac{1}{2}\times 5^2\times\frac{2\pi}{3} - \left(\frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3}\right)\) or Area of \(R = \frac{1}{2}\times\pi\times 5^2 - \left(\frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3} + \frac{1}{2}\times 5^2\times\frac{\pi}{3}\right)\) | dM1 | Fully correct strategy for area of \(R\). Follow through on incorrectly simplified areas but method must be correct. Reference area \(\approx 23.92\) (2 d.p.) likely implies B1M1dM1 |
| \(= \frac{25}{4}\sqrt{3} + \frac{25}{6}\pi \ (\text{cm}^2)\) | A1 | Correct expression in correct form e.g. \(\frac{50}{8}\sqrt{3} + \frac{75}{18}\pi\). Ignore absence of units. Do not apply isw if additional areas added/subtracted. Condone poor simplification e.g. \(\frac{25}{12}(\sqrt{3}+\pi)\) following correct answer |
## Question 11:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies angle $BAO = \frac{\pi}{3}$ or angle $BOC = \frac{2\pi}{3}$ | B1 | May be seen on diagram or embedded in formula. Implied by e.g. $\frac{1}{6}\times\pi\times 5^2$ for minor sector area |
| Area(segment) $= \frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3} \left\{= \frac{25\pi}{6} - \frac{25\sqrt{3}}{4}\right\}$ **or** Area$(AOB) = 2\times\frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3}$ | M1 | Uses correct process with angle $\frac{\pi}{3}$ radians or 60°. Use of 60° must be in correct formula in degrees. $0.5\times 5^2\times 60$ scores M0 |
| Area of $R = \frac{1}{2}\times 5^2\times\frac{2\pi}{3} - \left(\frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3}\right)$ **or** Area of $R = \frac{1}{2}\times\pi\times 5^2 - \left(\frac{1}{2}\times 5^2\times\frac{\pi}{3} - \frac{1}{2}\times 5^2\sin\frac{\pi}{3} + \frac{1}{2}\times 5^2\times\frac{\pi}{3}\right)$ | dM1 | Fully correct strategy for area of $R$. Follow through on incorrectly simplified areas but method must be correct. Reference area $\approx 23.92$ (2 d.p.) likely implies B1M1dM1 |
| $= \frac{25}{4}\sqrt{3} + \frac{25}{6}\pi \ (\text{cm}^2)$ | A1 | Correct expression in correct form e.g. $\frac{50}{8}\sqrt{3} + \frac{75}{18}\pi$. Ignore absence of units. Do not apply isw if additional areas added/subtracted. Condone poor simplification e.g. $\frac{25}{12}(\sqrt{3}+\pi)$ following correct answer |11.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{e116a86f-63e0-4e80-b49c-d9f3c819ce15-28_451_899_239_584}
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\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows the design of a badge.\\
The shape $A B C O A$ is a semicircle with centre $O$ and diameter 10 cm .\\
$O B$ is the arc of a circle with centre $A$ and radius 5 cm .\\
The region $R$, shown shaded in Figure 4, is bounded by the arc $O B$, the arc $B C$ and the line $O C$.
Find the exact area of $R$.\\
Give your answer in the form $( a \sqrt { 3 } + b \pi ) \mathrm { cm } ^ { 2 }$, where $a$ and $b$ are rational numbers.
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q11 [4]}}