| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Triangle with circular sector |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing basic circle theorems (arc length s=rθ) and sector area formulas. Part (a) is direct substitution, part (b) requires setting up and simplifying an area equation, and part (c) involves solving a quadratic and adding perimeter components. All steps are standard bookwork applications with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
9.
Diagram NOT accurately drawn
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-24_581_1491_340_296}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the plan view of the area being used for a ball-throwing competition.\\
Competitors must stand within the circle $C$ and throw a ball as far as possible into the target area, $P Q R S$, shown shaded in Figure 3.
Given that
\begin{itemize}
\item circle $C$ has centre $O$
\item $P$ and $S$ are points on $C$
\item $O P Q R S O$ is a sector of a circle with centre $O$
\item the length of arc $P S$ is 0.72 m
\item the size of angle $P O S$ is 0.6 radians
\begin{enumerate}[label=(\alph*)]
\item show that $O P = 1.2 \mathrm {~m}$
\end{itemize}
Given also that
\begin{itemize}
\item the target area, $P Q R S$, is $90 \mathrm {~m} ^ { 2 }$
\item length $P Q = x$ metres
\item show that
\end{itemize}
$$5 x ^ { 2 } + 12 x - 1500 = 0$$
\item Hence calculate the total perimeter of the target area, $P Q R S$, giving your answer to the nearest metre.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q9 [7]}}