Moderate -0.8 This is a straightforward application of the standard formula for segment area (sector area minus triangle area). With the angle and radius given directly, it requires only substituting into memorized formulas: sector area = ½r²θ and triangle area = ½r²sin(θ). The algebra is simple and the question explicitly guides students to the final form, making it easier than average.
The diagram below shows a sector of a circle \(OAB\).
The chord \(AB\) divides the sector into a triangle and a shaded segment.
Angle \(AOB\) is \(\frac{\pi}{6}\) radians.
The radius of the sector is 18 cm.
\includegraphics{figure_5}
Show that the area of the shaded segment is
$$k(\pi - 3) \text{cm}^2$$
where \(k\) is an integer to be found.
[3 marks]
The diagram below shows a sector of a circle $OAB$.
The chord $AB$ divides the sector into a triangle and a shaded segment.
Angle $AOB$ is $\frac{\pi}{6}$ radians.
The radius of the sector is 18 cm.
\includegraphics{figure_5}
Show that the area of the shaded segment is
$$k(\pi - 3) \text{cm}^2$$
where $k$ is an integer to be found.
[3 marks]
\hfill \mbox{\textit{AQA Paper 3 2024 Q5 [3]}}