Resources for Area of Sectors
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Area of Sectors Theory
![\begin{minipage}[c]{10cm} \begin{center} Area of a sector \(=\dfrac{\theta}{360^{\circ}} \times \pi r^2\) \end{center} \end{minipage} \begin{minipage}[c]{3cm} \begin{tikzpicture}[scale=1,line width=1pt] \coordinate[label=above:C] (O) at (0,0); \coordinate[label=below right:B] (B) at (-45:2); \coordinate[label=below left:A] (A) at (225:2); \path[fill=blue!20] (O)--(B) arc (-45:-135:2)--cycle; \draw (O)--(B) node[right,midway]{\(r\)} arc (-45:-135:2)node[below,midway]{}--(O); \fill (O) circle (2.5pt); \pic [draw=black,line width=1pt,angle radius=0.4cm,angle eccentricity=1.6,"\(\theta^{\circ}\)"] {angle=A--O--B}; \end{tikzpicture} \end{minipage}\\ \begin{multicols}{2} \textbf{Example 1}\\ Find the exact area of a sector given that \(\theta=60^{\circ}\) and \(r=6 \text{~cm}\).\\ \textbf{Example 1 solution}\\ \(\begin{aligned} \text { Area } & =\frac{60^{\circ}}{360^{\circ}} \times \pi \times 6^2 \\ & =\frac{1}{6} \times \pi \times 6^2 \\ & =6 \pi \text{ cm}^2 \end{aligned}\) \columnbreak \textbf{Example 2}\\ The exact area of a sector is \(24 \pi \text{ cm}^2\). The radius of the circle is \(8 \text{~cm}\). Find the angle of the sector.\\ \textbf{Example 2 solution}\\ \(\begin{aligned} \text { Area } & =\frac{\theta}{360^{\circ}} \times \pi \times r^2 \\ 24 \pi & =\frac{\theta}{360} \times \pi \times 8^2 \\ 24 & =\frac{\theta}{360} \times 64 \\ \theta & =\frac{24}{64} \times 360 \\ \theta & =\frac{3}{8} \times 360^{\circ} \\ \theta & =135^{\circ} \end{aligned}\) \end{multicols}](/media/m2ofxl2x/7963.png)
NSW Y8 Maths Area and Perimeter Area of Sectors
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Area of Sectors - Video - Area of Sectors 1
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Area of Sectors - Video - Area of Sectors 2
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