Resources for Arc Lengths
-
Questions
10
With Worked SolutionClick Here -
Video Tutorials
1
Click Here
Arc Lengths Theory
![\begin{minipage}[c]{11cm} The arc length of a circle with radius \(r\) and the angle at the centre is given by: $$ l=\frac{\theta}{360^{\circ}} \times 2 \pi r $$ \end{minipage} \quad \begin{minipage}[c]{3cm} \begin{center} \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); \draw (O)--(B) node[right,midway]{\(r\)} arc (-45:-135:2)node[below,midway]{\(l\)}--(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{center} \end{minipage} \hfill \linebreak \begin{multicols}{2} \textbf{Example 1}\\ Find the are length \(AB\) given that the radius is \(6 \text{~cm}\) and \(\theta=120^{\circ}\).\\ \begin{center} \begin{tikzpicture}[scale=1,line width=1pt] \coordinate[label=above:C] (O) at (0,0); \coordinate[label=below right:B] (B) at (-30:2); \coordinate[label=below left:A] (A) at (210:2); \draw (O)--(B) node[above right,midway]{\(6\text{ cm}\)} arc (-30:-150:2)node[below,midway]{}--(O); \fill (O) circle (2.5pt); \pic [draw=black,line width=1pt,angle radius=0.4cm,angle eccentricity=1.6,"\(120^{\circ}\)"] {angle=A--O--B}; \end{tikzpicture} \end{center} \textbf{Example 1 solution}\\[3pt] \(\begin{aligned} l & =\frac{120^{\circ}}{360^{\circ}} \times 2 \pi \times 6 \\ & =\frac{1}{3} \times 2 \pi \times 6 \\ & =4 \pi \text{ cm}. \end{aligned}\) \columnbreak \textbf{Example 2}\\ Find the perimeter of the sector given that \(\theta=60^{\circ}\) and \(r=4 \text{~cm}\).\\ \begin{center} \begin{tikzpicture}[scale=1,line width=1pt] \coordinate[label=above:C] (O) at (0,0); \coordinate[label=below right:B] (B) at (-60:2); \coordinate[label=below left:A] (A) at (240:2); \draw (O)--(B) node[above right,midway]{\(4\text{ cm}\)} arc (-60:-120:2)node[below,midway]{}--(O); \fill (O) circle (2.5pt); \pic [draw=black,line width=1pt,angle radius=0.5cm,angle eccentricity=1.6,"\(60^{\circ}\)"] {angle=A--O--B}; \end{tikzpicture} \end{center} \textbf{Example 2 solution}\\[3pt] \(\begin{aligned} l & =\frac{60^{\circ}}{360^{\circ}} \times 2 \pi \times 4 \\ & =\frac{1}{6} \times 8 \pi \\ & =\frac{4 \pi}{3} \end{aligned}\)\\ \(\begin{aligned} \therefore \text { Perimeter } & =4+4+\frac{4 \pi}{3} \\ & =\left(8+\frac{4 \pi}{3}\right) \mathrm{cm} .\end{aligned}\) \end{multicols}](/media/k2sfycpt/32354.png)
NSW Y8 Maths Area and Perimeter Arc Lengths
Videos
Videos relating to Arc Lengths.
-
Arc Lengths - Video - Arc Lengths - Video
Plans & Pricing
With all subscriptions, you will receive the below benefits and unlock all answers and fully worked solutions.
-
Teachers TutorsFeaturesFreeProAll ContentAll courses, all topicsQuestionsAnswersWorked SolutionsSystemYour own personal portalQuizbuilderClass ResultsStudent ResultsExam RevisionRevision by TopicPractise ExamsAnswersWorked SolutionsAwesome StudentsFeaturesFreeProContentAny course, any topicQuestionsAnswersWorked SolutionsSystemYour own personal portalBasic ResultsAnalyticsStudy RecommendationsExam RevisionRevision by TopicPractise ExamsAnswersWorked Solutions