Question

In a circle of radius $21\mathrm{cm}$, an arc subtends an angle of $60°$ at the center. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord

Answer 1

Step 1: Find the length of the arc.

Given: a radius $=21\mathrm{cm}$ and an arc angle $=60°$

$$\begin{gathered} \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{arc}=2\mathrm{\pi r}\frac{\mathrm{\theta }}{360°} \\ =2\times \frac{22}{7}\times 21\times \frac{60°}{360°}\mathbf{[}\mathbf{\because }\mathbf{\pi }\mathbf{=}\frac{\mathbf{22}}{\mathbf{7}}\mathbf{]} \\ =22\mathrm{cm} \end{gathered}$$

Step 2: Find the area of the sector.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{sector}=\frac{\mathrm{\theta }}{360°}{\mathrm{\pi r}}^2 \\ =\frac{60°}{360°}\times \frac{22}{7}\times 21\times 21\mathbf{[}\mathbf{\because }\mathbf{\pi }\mathbf{=}\frac{\mathbf{22}}{\mathbf{7}}\mathbf{]} \\ =231{\mathrm{cm}}^2 \end{gathered}$$

Step 3: Find the area of the segment.

Area of the segment APB $=$Area of sector OAPB $-$ Area of triangle OAB

In $∆\mathrm{AOB}$ two sides are equal. Hence angles opposite to them are also equal.

Let $\angle \mathrm{OAB}=\angle \mathrm{OBA}=\mathrm{x}$

$$\begin{gathered} \angle \mathrm{AOB}+\angle \mathrm{OAB}+\angle \mathrm{ABO}=180°\left(\mathrm{sum}\mathrm{angle}\mathrm{property}\mathrm{of}\mathrm{triangle}\right) \\ \Rightarrow 60°+\mathrm{x}+\mathrm{x}=180° \\ \Rightarrow 60°+2\mathrm{x}=180° \\ \Rightarrow 2\mathrm{x}=180°-60° \\ \Rightarrow 2\mathrm{x}=120° \\ \Rightarrow \mathrm{x}=60° \end{gathered}$$

Therefore $\angle \mathrm{OAB}=\angle \mathrm{OBA}=60°$

Hence $∆\mathrm{AOB}$ is an equilateral triangle.

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2$

Area of the segment $=231-\frac{\sqrt{3}}{4}\times {21}^2$

$$\begin{gathered} =231-190.95 \\ =40.04{\mathrm{cm}}^2 \end{gathered}$$

Therefore, the length of the arc $=22\mathrm{cm}$ area of sector$=231{\mathrm{cm}}^2$, area of segment $=40.04{\mathrm{cm}}^2$.