Question

Find the perimeter of the sector of a circle of radius 14cm and angle ${45}^0$.

Answer 1

Hint: Perimeter of a sector is the total length of the circumference of the circle subtended within the angle$$\theta$$. Perimeter is the sum of the total length of the arc and the two radii.
The length of the arc of a circle is a part of the total circumference of the circle given by$2\pi r$. In the given question, we have to find how much of the total circumference of the circle is subtended by the sector. The formula used for the length of the arc of a circle of radius $r$ and subtending $\theta$ degrees at the center of the circle is$P_{arc}=2\pi r\times \left({\displaystyle \frac{\theta }{360}}\right)$ . Adding the two radii arm with the length of the arc results in the total perimeter of the sector that is equivalent to $P_s=2\pi r\times \left({\displaystyle \frac{\theta }{360}}\right)+2r$. The formula can be used to determine the perimeter of any part of the circle (for all the sectors of a circle) depending on the angle subtended in the center.

Complete step by step answer:
Substitute $r=14\text{ cm}$ and $\theta =45$ in the formula $P_s=2\pi r\times \left({\displaystyle \frac{\theta }{360}}\right)+2r$to determine the perimeter of the sector subtending ${45}^0$ of the angle at the center.

 $$\begin{gathered} P_s=2\pi r\times \left({\displaystyle \frac{\theta }{360}}\right)+2r \\ =2\times {\displaystyle \frac{22}{7}}\times 14\times \left({\displaystyle \frac{45}{360}}\right)+2\times 14 \\ =2\times 22\times 2\times {\displaystyle \frac{1}{8}}+28 \\ =11+28 \\ =39\text{ cm} \end{gathered}$$

Hence, $39\text{ cm}$is the perimeter of the sector subtending ${45}^0$ of the angle at the centre of the circle whose radius is$14\text{ cm}$.

Note: Since the sector is just a part of the circle subtending an angle$\theta$ at the center, first find out by what factor of the full circle is covered by the sector using$\left({\displaystyle \frac{\theta }{360}}\right)$.