Question

Question 16
The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively ${120}^{\circ }$ and ${40}^{\circ }$. Find the areas of the two sectors as well as the lengths of the corresponding arcs, what do you observe?

Answer 1

Let the lengths of the corresponding arc be $l_1$ and $l_2$

Given that, radius of sector $P{O}_{1}QP=7cm$

And radius of sector $A{O}_{2}BA=21cm$

And central angle of the sector $A{O}_{2}BA={40}^{\circ }$

$\therefore$ Area of the sector with central angle $O_1$

$=\frac{\pi r^2}{{360}^{\circ }}\times \theta =\frac{\pi (7{)}^2}{{360}^{\circ }}\times {120}^{\circ }$

$=\frac{22}{7}\times \frac{7\times 7}{{360}^{\circ }}\times 120$

$=\frac{22\times 7}{3}=\frac{154}{3}c{m}^2$

and area of the sector with central angle $O_2$

$=\frac{\pi r^2}{{360}^{\circ }}\times \theta =\frac{\pi (21{)}^2}{{360}^{\circ }}\times {40}^{\circ }$

$=\frac{22}{7}\times \frac{21\times 21}{{360}^{\circ }}\times {40}^{\circ }$

$=\frac{22\times 3\times 21}{9}=22\times 7=154c{m}^2$

Now, corresponding arc length of the sector $P{O}_{1}QP$

= $\frac{\theta }{{360}^{\circ }}\times 2\pi r$

= $\frac{{120}^{\circ }}{{360}^{\circ }}\times 2\frac{22}{7}\times 7$

= $\frac{44}{3}cm$

Now, corresponding arc length of the sector $A{O}_{2}BA$

= $\frac{\theta }{{360}^{\circ }}\times 2\pi r$

= $\frac{{40}^{\circ }}{{360}^{\circ }}\times 2\frac{22}{7}\times 21$

= $\frac{44}{3}cm$

Hence, we observe that arc lengths of two sectors of two different circles may be equal but their area need not be equal.