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?

Let the lengths of the corresponding arc be $l_1$ and $l _{2}$

Given that, radius of sector $P O_1 Q P=7\, cm$

and radius of sector $A O_{2} B A=21\, cm$

Central angle of the sector $P O, Q P=120^{\circ}$0

and central angle of the sector $A O_{2} B A=40^{\circ}$

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

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

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

$=\frac{22 \times 7}{3}=\frac{154}{3}\, cm ^{2}$

and area of the sector with central angle $O _{2}$

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

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

$=\frac{22 \times 3 \times 21}{9}=22 \times 7=154 \,cm ^{2}$

Now, corresponding arc length of the sector $PO_1Q$ P $=$ Central angle $\times$ Radius of the sector

$=120^{\circ} \times 7 \times \frac{\pi}{180^{\circ}}$ $\left[\because \theta=\frac{l}{r}\right.$ and $\left.1^{\circ}=\frac{\pi}{180^{\circ}} R\right]$

$=\frac{2}{3} \times 7 \times \frac{22}{7}$

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

and corresponding arc length of the sector $A O_{2} EA$ $=$ Central angle $\times$ Radius of the sector

$=40^{\circ} \times 21 \times \frac{\pi}{180^{\circ}}$ $\left[\therefore \theta=\frac{l}{r}\right.$ and $\left.1^{\circ}=\frac{\pi}{180^{\circ}} R\right]$

$=\frac{2}{9} \times 21 \times \frac{22}{7}$

$=\frac{2}{3} \times 22=\frac{44}{3}\, cm$

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