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?

(N/A) Let the radii of the two sectors be $r_1 = 7 \, cm$ and $r_2 = 21 \, cm$,and their central angles be $\theta_1 = 120^{\circ}$ and $\theta_2 = 40^{\circ}$.
Area of a sector is given by $A = \frac{\theta}{360^{\circ}} \times \pi r^2$.
For the first sector:
$A_1 = \frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (7)^2 = \frac{1}{3} \times \frac{22}{7} \times 49 = \frac{22 \times 7}{3} = \frac{154}{3} \approx 51.33 \, cm^2$.
For the second sector:
$A_2 = \frac{40^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (21)^2 = \frac{1}{9} \times \frac{22}{7} \times 441 = \frac{22 \times 63}{9} = 22 \times 7 = 154 \, cm^2$.
Length of an arc is given by $l = \frac{\theta}{360^{\circ}} \times 2\pi r$.
For the first sector:
$l_1 = \frac{120^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 7 = \frac{1}{3} \times 44 = \frac{44}{3} \approx 14.67 \, cm$.
For the second sector:
$l_2 = \frac{40^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{9} \times 2 \times 22 \times 3 = \frac{132}{9} = \frac{44}{3} \approx 14.67 \, cm$.
Observation: The arc lengths of the two sectors are equal,but their areas are not equal.