Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

#### Solution

Given that, radius of the circle (r) = 21 cm and central angle of the sector AOBA (θ) = 120°

So, area of the circle = `pir^2 = 22/7 xx (21)^2 = 22/7 xx 21 xx 21`

= `22 xx 3 xx 21`

= 1386 cm^{2}

Now, area of the minor AOBA with central angle 120°

= `(pir^2)/360^circ xx theta = 22/7 xx (21 xx 21)/360^circ xx 120`

= `(22 xx 3 xx 21)/3`

= `22 xx 21`

= 462 cm^{2}

∴ Area of the amjor sector ABOA = Area of the circle – Area of the sector AOBA

=1386 – 462

= 924 cm^{2}

∴ Difference of the areas of a sector AOBA and its corresponding major sector ABOA

= |Area of major sector ABOA – Area of minor sector AOBA|

= |924 – 462|

= 462 cm^{2}

Hence, the required difference of two sectors is 462 cm^{2}