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 cm2
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 cm2
∴ Area of the amjor sector ABOA = Area of the circle – Area of the sector AOBA
=1386 – 462
= 924 cm2
∴ 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 cm2
Hence, the required difference of two sectors is 462 cm2