Question

Question 20
Find the difference of the areas of a sector of angle ${120}^{\circ }$ and its corresponding major sector of a circle of radius 21 cm.

Answer 1

Given that, radius of the circle (r) = 21 cm and central angle of the sector AOBA $\left(\theta \right)={120}^{\circ }$

So, area of the circle $=\pi r^2=\frac{22}{7}\times (21{)}^2=\frac{22}{7}\times 21\times 21$

$=22\times 3\times 21=1386c{m}^2$

Now, area of the minor sector AOBA with central angle ${120}^{\circ }$

$=\frac{\pi r^2}{{360}^{\circ }}\times \theta =\frac{22}{7}\times \frac{21\times 21}{{360}^{\circ }}\times 120$

$=\frac{22\times 3\times 21}{3}=22\times 21=462c{m}^2$

$\therefore$ Area of the major sector ABOA

= Area of the circle – Area of the sector AOBA

$=1386-462=924c{m}^2$

|Area of major sector AOBA and its corresponding major sector ABOA|

$=|924-462|=462c{m}^2$

Hence, the required difference of two sectors is $462c{m}^2$.