In Figure 3, two concentric circles with centre 0, have radii 21cm and 42 em. If ∠ AOB = 60°, find the area of the shaded region. - Mathematics

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In Figure , two concentric circles with centre O, have radii 21cm and 42 cm. If ∠ AOB = 60°, find the area of the shaded region. [use π=22/7]

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Solution 1

Solution:
Given :
Radii of inner circle = 21 cm = r
Radii of outer circle = 42 cm = R
∠AOB = θ = 60°

Also,
Area of ring = π(R2-r2)

Area of a sector `= theta/360 pir^2`

The area of shaded region = Area of ring – Area of ABCD
= Area of ring – Area of sector of Outer Circle - Area of sector of Inner Circle

`=pi(R^2-r^2)-(pi(R^2-r^2))/1 xx theta/360`

`=pi(R^2-r^2)[1-theta/360]`

`=22/7(42^2-21^2)(1-60/360)`

`=3465 cm^2`

Solution 2

Solution:
Given :
Radii of inner circle = 21 cm = r
Radii of outer circle = 42 cm = R
∠AOB = θ = 60°

Also,
Area of ring = π(R2-r2)

Area of a sector `= theta/360 pir^2`

The area of shaded region = Area of ring – Area of ABCD
= Area of ring – Area of sector of Outer Circle - Area of sector of Inner Circle

`=pi(R^2-r^2)-(pi(R^2-r^2))/1 xx theta/360`

`=pi(R^2-r^2)[1-theta/360]`

`=22/7(42^2-21^2)(1-60/360)`

`=3465 cm^2`

Solution 3

Given: Radius of the inner circle with radius OC, r = 21 cm

Radius of the inner circle with radius OA, R = 42 cm

∠AOB = 60°

Area of the circular ring

`= piR^2 - pir^2`

`=pi[R^2 - r^2]`

`=pi[42^2 - 21^2] cm^2`

Area of ACDB = area of sector AOB − area of COD

`= 60/360 xx pi xx R^2 - 60/300 xx pi xx r^2`

`= 60/360 xx pi[R^2 - r^2]`

`= 60/360 xx pi[42^2 - 21^2]`

Area of shaded region = area of circular ring − area of ACDB

`= pi[42^2 - 21^2] - 60/360 pi [42^2 - 21^2]`

`= pi[42^2 - 21^2][1 - 60/360]`

`= 22/7 (42 - 21) (42 + 21) xx 300/360`

`=3465 cm^2`

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2013-2014 (March) All India Set 2

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