(Q5) Find the area of the segment AYB shown in the adjacent figure. It is given that the radius of the circle is 21 cm and ∠AOB =120°( use π value)

ABYO21 cm21 cm120o

Area of the segment AYB = Area of sector OAYB - Area of △OAB

Now, area of the sector OAYB =
120°
360°
 ×
22
7
 ×21×21 cm2 = 462cm2-----(1)

For finding the area of △OAB , OM ⟂ AB as shown in the figure

Note: OA = OB. Therefore by RHS congruence △AMO similarly △ BMO

So, M is the midpoint of AB and ∠AOM = ∠BOM =
1
2
 ×120° = 60°

Let , OM = xcm

So, from △OMA ,
OM
OA
 = cos 60°
x
21
 =
1
 ( cos 60° =
1
 )
x=
21
So, OM =
21
 cm
Also,
AM
OA
 = sin 60°
AM
21
 =
√
2
 ( sin 60° =
√
2
 )
So, AM =
21√
2
 cm
Therefore AB = 2AM =
2×21√
2
 cm = 21√ cm
So, area of △OAB =
1
2
 ×AB×OM
        =
1
2
 ×21√ ×
21
 cm2
        =
441
 √ cm2----(2)

[From (1) and (2)]

Therefore area of the segment AYB = (462 -
441
 √)cm2
 =
21
 (88 - 21√)cm2

 = .047cm2