Question

please answer immediately

Answer 1

hey dear

here is your answer

Solution

Area of the segment AyB

= Area of sector OAYB - Area of OAB

next

area of sector OAYB = 120/360 *22/7 21*21 cmsq

we simplify this

then it come = 462 cmsq

OM perpendicular to Ab

OA = OB

so

AMO congruent BMO. ( by RHS criteria)

so
M is the mid point of AB

AOM = BOM

1/2 *120 = 60 degree


Let

OM = x cm

OM / OA = cos 60 degree

x/21 = 1/2. ( cos 60 = 1/2)

x = 21/2

OM = 21/2cm

AM / OA = sin60 = root 3/2

AM = 21 root 3/ 2 cm

AB = 2 Am = 2* 21 root 3/2 cm

= 21/ root 3 cm

area of triangle OAB

= 1/2 Ab * Om

= 1/2 *21 root 3 *21/2 cmsq

441/4 root 3 cmsq


area of segment AYB = ( 462 -441/4 root 3 ) cmsq

from 1 ,2 and 3


21/4 ( 88-21/ root 3) cmsq


this is the answer


hope it helps

thank you