# given a circle of radius 9 cm , and the length of the chord AB of a circle is 9 root 3 cm , find the area of the sector formed by the arc AB

we form our diagram from given information , As :

Here OA  =  OB  =   9  cm ( Given radius )                         ----------- ( 1 )
And
AB  =  9$\sqrt{3}$

And we draw a perpendicular OM to AB , SO we know when we draw a perpendicular from centre to chord that bisect the chord , So

AM  =  BM  =                                                       ---------------- ( 2 )

SO, In $∆$ OAM and $∆$ OBM

OA  =  OB                                                      ( From equation 1 )

AM  =  BM                                                   ( From equation 2 )
And
MO  =  MO                                                   ( Common side )
Hence
$∆$ OAM  $\cong$ $∆$ OBM                       (  by SSS rule )
So,
$\angle$ AOM  = $\angle$ BOM                             ( by CPCT )

And we know

Sin

In $∆$ OAM , we know

So,
$\angle$ AOM  = $\angle$ BOM   = 60$°$
And
$\angle$ AOB  = $\angle$ AOM  +  $\angle$ BOM   = 60$°$ + 60$°$  =  120$°$
So,
Area of sector OAB formed by AB , have Area

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