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

Answer :

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  =  $\frac{AB}{2} = \frac{9\sqrt{3}}{2} = 4.5\sqrt{3}$                                                     ---------------- ( 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 $\theta = \frac{perpendicular}{Hypotenuse}$

In $∆$ OAM , we know

$$\begin{gathered} \Rightarrow Sin \angle AOM = \frac{AM}{OA} \\ \Rightarrow Sin \angle AOM = \frac{4.5\sqrt{3}}{9} \\ \Rightarrow Sin \angle AOM = \frac{\sqrt{3}}{2} \\ \Rightarrow Sin \angle AOM = Sin 60° ( As we know Sin 60°= \frac{\sqrt{3}}{2} ) \\ \Rightarrow \angle AOM =60° \end{gathered}$$
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

$= \frac{\angle AOB}{360°} \mathrm{\pi } {\mathrm{r}}^2 = \frac{120°}{360°} \times \frac{22}{7}\times 9 \times 9 = \frac{1}{3}\times \frac{22}{7}\times 9 \times 9 = \frac{22}{7}\times 9 \times 3 =\frac{594}{7} = \mathbf{84}\mathbf{.}\mathbf{857}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$