Find area of isosceles triangleabc in …

Let O be the centre of the circle and let P be the mid-point of BC. Then ,$O P \perp B C$
Since ∆ ABC is isosceles and P is the mid-point of BC. Therefore, $A P \perp B C$ as median from the vertex in an isosceles triangle is perpendicular to the base.
Let A P = x and PB = CP = y.
Applying Pythagoras theorem in ∆ APB and ∆ OPB, we have
         AB² = BP² + AP²
and, OB² = OP² + BP²
$\Rightarrow$    36 = y² + x² ..........(i)
and, 81 = (9 - x)$^2$ + y$^2$ ...(ii)
$\Rightarrow$ 81 - 36 = [(9 - x)² + y²  ]- (y² + x² )[Subtracting (i) from (ii)]
$\Rightarrow$ 45 = 81 - 18x
$\Rightarrow$ x = 2 cm, therefore AP = x  = 2 cm .....(iii)
Putting x = 2 in (i), we get
36 = y² + 4 $\Rightarrow$ y² = 32 $\Rightarrow$ y = $4 \sqrt { 2 }$  cm
Therefore,BC = 2BP = 2y = $8 \sqrt { 2 }$ .cm .....(iv)
Therefore,

Area of ∆ ABC

= $\frac 1 2$BC × AP

= $\frac 1 2$× 2y × x  [ from (iii) & (iv) ]

= xy

= 2 × 4√2 = 8√2 cm^2.