Find area of isosceles triangleabc in …

Let O be the centre of the circle and let P be the mid-point of BC. Then ,{tex}O P \perp B C{/tex}
Since ∆ ABC is isosceles and P is the mid-point of BC. Therefore, {tex}A P \perp B C{/tex} 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
{tex}\Rightarrow{/tex}    36 = y² + x² ..........(i)
and, 81 = (9 - x){tex}^2{/tex} + y{tex}^2{/tex} ...(ii)
{tex}\Rightarrow{/tex} 81 - 36 = [(9 - x)² + y²  ]- (y² + x² )[Subtracting (i) from (ii)]
{tex}\Rightarrow{/tex} 45 = 81 - 18x
{tex}\Rightarrow{/tex} x = 2 cm, therefore AP = x  = 2 cm .....(iii)
Putting x = 2 in (i), we get
36 = y² + 4 {tex}\Rightarrow{/tex} y² = 32 {tex}\Rightarrow{/tex} y = {tex}4 \sqrt { 2 }{/tex}  cm
{tex}{/tex}Therefore,BC = 2BP = 2y = {tex}8 \sqrt { 2 }{/tex} .cm .....(iv)
Therefore,

Area of ∆ ABC

= {tex}\frac 1 2{/tex}BC × AP

= {tex}\frac 1 2{/tex}× 2y × x  [ from (iii) & (iv) ]

= xy

= 2 × 4√2 = 8√2 cm^2.