#### If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Answer $8\sqrt{2}\, cm^{2}$
Solution
According to question

In $\bigtriangleup ABD$ and $\bigtriangleup ACO$
AB = AC        [Given]
AO = AO        [Common side]
$\therefore \, \bigtriangleup ABO\cong \bigtriangleup ACO$ [By SSS congruence Criterion]
$\angle Q_{1}= \angle Q_{2}$ [CPCT]
In $\bigtriangleup ABD$ and $\bigtriangleup ACD$
AB = AC        [given]
$\angle Q_{1}= \angle Q_{2}$
$\therefore \bigtriangleup ABD\cong \bigtriangleup ACD$    [By SAS congruence Criterion]
$\angle ADB= \angle ADC$  …..(i)   [CPCT]
$\angle ADB= \angle ADC= 180^{\circ}$ …..(ii)
From (i) and (ii)
$\angle ADB= 90^{\circ}$
OA is a perpendicular which bisects chord BC
Let AD = x, then OD = 9 – x     $\left ( \because OA= 9\, cm \right )$
Use Pythagoras in $\bigtriangleup$ADC
$\left ( AC \right )^{2}= \left ( AD \right )^{2}+\left ( DC \right )^{2}$
$\left ( 6 \right )^{2}= x^{2}+\left ( DC \right )^{2}$
$\left ( AC \right )^{2}= 36-x^{2}$ …..(iii)
In $\bigtriangleup$ODC using Pythagoras theorem
$\left ( OC \right )^{2}= \left ( OD \right )^{2}+\left ( DC \right )^{2}$
$\left ( DC \right )^{2}= 81-\left ( 9-x \right )^{2}$
…..(iv)
From (iii) and (iv)
$36-x^{2}= 81-\left ( 9-x \right )^{2}$
$36-x^{2}- 81+\left ( 81+x^{2} -18x\right )= 0$
$\left [ \because \left ( a-b \right )^{2} +a^{2}+b^{2}-2ab\right ]$
$36-x^{2}-81+81+x^{2}-18x= 0$
$18x= 36$
$x= 2$
i.e., AD = 2 cm, OD = 9 – 2 = 7 cm
Put value of x in (iii)
$\left ( DC \right )^{2}= 36-4$
$\left ( DC \right )^{2}= 32$
$DC= 4\sqrt{2}\, cm$
$BC= BD+DC$
$BC= 2DC$    $\left [ \because BD= DC \right ]$
$BC= 8\sqrt{2}\, cm$
Areao of $\bigtriangleup ABC= \frac{1}{2}\times base\times height$
$= \frac{1}{2}\times \times 8\sqrt{2}\times 2= 8\sqrt{2}\, cm$