If an isosceles $\triangle ABC$ in which $AB=AC=6$ cm is inscribed in a circle of radius $9$cm. Find area of the triangle.

For the triangle with sides AB and AC equal to 6 cm inscribed in a circle of radius 9 cm, we use the extended sine rule which states that the side opposite an angle divided by the sine of that angle equals twice the circumradius. Therefore, 6 divided by the sine of the base angle B equals 2 times 9, meaning the sine of angle B is 1/3. Using the Pythagorean identity, the cosine of angle B is 2 times the square root of 2 divided by 3. Applying the sine double angle formula gives the sine of angle B as 4 times the square root of 2 divided by 9. The area of triangle ABC is then 0.5 times 6 times 6 multiplied by 4 times the square root of 2 divided by 9, resulting in 8 times the square root of 2 square centimeters.

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