Question

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 1

CONCEPT:

i) In an Isosceles triangle, the altitude, median and angle bisector drawn from the vertical angle to the base (non equal side), as well the perpendicular bisector to the side are one and the same. So, the center of the circumcircle lies on this line.

ii) Area of a triangle = (1/2)*(Product of the two sides and the sine of the angle included between them)

Solution:

i) Let the isosceles triangle ABC, with AB = AC = 6 cm, be inscribed in the circle C(O,9) [Center O and radius = 9 cm]. Hence from the concepts described above, OA is the bisector of angle BAC.

ii) Applying cosine law to the triangle OAB,

cos(<BAO) = (AB² + OA² - OB²)/{2*(AB)*(OA)}

= (6² + 9² - 9²)/(2*6*9) = 1/3

iii) So, sin(<BAC) = sin{2*(<BAO)} = 2*sin(<BAO)*cos(<BAO)
[Applying sin(2x) 2sin(x)cos(x)]

= 2*√{1 - (1/3)²}*(1/3) = (4√2)/9

iv) So area of triangle ABC = (1/2)(AB)(AC)*sin(<BAC)

= (1/2)(6)(6)*{(4√2)/9}

= 8√2 cm² = 11.312 cm²