Question

If an isosceles triangle ABC , in which AB = AC = 16 cm , is inscribed in a circle of

radius 9 cm, Find the area of the triangle.​

Answer 1

Answer:

Let the center of the circle be O. Join AO and let the intersection of BC and AO be P.

Triangle ABC is isosceles and so is triangle BOC. OB = OC = 9 cm, being radii. For finding the area of Triangle ABC we need to know the length BC and the altitude from A to BC.

Let AP = x so that OP = 9-x.

In right angled triangle ABP, AB^2 - AP^2 = BP^2 or 6^2 - x^2 = BP^2

In right angled triangle OBP, OB^2 - OP^2 = BP^2 or 9^2 - (9-x)^2 = BP^2

Equating the value of BP^2 from both the triangles we get,

BP^2 = 6^2 - x^2 =9^2 - (9-x)^2, or

36 - x^2 = 81 - (81 - 18x + x^2) = 81 - 81 + 18x - x^2 , or

36 = 18 x or x = 2 cm i.e. AP, then BP^2 = 36 - 4 = 32, or BP = 4*2^0.5, and BC = 2*4*2^0.5 = 8*2^0.5.

Thus, the area of triangle ABC = 8*2^0.5*2/2 = 8*2^0.5 sq cm,