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If an isosceles triangle ABC , in which AB = AC = 16 cm , is inscribed in a circle ofradius 9 cm, Find the area of the triangle. |
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Answer» 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,
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