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ABC is an isosceles triangle with AB=BC. The circumcircle of ABC has radius 8. Given that AC is a diameter of the circumcircle, what is the area of triangle ABC?

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|dw:1375617084417:dw|Note that angle B will be a right angle (which can be proved using the 'Angle at the center theorem'). We also note that this triangle is isosceles and and it's base is the diameter of the circle \(\bf AC=16\). And the height of the triangle is just the radius of the circle which is given to us as 8. Hence of the triangle is:\[\bf Area \ of \ \triangle ABC=\frac{ 16 \times 8 }{ 2 }=64 \ units^2\]

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