The diameter of a circle is AB and C is anoth | vedclass

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(A) Let the diameter $AB$ be the base of the triangle. The length of the base $AB$ is constant.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base

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Maximum,if the triangle is isosceles

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(A) Let the diameter $AB$ be the base of the triangle. The length of the base $AB$ is constant.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes h$.Since $AB$ is constant,the area is maximum when the height $h$ is maximum.The height $h$ is the perpendicular distance from point $C$ to the diameter $AB$. This distance is maximum when $C$ is at the highest point of the semicircle,which occurs when $AC = BC$.Therefore,the triangle $ABC$ is an isosceles triangle when its area is maximum.

The diameter of a circle is $AB$ and $C$ is another point on the circle. Then the area of triangle $ABC$ will be:

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