scarish wrote:

Area of traingle = 1/2 (bh)

So we have to maximise the base and height. The base and height of inscribed triangle with one vertex at the center are maximum when both are equal to the radius = 1.

Therefore, the maximum area would be 1/2 (1*1) = 1/2

Although you got the correct answer, your explanation is incorrect. Since the triangle has a vertex at the center and the other two are on the circle with radius =1, at most all three sides can equal 1. BUT the area of a triangle is 1/2 *b*h. Thus you need to find h when the three sides are equal to 1. You will quickly realize your answer is no longer 1/2!

The correct way to reach 1/2 as an answer is to understand that an isosceles triangle has the max area when it is a right angled triangle. Now you know that the two sides are 1 and the angle between them at the center of the circle is 90. Thus the base or hypotenuse of the triangle will be 1/sqr2. This gives the area of 1/2 * b * h = 1/2 * 1/sqr2 * sqr2 = 1/2

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-DK

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