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What is the greatest possible area of a triangular region [#permalink]
16 Jun 2006, 17:10

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What is the greatest possible area of a triangular region with one vertex at the center of the circle of a radius of 1 and the other two vertices on the circle?

So in order to maximize the area we should attempt to choose the highest possible values for base and height.
In a right triangle, Base=1 and Height=1.

Height will be maximum when it is equal to radius. Imagine a rt. traingle with vertex at center o.

You could go with a higher base, as close to the length of the diameter as possible but only lesser however (imagine a line almost perpendicular cutting the circle into half, but not quite perpendicular). In this case the height will be small.

If we know that the two sides (a and b) are always going to be 1, then we could perhaps use this area formula: S=1/2absinO, where O is the angle formed by the two sides. Then we can see that sin90=1, which gives us the greatest value. In that case S=1/2. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

It`s even simpler than that. With the center of the circle serving as the vertice, then the largest inscribed triangle must be 90-45-45. The two shorter sides, also the length and width, are given as 1.

Imagine the largest rectangle in a circle - it would be a square. So if we fit in a square to the circle of radius 1, then it's sides would be root(2), and therefore area is 2.

Area of square = 2 (maximizes area), and the triangle in question is exactly (1/4) the area of the square, so the area of triangle is

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