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krishan
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GMATIGER, could you please explain in bit more detail

You will maximize the area of a triangle if you make it a isosceles right triangle.

that's why the answer is 1/2. 1*1/2
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B.

In this case for maximum area, triangle needs to be right angle triangle. This can be proved by simply drawing it on paper - we know that lengths of two sides are fixed here i.e. 1. Draw one side as a base and try to fix other side with base for maximum area so that hight of triangle is max. Thus maximum height can be achieved only when other side is perpendicular to it.

Initially I took a long approach and did this with dy/dx formulas from calculus. :)
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I find it easiest to see the answer here if you draw the circle in the x-y plane. Put the centre at (0,0), and put one vertex at (1,0). Let the line connecting these be the base of the triangle- its length is 1. Then the height of the triangle is equal to the y-co-ordinate of the third point, so clearly we get the maximum height (and therefore the maximum area) if the third point is at (0,1) or (0,-1), and more importantly, the maximum height is 1. So the maximum area is 1/2.
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like ian's approach...

...calculus method is pretty quick too and gives the same answer
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I agree with the maximum height concept.

Area of triangle is (1/2) (b X h) we know b=1 and we get max area when h is max. h will be max when the angle is right angle between the center and the third vertex. The height will be less when we have an acute angle or obtuse angle.

Hence it is 1/2
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HG
GMATIGER, could you please explain in bit more detail


Yeah, it is a bit tricky.


Hope you already got it as many explained above.
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its 1/2
nice problem
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IanStewart
I find it easiest to see the answer here if you draw the circle in the x-y plane. Put the centre at (0,0), and put one vertex at (1,0). Let the line connecting these be the base of the triangle- its length is 1. Then the height of the triangle is equal to the y-co-ordinate of the third point, so clearly we get the maximum height (and therefore the maximum area) if the third point is at (0,1) or (0,-1), and more importantly, the maximum height is 1. So the maximum area is 1/2.

I believe the vertex of the triangle must be at (0,0)
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IanStewart
I find it easiest to see the answer here if you draw the circle in the x-y plane. Put the centre at (0,0), and put one vertex at (1,0). Let the line connecting these be the base of the triangle- its length is 1. Then the height of the triangle is equal to the y-co-ordinate of the third point, so clearly we get the maximum height (and therefore the maximum area) if the third point is at (0,1) or (0,-1), and more importantly, the maximum height is 1. So the maximum area is 1/2.

I believe the vertex of the triangle must be at (0,0)

Yes, one vertex must be at the centre of the circle, at (0,0). We then have two other vertices to place. We can put one of these two vertices at (1,0), and then easily see that we get the largest height, and therefore the largest area, if the third vertex is at (0,1) or (0,-1).



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