Hard-- Area of Triangle

So, we know one point is at the center. We may think of the other two points sliding freely around the circle, but for simplicity's sake, let's "nail" one of them down to the 0 degree point on a standard unit circle. The last point's position can be thought of relative to this point, so it does not matter where we placed it.

Let us also constrain our thinking to 0 through 180 degrees (0 though pi radians), since moving beyond that will merely create a mirror image of a triangle previously obtained.

So, what's the formula for any triangle? 1/2*b*h. We know the base: the line from the center to the 0 degree point; in other words, the radius of the circle, 1. So, our job becomes maximizing h. Well, what is h? h is the y-coordinate of the final point that we're swinging around, the height of the triangle. It becomes fairly obvious that this height is achieved at 90 degrees, where the y-coordinate is 1.

So, 1/2*1*1 = 1/2. B.

I agree with Feep's answer and logic; I looked at this on a somewhat simpler level though:

We know that one vertex is at the center and that the two others are on the circumference. Maximum base & height measurements would be taken if the two radii formed a right angle (One of the radii will always be the base, therefore the maximum height will be when the other forms a right angle with it).

Therefore, since the radius is one, [spoiler](1*1)/2 = 0.5[/spoiler]

Please refer to the diagram

I think it will clear the doubt


Since for the area to be max, the triangle has to be right-angled triangle

here base and altitude are both 1( equal to radius)


So area is 1/2*Base * altitude =1/2*1*1 = 1/2

You can extrapolate the triangle into an square with diagonal 2(twice the one side of triangle), if so the area of such square be:



Diagonal = 2 so each side = 2^1/2 and so area of square will be 2.

There are four such triangle so area of each triangle will be 1/2

You can extrapolate the triangle into an square with diagonal 2(twice the one side of triangle), if so the area of such square be:

looking at this at first I thought, why use a complex method to solve a problem which otherwise can be solved by normal logic.

But then a moment later I thought that a problem can be solved in a variety of ways,what's important is that method clicks during the time you are being tested.

Bottom line is to get the answers correct,whichever method you use.

I think everybody will agree with me on this