GMAT prep-geometry

The max B+h+1 because the radius is given is 1. since one vertex is hte centre of the circle and the other two are on the circle, we get two line segments which are equal to the raduis of the circle and these two segments correspond to the base and the height of the triangle respectively.
hope this helps.

for longer but more linear explanation use calculus. first note that any triangle described above can be broken into two "easier" triangles with a hyp of 1, hight of h and a base of (1-h^2)^(0.5).

so the area of the trinagle would be 2 x (1/2)(h)(1-h^2)^(.5). ---- (1)
this is tricky to differenciate, but just take a square (as what ever maximized the area will aslo maximize area squared)
you get h^2 + h^4 = A^2
differenciate and set to zero
2h+4h^3 = 0 follows that h = (1/2) and plug into (1) and you get 1/2

I still do not get it..... can anyone post a image with the circle and its triangle inscribed... Thanks!!

First..In this question the triangle is not inscribed in the Circle. "Inscribed" in Planar geometry means the end points of the triangle must touch the circumference of the circle.
In this case:
Imagine a circle with Centre O. There are 2 points on the circle A and B. There you have your triangle, OAB. since OA and OB are the radii of the circle both are equal to 1.
Now we have to find out the maximum possible area of this Triangle.

Area = 1/2 * Base * Height.

If you take OA as the base, we have to find out what is the maximum height possible for this triangle with B on the Circle.
By simple imagination it can be found out that the max hieght is the radius itself, in any other case, the hieght will be less than the radius.

Hence Area = 1/2 * OA * OB.
or 1/2 * 1 * 1
or 1/2
Option B.

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