HI! Guys, Is there any particular rule for the maximum area when triangle is isosceles,

hi Goyalsau, I had ended up with E last time, however OA is C. I am including an expert's reply at https://www.beatthegmat.com/gprep-co-ord ... 48136.html

Triangle has greatest possible area when the triangle is right-angled-triangle. In question one vertex is at center other two on periphery of the circle, to get maximum area triangle is isosceles right angled triangle (as two legs are radius) Area = 1/2*b*h b and h = radius = 1 Area = 1/2 Is OA B ?

Hi there! Let me try to help you... Let O be the center of the given circle (of radius 1), and let choose ANY point in the circumference of the circle to be (say) point A. For the sake of visualizing the next arguments, I suggest you put point A exactly EAST (therefore OA is horizontal, O at the left of A, A at the circumference of the circle). Now we must choose (say) point B, also in the circumference of the circle, to MAXIMIZE the area of triangle OAB... please note that the area of this triangle may be calculated as half the product of (length of) OA times the height relative to this segment... therefore we are looking for maximizing the height, because everything else is fixed!! You do not take more than 10 sec to understand that B must be such that OB and OA are perpendicular, because choosing so you get maximum height (equal to 1), therefore the area we are looking for is half 0A = 1 times OB = 1, therefore 1/2. All this should be enough for you to understand that beat_gmat_09 is correct. Regards, Fabio.