imo B, what is OA please

The maximum possible area of the triangle must be 1/2.



Let O be the centre of the circle and the other vertices of the triangle be A and B respectively. Now we have to find the maximum possible area of the triangle OAB. Clearly OA = OB = 1 (radius of the circle). Hence triangle OAB is isosceles triangle. Now area of OAB is given by 1/2 x OA x OB X Sinθ. where θ is angle AOB.

This will be maximum if Sinθ is maximum and hence θ must be 90. So the area of this isosceles triangle is maximum it it is right angled isosceles triangle.

Thus the maximum area of triangle OAB = 1/2 x 1 x1 = 1/2

The largest triangle can be formed within 1/4 of the circle (if 2 sides need to be radii).
So right away that eliminates choice D

Now since the triangle in the quarter of the circle has two of its edges as radii of the circle,
Use the theorem for right-angled triangles to determine the 3rd side
sqrt(1^2 + 1^2) = sqrt(2)

using this the height of the triangle can be found
x^2+(sqrt(2)/2)^2 = 1^2
x=1/sqrt(2)

Using the height and base(3rd side) area of triangle can be determined,
1/2 * 1/sqrt(2) * sqrt(2) = 1/2

I approached it a little differently, though arriving at the same answer.
Drop a perpendicular from O onto AB meeting AB at P.
Since OAB is an isosceles triangle, Angle OAB = Angle OBA

Area of the triangle AOB
= 1/2 * AB * OP
= 1/2 * (OB * CosB + OA * CosA) * OB * SinB
= 1/2 * 2 (OB * Cos B) * OB * Sine B [Since LA = LB]
= Cos B * Sin B [Since OB = 1 = radius of the circle]

Now knowing that for angles
0 30 45 60 90
Sin 0 0.5 0.707 0.866 1
Cos 1 0.866 0.707 0.5 0

The greatest product of Sin and Cos ratios will be when Angle B = Angle A = 45degrees.
Area of triangle AOB = 0.705 * 0.707 ~ 0.5 = 1/2[/img]

Largest triangle is going to one that has two radii as sides:

area of triangle = 1/2 *base * height
= 1/2 *1 *1 = 1/2

For those who are not very familiar with GMAT quant scope, knowledge of trigonometry (sine, cosine...) is not required by ANY GMAT question.

-Patrick