Here's how this question can be solved from the first principles, by making use of only the most basic properties:

The first step is to draw a diagram to visualize the given information. Here it is:

We've dropped a perpendicular OP on the base AB.

So,

the area of the triangle OAB = \(\frac{1}{2}\)*AB*OP . . . (1)Now, let's assume the angle AOP to be x degrees.

So, OP = OAcosx = 1*cosx = cosx . . . (2)

Similarly, AP = OAsinx = 1*sinx = sinx . . . (3)

Now, we know that the perpendicular drawn from the center of a circle to a chord bisects the chord.

So, perpendicular OP bisects the chord AB.

Therefore, AP = BP = AB/2 . . . (4)

Combining (3) and (4), we get:

AB = 2sinx . . . (5)

Using (2) and (5) in (1), we get:

Area of triangle OAB = \(\frac{1}{2}\)*(2sinx)*(cosx)

So,

Area of triangle OAB = (sinx)*(cosx)We have to

maximize the area, and therefore, we've to

maximize the value of (sinx)*(cosx)When sinx is maximum (for x = 90 degrees), cosx = 0, and so, the product (sinx)*(cosx) = 0

When cosx is maximum (for x = 0 degrees), sinx = 0, and so, the product (sinx)*(cosx) = 0

It's easy to see that the product

(sinx)*(cosx) will be maximum when sinx = cosx. This happens for x = 45 degrees. At this point, sinx = cosx = \(\frac{1}{\sqrt{2}}\)

Therefore,

maximum area = (\(\frac{1}{\sqrt{2}}\))*(\(\frac{1}{\sqrt{2}}\)) = \(\frac{1}{2}\)As you can see, in solving this question, we have used only three basic properties from the concepts of Triangles, Circles and Trigonometry respectively. All students preparing for the GMAT already know these three basic properties:

1. The formula for area of triangle

2. The property that the perpendicular drawn from the center of a circle to a chord bisects the chord

3. The values of sin x and cos x for x = 0 degrees, 45 degrees and 90 degrees

Takeaway from the discussionThe questions on GMAT do not test your knowledge of esoteric properties, but your ability to apply basic concepts from a variety of topicsHope this helped!

Regards

Japinder

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