Let's start with the left circle. If x = 45deg, this is an isosceles triangle with the angle at the centre point = 90deg.

Given that the radius = 1 and it is the hypotenuse, the base AC and height OC must be 1 / sqrt 2. The area of this AOB triangle = 1/2 * 1 / sqrt2 * 1/sqrt 2 * 2 = 1/2 = 0.5

For the right circle, if x = 60deg, this is an equilateral triangle with angle at the centre point = 60deg.

Again, given that the radius is 1, and it is the hypotenuse, the base AC = 1 / 2 and height OC = sqrt 3 / 2. The area of this AOB triangle = 1/2 * 1/2 * sqrt 3 / 2 * 2 = sqrt 3 / 4 = ~ 1.7 / 4 = ~ 0.4.

Therefore the answer is

BI solve the above using these rules:

For 45-45-90 triangle (isosceles triangle)

> 45deg : 45 deg : 90deg has a ratio of 1 : 1 : sqrt 2

For 30-60-90 triangle (equilateral triangle)

> 30deg : 60 deg : 90deg has a ratio of 1 : sqrt3 : 2

This takes away the need to calculate sin and cosin etc. You dont have that luxury in GMAT

Attachments

file.jpg [ 10.92 KiB | Viewed 672 times ]

_________________

Jimmy Low, Frankfurt, Germany

Blog: http://mytrainmaster.wordpress.com

GMAT Malaysia: http://gmatmalaysia.blogspot.com