[spoiler]Ans:D - 7/5[/spoiler]




Angle ABY = 360 - Angle (YXA+ XAB+XYB) --- (1)

Since both tringle XAB and triangle YXB are isoceles.

We have:
Angle BXA=Angle BAX
Angle BXY =Angle BYX

Now, BXY + BXA = Angle YXA = 105 (given) = BYX+BAX

Substituting this in (1)..we get Angle ABY = 360-(105+105) = 150

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Let the radius of the Big circle be .2R
Area of the BIG circle = � (2R)² = 4�R²
Area of BIG semicircle = 2�R²
Are of the Sector AYB = (150/180) * 2�R² = (5/3) �R²
Area of sector YCB = (30/180)* 2�R² = (1/3) �R²
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Raius of the semicircles inside the BIG circle= R
Area of the semicircle = (�R²)/2
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Area of shaded portion above YB = (5/3) �R² - (�R²)/2
Are of shaded region below YB =(1/3) �R² + (�R²)/2


Ratio = ((5/3)-(1/2))/((1/3)+(1/2)) =[spoiler] 7/5.[/spoiler]

Thanks,
Jaxis.

i think angle a B can be very easily calculated, compared to cumbersome tchniques above, using circle formulae of angle is twice at center than at any other point, since B is minor angle it will be : 360-(105*2)