Deepthi Subbu wrote:The length of arc AB is 2 pie and the circumference of the circle is 6 pie . What is the minimum possible distance between the length of the chord AB and the centre of the circle?

1. sqr root (3) / 2
2.2
3.3/2
4.sqr root (3 pie) / 2
5.1

Circumference = 6 pie => Radius = 3
Length of arc AB = 2 pie

Therefore, the angle subtended by arc AB at the center = (360)*(2)/6 degrees = 120 degrees

Now refer to the image below,


The minimum possible distance between the length of the chord AB and the centre of the circle, O is the length of the perpendicular drawn from O on AB, i.e. the length of OM.

Now, OM bisects the angle AOB.
Thus, angle AOM = angle BOM = 60 degrees

Thus the right-angled triangle AOM is a 30-60-90 triangle.
Hence, OM = OA/2 = Radius/2 = 3/2

The correct answer is C.

ankurmit wrote:Nice axplanation Anurag.

Can you elaborate how you calculated the angle subtended by arc at the centre.

Circumference subtends an angle of 360 degrees at the center. Now, circumference = 6 pie.

Therefore,

• 6 pie subtends an angle of 360 degrees at the center
  1 pie subtends an angle of (360/6) degrees at the center
  2 pie subtends an angle of (360*2/6) degrees at the center

Hope it helps.

nikhilsrl wrote:Hello Anurag,

How did you get OM = OA/2?

In any 30-60-90 triangle, the ratio of three sides is as follows,

• Smallest : Medium : Hypotenuse = 1:√3:2

Thus, Smallest = (Hypotenuse)/2

And in any triangle the smallest side is the one opposite to the smallest angle. In the right-angled triangle AOM, smallest is angle OAM = 30 degrees. Hence smallest side is the side opposite to angle OAM, i.e. OM. Hence, OM = (Hypotenuse)/2 = OA/2.

Hope it is clear now.

Note: The figure provided with the problem was not to scale. I just made constructions on the same figure. Hence mine was not to scale too.[/list]