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distance of the chord

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by Deepthi Subbu » Thu Jan 13, 2011 4:43 am
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

OA 3
Image
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by Anurag@Gurome » Thu Jan 13, 2011 5:18 am
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,
Image

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.
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by ankurmit » Thu Jan 13, 2011 10:01 pm
Nice axplanation Anurag.

Can you elaborate how you calculated the angle subtended by arc at the centre.
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by Anurag@Gurome » Thu Jan 13, 2011 11:31 pm
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.
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by nikhilsrl » Fri Jan 14, 2011 12:35 am
Hello Anurag,

How did you get OM = OA/2?

Regards,
Nikhil
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by nikhilsrl » Fri Jan 14, 2011 12:36 am
Hello Anurag,

How did you get OM = OA/2?

Regards,
Nikhil
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by nikhilsrl » Fri Jan 14, 2011 12:37 am
Hello Anurag,

How did you get OM = OA/2?

Regards,
Nikhil
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by Anurag@Gurome » Fri Jan 14, 2011 12:44 am
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]
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