distance of the chord

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Deepthi Subbu: Master | Next Rank: 500 Posts; Posts: 298; Joined: Tue Feb 16, 2010 1:09 amFollowed by: 1 member; Thanked: 2 times

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

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Source: Beat The GMAT — Problem Solving | Thu Jan 13, 2011 4:43 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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,

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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ankurmit: Legendary Member; Posts: 516; Joined: Mon Nov 02, 2009 6:42 amFollowed by: 1 member; Location: Mumbai; Thanked: 14 times; GMAT Score:710

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.

--------
Ankur mittal

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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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nikhilsrl: Master | Next Rank: 500 Posts; Posts: 109; Joined: Mon Jan 10, 2011 3:05 amFollowed by: 7 members; Thanked: 8 times; GMAT Score:760

by nikhilsrl » Fri Jan 14, 2011 12:35 am

Hello Anurag,

How did you get OM = OA/2?

Regards,
Nikhil

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nikhilsrl: Master | Next Rank: 500 Posts; Posts: 109; Joined: Mon Jan 10, 2011 3:05 amFollowed by: 7 members; Thanked: 8 times; GMAT Score:760

by nikhilsrl » Fri Jan 14, 2011 12:36 am

Hello Anurag,

How did you get OM = OA/2?

Regards,
Nikhil

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nikhilsrl: Master | Next Rank: 500 Posts; Posts: 109; Joined: Mon Jan 10, 2011 3:05 amFollowed by: 7 members; Thanked: 8 times; GMAT Score:760

by nikhilsrl » Fri Jan 14, 2011 12:37 am

Hello Anurag,

How did you get OM = OA/2?

Regards,
Nikhil

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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:âˆš3:2

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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