BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

distance of the chord

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 298
Joined: Tue Feb 16, 2010 1:09 am
Thanked: 2 times
Followed by:1 members

distance of the chord

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
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
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,
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.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
Legendary Member
Posts: 516
Joined: Mon Nov 02, 2009 6:42 am
Location: Mumbai
Thanked: 14 times
Followed by:1 members
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
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
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.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Master | Next Rank: 500 Posts
Posts: 109
Joined: Mon Jan 10, 2011 3:05 am
Thanked: 8 times
Followed by:7 members
GMAT Score:760

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

How did you get OM = OA/2?

Regards,
Nikhil
Top
Master | Next Rank: 500 Posts
Posts: 109
Joined: Mon Jan 10, 2011 3:05 am
Thanked: 8 times
Followed by:7 members
GMAT Score:760

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

How did you get OM = OA/2?

Regards,
Nikhil
Top
Master | Next Rank: 500 Posts
Posts: 109
Joined: Mon Jan 10, 2011 3:05 am
Thanked: 8 times
Followed by:7 members
GMAT Score:760

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

How did you get OM = OA/2?

Regards,
Nikhil
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
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:√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]
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Post new topic Post Reply
• Page 1 of 1