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

Redeem

Circle Question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Tue Sep 20, 2011 7:58 am
Thanked: 1 times
Followed by:1 members

Circle Question

by sparkle6 » Mon Sep 26, 2011 8:07 am
In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? (Please refer to image for answer choices and diagram)

Image


[spoiler]Answer: D, but I got B. Can someone explain please? [/spoiler][/img]
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 158
Joined: Sat Sep 03, 2011 10:31 am
Thanked: 29 times
Followed by:2 members

by gmatclubmember » Mon Sep 26, 2011 8:24 am
sparkle6 wrote:In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? (Please refer to image for answer choices and diagram)

Image


[spoiler]Answer: D, but I got B. Can someone explain please? [/spoiler][/img]

The angle subtended at B is 60. so the angle subtended at center will be 120. so the length of curve part of the figure will be 2PI*5/3=10PI/3.
Coming to the BC and CE. They both are equal (look at the various angles 'x').
Angle ACB is 90 and hence in this right angle triangle BC will be 5sqrt3. (cos60=BC/10).
BC+BE = 10sqrt3.
Perimiter will be
10sqrt3+10PI/3

D.

Is this the OA.

Cheers
Ami/-
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 » Mon Sep 26, 2011 6:37 pm
sparkle6 wrote:In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? (Please refer to image for answer choices and diagram)

Image


[spoiler]Answer: D, but I got B. Can someone explain please? [/spoiler][/img]
It can be seen in the figure that a parallel line is cut by a transversal, so x = 30º
Central angle = 2 * Inscribed angle
So, central angle = 2 * 60 = 120º
Sides BC and BE are equal and also angle B is equal to 60º, so the other two angles are 60º each.
Using the 30-60-90 rule, BC = 5√3
Perimeter of CAE = (120 / 360) * 2 * pi * 5 = 10pi/3

Therefore, required perimeter = 10pi/3 + 5√3 + 5√3 = 10pi/3 + 10√3

The correct answer is D.
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
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sun Jul 15, 2012 9:07 am

by zorawar.mann » Tue Sep 18, 2012 3:47 am
How do you know that length BE or CB is ?? If the triangle is formed then it intersects the diametre at an unknown point and not exactly in the centre at O. Which means that we do not know how much is the exact length where the 30-60-90 rule can be used. If the base of the triangle intersects at P, then the height is less then 10. This whole question is incorrect.

Even if we consider half a triangle through the centre where the base intersects the centre, even then we do not know the ratio by which we can add the remaining part to complete the hypotenuse of the triangle to get the exact figure.

Please explain the answer with more valid reason. Height cannot be considered 5 or 10 for the triangle. How did we calculate this.


Image
Top
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sun Jul 15, 2012 9:07 am

by zorawar.mann » Tue Sep 18, 2012 4:22 am
Ans: 10pie/3 + 10rt3

Explanation:

1) Look at the figure, Inside the circle, OC=OB=5 (Radius). Thus angle OCB = 30 (Isoceles Triangle). Draw a perpendicular, and using the rule 30-60-90, thus one half is 2.5rt3 and the complete length is 2 x 2.5rt3 = 5rt3.
Therefore the length of the straight lines together is 10rt3.

2) Angle COE is 120. Thus the length of the arc is 120/360 x 2pie5 = 10pie/3

Add 1 and 2 = 10pie/3 + 10rt3

Thanks.

Reply if there is any discrepancy with the explanation.



Image
Top
Post new topic Post Reply
• Page 1 of 1