Find all School-related info fast with the new School-Specific MBA Forum

It is currently 31 Jul 2014, 19:54

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A regular pentagon is inscribed in a circle. If A and B are

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2770
Location: New York City
Followers: 6

Kudos [?]: 201 [1] , given: 4

GMAT Tests User
A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 23 Oct 2007, 13:04
1
This post received
KUDOS
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

47% (01:42) correct 53% (00:43) wrong based on 150 sessions
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and O is the center of the circle, what is the value of ∠OAB ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Jul 2013, 00:16, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
Director
Director
User avatar
Joined: 03 May 2007
Posts: 899
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 40 [0], given: 6

GMAT Tests User
Re: Challenge - Pentagon inscribed in a circle [#permalink] New post 23 Oct 2007, 13:27
bmwhype2 wrote:
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and 0 is the center of the circle, what is the value of angle OAB?

48
54
72
84
108


Can someone draw this out?



its a regular pentagon so divide the globe by 5.
= 360/5
= 72
Director
Director
User avatar
Joined: 20 Aug 2007
Posts: 853
Location: Chicago
Schools: Chicago Booth 2011
Followers: 9

Kudos [?]: 82 [0], given: 1

GMAT Tests User
Re: Challenge - Pentagon inscribed in a circle [#permalink] New post 23 Oct 2007, 17:29
Fistail wrote:
bmwhype2 wrote:
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and 0 is the center of the circle, what is the value of angle OAB?

48
54
72
84
108


Can someone draw this out?



its a regular pentagon so divide the globe by 5.
= 360/5
= 72


72 would be the angle at O.
Angles at A and B are equivalent, so

72 + 2x = 180
2x = 108
x = 54

Angles OAB and OBA will be 54 degrees each.
B
Director
Director
User avatar
Joined: 03 May 2007
Posts: 899
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 40 [0], given: 6

GMAT Tests User
Re: Challenge - Pentagon inscribed in a circle [#permalink] New post 23 Oct 2007, 18:05
sonibubu wrote:
Fistail wrote:
bmwhype2 wrote:
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and 0 is the center of the circle, what is the value of angle OAB?

48
54
72
84
108

Can someone draw this out?


its a regular pentagon so divide the globe by 5.
= 360/5
= 72


72 would be the angle at O.
Angles at A and B are equivalent, so

72 + 2x = 180
2x = 108
x = 54

Angles OAB and OBA will be 54 degrees each.
B


oh, its the edge not the center. ok, lol.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5097
Location: Singapore
Followers: 16

Kudos [?]: 131 [0], given: 0

GMAT Tests User
 [#permalink] New post 23 Oct 2007, 19:25
OA and OB are radius of a circle, so we have an isoceles triangle. Also,360degrees in the centre of circle will be divided up equally into 5 parts, each measuring 360/5 = 72 degrees.

So 72 + 2x = 180 where x = OAB = OBA

x = 54
CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2770
Location: New York City
Followers: 6

Kudos [?]: 201 [0], given: 4

GMAT Tests User
Re: Challenge - Pentagon inscribed in a circle [#permalink] New post 23 Oct 2007, 21:40
bmwhype2 wrote:
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and 0 is the center of the circle, what is the value of angle OAB?

48
54
72
84
108


Can someone draw this out?


OA is B 54.

1/5 of 360 is 72.

180 - 72 = 102

102/2 isos angles = 54
Attachments

Untitled-1.jpg
Untitled-1.jpg [ 29.1 KiB | Viewed 2904 times ]

Expert Post
4 KUDOS received
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1425
Location: India
Concentration: Finance, Marketing
GPA: 3.75
Followers: 126

Kudos [?]: 570 [4] , given: 62

GMAT ToolKit User GMAT Tests User Premium Member
Re: A regular pentagon is inscribed in a circle [#permalink] New post 25 Dec 2012, 01:43
4
This post received
KUDOS
Expert's post
Since it is mentioned that the pentagon is regular, hence its all the sides are equal. In the diagram attached, if you join the vertices of the pentagon with the centre, then you will get 4 isosceles triangle. Each triangle will have a central angle i.e. angle at the centre of the circle and two equal angles, both at the vertex of the pentagon.
So there will be 5 central angles. Name them as x. So as per the property of the circle, the sum of the entire central angle will be 360.
Hence 5x=360 or x=72.
Its now clear that the value of central angle of each triangle will be 72. Therefore the sum of the rest of the angles will be 180-72=108
108 is the sum of two equal angles, hence each angle will be equal to 54.
The value of the angle OAB is 54.
Attachments

GC solution.png
GC solution.png [ 6.68 KiB | Viewed 2657 times ]


_________________

Prepositional Phrases Clarified|Elimination of BEING| Absolute Phrases Clarified
Rules For Posting
www.Univ-Scholarships.com

Manager
Manager
avatar
Joined: 05 Nov 2012
Posts: 72
Concentration: International Business, Operations
Schools: Foster '15 (S)
GPA: 3.65
Followers: 1

Kudos [?]: 41 [0], given: 8

GMAT ToolKit User
Re: A regular pentagon is inscribed in a circle [#permalink] New post 25 Dec 2012, 08:29
Marcab wrote:
Since it is mentioned that the pentagon is regular, hence its all the sides are equal. In the diagram attached, if you join the vertices of the pentagon with the centre, then you will get 4 isosceles triangle. Each triangle will have a central angle i.e. angle at the centre of the circle and two equal angles, both at the vertex of the pentagon.
So there will be 5 central angles. Name them as x. So as per the property of the circle, the sum of the entire central angle will be 360.
Hence 5x=360 or x=72.
Its now clear that the value of central angle of each triangle will be 72. Therefore the sum of the rest of the angles will be 180-72=108
108 is the sum of two equal angles, hence each angle will be equal to 54.
The value of the angle OAB is 54.


Hey marcab gr8 explanation. Thx
_________________

___________________________________________
Consider +1 Kudos if my post helped

Intern
Intern
avatar
Joined: 23 Apr 2013
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 07 Jul 2013, 05:00
Another way of cracking this sum: the sum of angles of a polygon is 180 (n-2). n stands for the no. of sides of the polygon. So, here we have n=5. Therefore, sum of all angles= 180 (5-2)= 540. Since, this is a regular figure each angle= 540/5= 108. However, 108 is the complete angle. Thus, half the angle = 54. (The angle with the center of circle/point of intersection of all diagonals will always be 0.5 times of the complete angle in all regular figures).
Intern
Intern
avatar
Joined: 09 Jun 2012
Posts: 31
Followers: 0

Kudos [?]: 6 [0], given: 13

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 17 Jul 2013, 23:21
CIyer wrote:
Another way of cracking this sum: the sum of angles of a polygon is 180 (n-2). n stands for the no. of sides of the polygon. So, here we have n=5. Therefore, sum of all angles= 180 (5-2)= 540. Since, this is a regular figure each angle= 540/5= 108. However, 108 is the complete angle. Thus, half the angle = 54.


How do we assure that the radius is bisecting the angle? Isn't there a possibility that the radius will cut the angles into 2 unequal parts?

CIyer wrote:
(The angle with the center of circle/point of intersection of all diagonals will always be 0.5 times of the complete angle in all regular figures).


Do you mean for all regular polygons? Because if the figure is a regular hexagon, then the angle with the center will be 0.6 times the complete angle. Kindly clarify.
Intern
Intern
avatar
Joined: 23 Apr 2013
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 18 Jul 2013, 08:40
from a logical standpoint, in all regular figures the diagonals (drawn through the centre) will bisect the angles. Consider a regular hexagon ABCDEF. let O be the point of intersection of all diagonals. Here angle ABC= angle(OBA) + angle (OBC). angle (OBA) = angle (OBC) = 60. Thus, regular hexagon is made of 6 equilateral triangles. Hope it is clear. Perhaps, Bunnel can explain this in a better manner.
Senior Manager
Senior Manager
avatar
Joined: 02 Sep 2012
Posts: 292
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07-25-2013
GPA: 3.83
WE: Architecture (Computer Hardware)
Followers: 3

Kudos [?]: 65 [0], given: 99

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 19 Jul 2013, 01:59
I used the formula the sum of interior angles of pentagon 180(n-2) which gives 540 since pentagon 540/5=108 So this angle is <AOB right? And by using central angle theorem can we deduce that the answer is<OAB =108/2= 54? Is my understanding on this problem corrrect??can you please verify
?
_________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

1 KUDOS received
VP
VP
User avatar
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1125
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 108

Kudos [?]: 1100 [1] , given: 219

GMAT ToolKit User GMAT Tests User
Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 19 Jul 2013, 02:47
1
This post received
KUDOS
skamal7 wrote:
I used the formula the sum of interior angles of pentagon 180(n-2) which gives 540 since pentagon 540/5=108 So this angle is <AOB right? And by using central angle theorem can we deduce that the answer is<OAB =108/2= 54? Is my understanding on this problem corrrect??can you please verify
?


Nope. 540° is the sum of the angles of the pentagon, as you correctly say, so angles A, B would be 108, not angle AOB.

Angle <AOB is not \frac{540}{5} but it's \frac{360}{5} (a circle divided into 5 parts).

According to your calculus angle AOB is 108 and the other two angles are 54: that makes a total of a 216° triangle...

I solved like this:
Angle AOB is 72, and is part of an isosceles triangle, so the other two angles will be equal:
180-72=108 and each angle is \frac{108}{2}=54 (72+54+54=180)
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Senior Manager
Senior Manager
avatar
Joined: 02 Sep 2012
Posts: 292
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07-25-2013
GPA: 3.83
WE: Architecture (Computer Hardware)
Followers: 3

Kudos [?]: 65 [0], given: 99

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 19 Jul 2013, 03:04
So for pentagon what will be the central angle and interior angle?
_________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

1 KUDOS received
VP
VP
User avatar
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1125
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 108

Kudos [?]: 1100 [1] , given: 219

GMAT ToolKit User GMAT Tests User
Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 19 Jul 2013, 03:16
1
This post received
KUDOS
skamal7 wrote:
So for pentagon what will be the central angle and interior angle?


The central angle is 360° (it's a circle).
The interior angles (of a regular pentagon) are \frac{540}{5}=108 each.
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Intern
Intern
avatar
Joined: 09 Jun 2012
Posts: 31
Followers: 0

Kudos [?]: 6 [0], given: 13

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 19 Jul 2013, 07:48
CIyer wrote:
from a logical standpoint, in all regular figures the diagonals (drawn through the centre) will bisect the angles. Consider a regular hexagon ABCDEF. let O be the point of intersection of all diagonals. Here angle ABC= angle(OBA) + angle (OBC). angle (OBA) = angle (OBC) = 60. Thus, regular hexagon is made of 6 equilateral triangles. Hope it is clear. Perhaps, Bunnel can explain this in a better manner.


It makes sense as it matches the solution derived using another approach (isoceles triangle). Just wanted to check if there is any property that I should be aware of. Thanks for the explanation!
Intern
Intern
avatar
Joined: 09 Mar 2010
Posts: 40
Concentration: Technology, General Management
Schools: McCombs '14
Followers: 0

Kudos [?]: 4 [0], given: 17

GMAT ToolKit User
Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 19 Jul 2013, 07:56
The sum of interior angles of n sides : (n-2)*180 , here its 540
Any polygon inscribed in a circle is Regular polygon implies all the angles are equal , hence each angle is 540/5 = 108
OAB = 108/2 = 54.
SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1629
Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
WE: Corporate Finance (Investment Banking)
Followers: 10

Kudos [?]: 138 [0], given: 254

GMAT ToolKit User
Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 08 Oct 2013, 06:11
kalrac wrote:
The sum of interior angles of n sides : (n-2)*180 , here its 540
Any polygon inscribed in a circle is Regular polygon implies all the angles are equal , hence each angle is 540/5 = 108
OAB = 108/2 = 54.


Totally agree with explanation given by kalrac

Just to add there is other way

Find the central angle, that would be 360/5 = 72

Then 2x + 72 = 180

So x is also 54

Hope it helps!
Cheers!
J :)

Last edited by jlgdr on 12 Jan 2014, 12:12, edited 1 time in total.
Manager
Manager
avatar
Joined: 30 May 2013
Posts: 193
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 25 [0], given: 72

GMAT ToolKit User
Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 15 Oct 2013, 18:38
Understood the question wrongly and marked the angle as 72 which is the angle from center.
Intern
Intern
avatar
Joined: 14 Dec 2011
Posts: 17
Location: India
Concentration: Technology, Nonprofit
GMAT 1: 640 Q48 V29
GPA: 3.5
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 11 [0], given: 74

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink] New post 17 Oct 2013, 20:21
bmwhype2 wrote:
A regular pentagon is inscribed in a circle. If A and B are adjacent vertices of the pentagon and O is the center of the circle, what is the value of ∠OAB ?

A. 48 degrees
B. 54 degrees
C. 72 degrees
D. 84 degrees
E. 108 degrees


Other way of looking at the problem:

Sum of all angles in a pentagon = (n-2)*180 ; n=5 Therefore = (5-2) * 180 = 540

Each angle inside pentagon = 540/5=108.

Therefore radius of circle from each vertex of the pentagon to the centre of the circle bisects each angle of the pentagon = ∠OAB = 108/2 = 54 = Answer.
Re: A regular pentagon is inscribed in a circle. If A and B are   [#permalink] 17 Oct 2013, 20:21
    Similar topics Author Replies Last post
Similar
Topics:
A regular hexagon is inscribed in a circle. What is the area smartyman 1 02 Jul 2014, 00:03
1 Experts publish their posts in the topic If a regular hexagon is inscribed in a circle with a radius fozzzy 7 02 Jun 2013, 00:38
A regular pentagon is inscribed in a circle. If A and B are pikachu 0 25 Dec 2012, 08:29
Square A is inscribed in circle X. Square B is inscribed in macjas 6 06 Jun 2012, 02:25
2 Regular hexagon is inscribed in a circle. If the radius of t bigfernhead 6 28 Jan 2009, 15:44
Display posts from previous: Sort by

A regular pentagon is inscribed in a circle. If A and B are

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.