Author 
Message 
TAGS:

Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2685
Location: New York City

A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
Updated on: 19 Jul 2013, 01:16
Question Stats:
53% (00:59) correct 47% (01:02) wrong based on 484 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by bmwhype2 on 23 Oct 2007, 14:04.
Last edited by Bunuel on 19 Jul 2013, 01:16, edited 2 times in total.
Renamed the topic, edited the question and added the OA.




VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1251
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: A regular pentagon is inscribed in a circle
[#permalink]
Show Tags
25 Dec 2012, 02:43
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 \(18072=108\) \(108\) 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 [ 6.68 KiB  Viewed 12158 times ]
_________________
Prepositional Phrases ClarifiedElimination of BEING Absolute Phrases Clarified Rules For Posting www.UnivScholarships.com




Director
Joined: 03 May 2007
Posts: 829
Schools: University of Chicago, Wharton School

Re: Challenge  Pentagon inscribed in a circle
[#permalink]
Show Tags
23 Oct 2007, 14:27
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
Joined: 20 Aug 2007
Posts: 850
Location: Chicago
Schools: Chicago Booth 2011

Re: Challenge  Pentagon inscribed in a circle
[#permalink]
Show Tags
23 Oct 2007, 18:29
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
Joined: 03 May 2007
Posts: 829
Schools: University of Chicago, Wharton School

Re: Challenge  Pentagon inscribed in a circle
[#permalink]
Show Tags
23 Oct 2007, 19:05
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
Joined: 07 Jul 2004
Posts: 4938
Location: Singapore

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
Joined: 21 Jan 2007
Posts: 2685
Location: New York City

Re: Challenge  Pentagon inscribed in a circle
[#permalink]
Show Tags
23 Oct 2007, 22:40
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
Untitled1.jpg [ 29.1 KiB  Viewed 12413 times ]



Manager
Joined: 05 Nov 2012
Posts: 65
Concentration: International Business, Operations
GPA: 3.65

Re: A regular pentagon is inscribed in a circle
[#permalink]
Show Tags
25 Dec 2012, 09:29
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 \(18072=108\) \(108\) 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
Joined: 23 Apr 2013
Posts: 5

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
07 Jul 2013, 06:00
Another way of cracking this sum: the sum of angles of a polygon is 180 (n2). n stands for the no. of sides of the polygon. So, here we have n=5. Therefore, sum of all angles= 180 (52)= 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
Joined: 09 Jun 2012
Posts: 30

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
18 Jul 2013, 00:21
CIyer wrote: Another way of cracking this sum: the sum of angles of a polygon is 180 (n2). n stands for the no. of sides of the polygon. So, here we have n=5. Therefore, sum of all angles= 180 (52)= 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
Joined: 23 Apr 2013
Posts: 5

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
18 Jul 2013, 09:40
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.



Manager
Joined: 02 Sep 2012
Posts: 224
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07252013
GPA: 3.83
WE: Architecture (Computer Hardware)

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
19 Jul 2013, 02:59
I used the formula the sum of interior angles of pentagon 180(n2) 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



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1096
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
19 Jul 2013, 03:47
skamal7 wrote: I used the formula the sum of interior angles of pentagon 180(n2) 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: \(18072=108\) 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 III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



Manager
Joined: 02 Sep 2012
Posts: 224
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07252013
GPA: 3.83
WE: Architecture (Computer Hardware)

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
19 Jul 2013, 04:04
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



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1096
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
19 Jul 2013, 04:16
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 III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



Intern
Joined: 09 Jun 2012
Posts: 30

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
19 Jul 2013, 08:48
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
Joined: 09 Mar 2010
Posts: 32
Concentration: Technology, General Management

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
19 Jul 2013, 08:56
The sum of interior angles of n sides : (n2)*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
Joined: 06 Sep 2013
Posts: 1849
Concentration: Finance

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
Updated on: 12 Jan 2014, 13:12
kalrac wrote: The sum of interior angles of n sides : (n2)*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
Originally posted by jlgdr on 08 Oct 2013, 07:11.
Last edited by jlgdr on 12 Jan 2014, 13:12, edited 1 time in total.



Manager
Joined: 30 May 2013
Posts: 168
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
15 Oct 2013, 19:38
Understood the question wrongly and marked the angle as 72 which is the angle from center.



Intern
Joined: 14 Dec 2011
Posts: 16
Location: India
Concentration: Technology, Nonprofit
GMAT 1: 640 Q48 V29 GMAT 2: 660 Q45 V35
GPA: 3.5
WE: Information Technology (Computer Software)

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
Show Tags
17 Oct 2013, 21:21
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 = (n2)*180 ; n=5 Therefore = (52) * 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 &nbs
[#permalink]
17 Oct 2013, 21:21



Go to page
1 2
Next
[ 22 posts ]



