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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A regular pentagon is inscribed in a circle. If A and B are [#permalink]

Show Tags

23 Oct 2007, 14:04

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

52% (01:59) correct
48% (01:02) wrong based on 418 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

Re: A regular pentagon is inscribed in a circle [#permalink]

Show Tags

25 Dec 2012, 02:43

5

This post received KUDOS

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.

Re: Challenge - Pentagon inscribed in a circle [#permalink]

Show Tags

23 Oct 2007, 18:29

1

This post received KUDOS

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

Re: Challenge - Pentagon inscribed in a circle [#permalink]

Show Tags

23 Oct 2007, 22:40

1

This post received KUDOS

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?

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink]

Show Tags

07 Jul 2013, 06:00

1

This post received KUDOS

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).

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink]

Show Tags

19 Jul 2013, 03:47

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.

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

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

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.

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 \(180-72=108\) \(108\) 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

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 (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.

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.

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(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

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!

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 : (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.

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink]

Show Tags

08 Oct 2013, 07:11

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, 13:12, edited 1 time in total.

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 = (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.

gmatclubot

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
17 Oct 2013, 21:21

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...