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A regular pentagon is inscribed in a circle. If A and B are [#permalink]
23 Oct 2007, 13:04

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Question Stats:

51% (01:52) correct
49% (00:59) wrong based on 281 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

Re: A regular pentagon is inscribed in a circle [#permalink]
25 Dec 2012, 01:43

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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.

Re: A regular pentagon is inscribed in a circle. If A and B are [#permalink]
19 Jul 2013, 02:47

1

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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]
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

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

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

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

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

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

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

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

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

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

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

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

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

gmatclubot

Re: A regular pentagon is inscribed in a circle. If A and B are
[#permalink]
17 Oct 2013, 20:21

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