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# If a circle is circumscribed around a regular pentagon, as shown above

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If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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Updated on: 02 Nov 2014, 07:22
1
10
00:00

Difficulty:

65% (hard)

Question Stats:

55% (02:10) correct 45% (02:34) wrong based on 157 sessions

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File comment: Economist.png

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If a circle is circumscribed around a regular pentagon, as shown above, what is the value of x?

A) 72
B) 108
C) 135
D) 140
E) 144

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Kinjal

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Originally posted by kinjiGC on 02 Nov 2014, 07:18.
Last edited by Bunuel on 02 Nov 2014, 07:22, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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02 Nov 2014, 07:21
2
Each side subtends an angle of 360/5 = 72 at the center.
Then the angle of EBD = 72/2 = 36
lets assume the point where X is specified is named as F
So EBDF becomes a cyclic quadrilateral.

Hence Angle EBD and Angle EFD should be supplementary and hence X = 180 - 36 = 144.
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Kinjal

My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961

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If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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Updated on: 03 Nov 2014, 02:34
kinjiGC wrote:
Each side subtends an angle of 360/5 = 72 at the center.
Then the angle of EBD = 72/2 = 36
lets assume the point where X is specified is named as F
So EBDF becomes a cyclic quadrilateral.

Hence Angle EBD and Angle EFD should be supplementary and hence X = 180 - 36 = 144.

EDIT: this is the property of any cyclic quad (and not a kite), just looked it up. thanks!

Originally posted by usre123 on 03 Nov 2014, 00:26.
Last edited by usre123 on 03 Nov 2014, 02:34, edited 1 time in total.
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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03 Nov 2014, 00:30
kinjiGC wrote:
Each side subtends an angle of 360/5 = 72 at the center.
Then the angle of EBD = 72/2 = 36
lets assume the point where X is specified is named as F
So EBDF becomes a cyclic quadrilateral.

Hence Angle EBD and Angle EFD should be supplementary and hence X = 180 - 36 = 144.

Is EF = FD mentioned in the problem (or we have to assume from the diagram)?
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Manager
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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03 Nov 2014, 00:41
^ same question i was about to ask as well
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If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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03 Nov 2014, 04:57
PareshGmat wrote:
kinjiGC wrote:
Each side subtends an angle of 360/5 = 72 at the center.
Then the angle of EBD = 72/2 = 36
lets assume the point where X is specified is named as F
So EBDF becomes a cyclic quadrilateral.

Hence Angle EBD and Angle EFD should be supplementary and hence X = 180 - 36 = 144.

Is EF = FD mentioned in the problem (or we have to assume from the diagram)?

usre123 wrote:
^ same question i was about to ask as well

EF = FD is not required as chord will subtend the same angle in an arc whatever may be the point on the circle.

Circle properties

For example Angle EBD = Angle EAD = Angle ECD etc. Similarly below the chord as well.
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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20 Dec 2014, 03:05
kinjiGC wrote:
Attachment:
Economist.png
If a circle is circumscribed around a regular pentagon, as shown above, what is the value of x?

A) 72
B) 108
C) 135
D) 140
E) 144

360 * 4/5 = 72 * 4

[72*4] / 2 = 36 * 4 = 144

Ans E.
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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20 Dec 2014, 08:14
kinjiGC wrote:
Attachment:
Economist.png
If a circle is circumscribed around a regular pentagon, as shown above, what is the value of x?

A) 72
B) 108
C) 135
D) 140
E) 144

360 * 4/5 = 72 * 4

[72*4] / 2 = 36 * 4 = 144

Ans E.
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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12 Jan 2015, 01:14
1
1
Each arc is equal to 360/5=72 since it is a regular pentagon. Now the major arch ED is equal to 360-72=288 so angle substended by it at the circumferance 288/2=144.
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If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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09 Jul 2016, 14:31
look at x as an interior angle of a regular decagon
(10-2)(180°)=1440° total interior degrees
1440°/10=144° interior angle degrees
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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10 Jul 2016, 01:38
Hi guys,

I'm new here and this is my first post on the forum. Really struggling to get my head around this question and understand how it works. Would really appreciate if someone could show a clear step by step process of how they got the answer.

Thanks
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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30 Jul 2016, 23:32
1
Angle subtend by the arc ED at the center = 360/5 = 72.

As per the property of the circle: Angle subtended by an arc at any point on the circumference of a circle is 1/2 times the angle subtended by the same arc on the center.

Hence, angle EBD = (1/2)*72 = 36.
Now, Points E,B,D and the point where angle x is forming, makes a cyclic quadrilateral.
So, angle EBD + x = 180
x = 180 - 36 = 144.
Ans. E
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Re: If a circle is circumscribed around a regular pentagon, as shown above  [#permalink]

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31 Jul 2016, 16:34
Thanks shashank1tripathi really helpful reply! Got it now!
Re: If a circle is circumscribed around a regular pentagon, as shown above &nbs [#permalink] 31 Jul 2016, 16:34
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# If a circle is circumscribed around a regular pentagon, as shown above

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