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# By sharing adjacent sides , exactly n regular pentagons can

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Director
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By sharing adjacent sides , exactly n regular pentagons can [#permalink]  22 Jun 2008, 04:13
By sharing adjacent sides , exactly n regular pentagons can form a circular ring . What is the value of n= ?

A 8
B 9
C 10
D 12
E 16
Director
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Re: Regular Pentagon [#permalink]  22 Jun 2008, 09:19
1
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Walker Atleast u tell me how to solve this.
Senior Manager
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Re: Regular Pentagon [#permalink]  22 Jun 2008, 10:00
Ans is 10.
Each interior angle of pentagon =108, factors=6.6.3
In order to create the complete circular all angle must be equal to 360.
Now see the factors of 360=6.6.10
Now lets do the highest common factor=6.6, that means we need to split the each angle of pentagon into 3 parts and use those sides to form some other pentagon then only circular ring can be formed.
that means 10 pentagon are required to create circular ring.

Let me know the source of the question, its really difficult and quite difficult to solve in the exam. +1 to you mate.
Director
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Re: Regular Pentagon [#permalink]  22 Jun 2008, 19:41
Thanks . The Answer is 10

I got this in my friends GMAT notes. But couldnt understand the last step

Let me mention the technque

Interior angle of a pentagon = 108 deg by using formula ( n-2)x180 / n ; where n is no of sides

Now I didnt follow hence forth

360 - ( 108 x2 ) = 144 ( Why )

Hence interior angle of pentagon in middle is 144.

Now using same formula n-2 (180) /n = 144

we get n = 10.

Hence there are 10 pentagons needed
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Re: Regular Pentagon [#permalink]  23 Jun 2008, 01:46
The circular ring encloses a n-sided regular polygon. Since the sum of the angles of a k-sided polygon is (n-2)180º, each angle of a regular pentagon has a degree measure of 540/5 = 108º. Thus the degree measure of each angle of the n-sided regular polygon is 360 - 2(108) = 144º. But we know that the degree measure of each angle of the n-sided regular polygon is also 180(n-2)/n, so n=10.

It is also true that a 144º angle causes a change in direction of 36º. 10 such changes are needed to yield the entire 360º needed
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Director
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Re: Regular Pentagon [#permalink]  23 Jun 2008, 10:10
[quote="kevincan"]Thus the degree measure of each angle of the n-sided regular polygon is 360 - 2(108) = 144º.

Please explain this step.. Is it a formula.
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Re: Regular Pentagon [#permalink]  23 Jun 2008, 11:54
Perhaps the diagram will help!
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Re: Regular Pentagon [#permalink]  23 Jun 2008, 16:06
Well I am not sure if I could follow your solution. I used simpler method..
108=6.6.3, while 360=6.6.10, so if we create circular ring using pentagons then interior angle of pentagon must be able to divide 360. Thats possible only when we divide the pentagon into 3 parts, ie creating angle =36. In that case we need 10 such pentagons.
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Re: Regular Pentagon [#permalink]  29 Jun 2008, 10:43
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Expert's post
the line between centers of neighbor pentagons change its direction by 1/2*360/5=36. Therefore, we need 360/36=10 pentagons to complete a circle.
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7-t65948.png [ 2.49 KiB | Viewed 999 times ]

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Re: Regular Pentagon [#permalink]  29 Jun 2008, 11:49
1
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I think everyone is using the same formula but doing certain steps in their head and now writing the steps down on paper and that is what is confusing. Start with the basic formula.

If we want to find the value of an interior angle of any polygon the formula is:

\frac{180(n-2)}{n}

When n = 5 for a pentagon, you get

\frac{180(5-2)}{5} = 108

So in the picture below, A & B each = 108. If you think of A + B + C as being inside a complete circle (imaginary circle) you know the total must = 360. So 360 - 108 - 108 = 144.

So Angle C in the picture is 144 degrees. This is an interior angle of a polygon formed by all the pentagons being joined. We now have to answer the question: A polygon with how many sides has interior angles of 144? Now we know the answer, but we don't have n. Before we had n =5 (pentagon) but we didn't have the answer. This is basic alegbra.

\frac{180(N-2)}{n}=144
180(N-2) = 144N
180N - 360 = 144N
36N - 360 = 0
36N = 360
N = 10

So we have created a 10-sided polygon by joining all of those pentagons together.

Attachment:

InteriorAngle.gif [ 4.99 KiB | Viewed 978 times ]

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. Get the best GMAT Prep Resources with GMAT Club Premium Membership SVP Joined: 30 Apr 2008 Posts: 1893 Location: Oklahoma City Schools: Hard Knocks Followers: 30 Kudos [?]: 440 [2] , given: 32 Re: Regular Pentagon [#permalink] 29 Jun 2008, 12:15 2 This post received KUDOS Because I was bored this afternoon and I like Photoshop way too much Attachment: 10Pentagons.gif [ 25.91 KiB | Viewed 973 times ] _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Director
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Re: Regular Pentagon [#permalink]  02 Jul 2008, 01:08
+3 to all of you . Thanks
Director
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Re: Regular Pentagon [#permalink]  02 Jul 2008, 01:18
+3 to all of you . Thanks
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Re: Regular Pentagon [#permalink]  02 Jul 2008, 02:36
I've got it. Thanks! +Kudos for both of you.
Re: Regular Pentagon   [#permalink] 02 Jul 2008, 02:36
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