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# In the figure above, JKLMN is a regular pentagon. Find the

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Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4497

Kudos [?]: 8768 [0], given: 105

In the figure above, JKLMN is a regular pentagon. Find the [#permalink]

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29 Oct 2013, 10:49
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pentagon.JPG [ 19.74 KiB | Viewed 4836 times ]

In the figure above, JKLMN is a regular pentagon. Find the measure of ∠JLK.
A. 24°
B. 36°
C. 48°
D. 60°
E. 72°

For more on polygons & regular polygons, as well as the solution to this problem, see:
http://magoosh.com/gmat/2012/polygons-a ... -the-gmat/

Mike
[Reveal] Spoiler: OA

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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Kudos [?]: 8768 [0], given: 105

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Intern
Joined: 03 Oct 2013
Posts: 6

Kudos [?]: 23 [0], given: 0

Re: In the figure above, JKLMN is a regular pentagon. Find the [#permalink]

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29 Oct 2013, 14:06
Each interior angle of a regular polygon is given by 180*(n-2)/n, In this case this becomes 180*(5-2)/5=108
Therefore, <JKL = 108.
Since this is a regular pentagon, JK=KL and hence, <KJL = <KLJ = (180-108)/2 = 36

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Re: In the figure above, JKLMN is a regular pentagon. Find the [#permalink]

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25 Sep 2015, 09:21
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Kudos [?]: 283 [0], given: 0

Re: In the figure above, JKLMN is a regular pentagon. Find the   [#permalink] 25 Sep 2015, 09:21
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