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In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 07:42
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In the figure above, JKLMN is a regular pentagon. Find the measure of ∠KQL A. 45 B. 48 C. 51 D. 54 E. 60
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In the figure above, JKLMN is a regular pentagon. Find the measure of
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Updated on: 26 Apr 2018, 12:21
Sum of angles in any polygon = 180*(n2)=> Sum of angles of pentagon =180*(52)= 540. Each angle = 540/5=108 => J = 108. Since all sides of pentagon are equal then other 2 angles in equilateral triangle JKN = (180  108)/2 = 36. Triangle QRN is right triangle => Angle RQN = 180  90 36 = 54, hence Angle KQL = 54. Answer (D)
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Originally posted by Hero8888 on 26 Apr 2018, 07:55.
Last edited by Hero8888 on 26 Apr 2018, 12:21, edited 1 time in total.



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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 08:02
Hero8888Quote: Sum of angles in any polygon = 180*(n2)=> Sum of angles of pentagon =180*(52)= 540. Each angle = 540/5=108 => J = 108. Since all sides of pentagon are equal then other 2 angles in equilateral triangle JKN = (180  108)/2 = 36. Triangle QRN is right triangle => Angle RQN = 180  90 36 = 54, henceAngle KQN = 54. I could not understand if you applied concept of vertically opposite angles? If yes, what are the parallel lines you considered?
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 08:25
adkikani wrote: Hero8888Quote: Sum of angles in any polygon = 180*(n2)=> Sum of angles of pentagon =180*(52)= 540. Each angle = 540/5=108 => J = 108. Since all sides of pentagon are equal then other 2 angles in equilateral triangle JKN = (180  108)/2 = 36. Triangle QRN is right triangle => Angle RQN = 180  90 36 = 54, henceAngle KQN = 54. I could not understand if you applied concept of vertically opposite angles? If yes, what are the parallel lines you considered? adkikaniConsider Line segment LR, \(\angle\)RQN+ \(\angle\)NQL = 180\(^{\circ}\) ,Angles on one side of a straight line always add to 180\(^{\circ}\) Now consider Line segment KN, \(\angle\)KQL+ \(\angle\)NQL = 180\(^{\circ}\) \(\angle\)KQL+ \(\angle\)NQL = 180\(^{\circ}\)=\(\angle\)RQN+ \(\angle\)NQL \(\angle\)KQL+ \(\angle\)NQL=\(\angle\)RQN+ \(\angle\)NQL \(\angle\)KQL+ \(\angle\) NQL=\(\angle\)RQN+ \(\angle\) NQL\(\angle\)KQL=\(\angle\)RQN
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In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 08:38
Thanks Princbumping VeritasPrepKarishma Bunuel chetan2u niks18 gmatbusters for an alternative solution.
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 09:46
adkikani wrote: Hero8888Quote: Sum of angles in any polygon = 180*(n2)=> Sum of angles of pentagon =180*(52)= 540. Each angle = 540/5=108 => J = 108. Since all sides of pentagon are equal then other 2 angles in equilateral triangle JKN = (180  108)/2 = 36. Triangle QRN is right triangle => Angle RQN = 180  90 36 = 54, henceAngle KQN = 54. I could not understand if you applied concept of vertically opposite angles? If yes, what are the parallel lines you considered? HI adkikaniI think this is one of the most concise method to solve this question. For vertically opposite angles, you need two line intersecting at a point. In this question line KL & LR intersect at point Q, creating two vertically opposite angles KQL & RQN



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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 09:46
adkikani wrote: adkikaniOA: D Let me try another way Sum of internal angle of a polygon is given (n2)* 180\(^{\circ}\) for Pentagon, n = 5 so sum of internal angle would be =(52)*180\(^{\circ}\) = 540\(^{\circ}\) For regular pentagon , all internal angle would be equal to \(\frac{540}{5}\)= 108\(^{\circ}\) In Triangle KJN,KJ=JN as KJ and JN are sides of regular pentagon. So \(\angle\)JKN = \(\angle\)JNK \(\angle\)JKN+\(\angle\)JNK+\(\angle\)KJN = 180\(^{\circ}\) 2\(\angle\)JKN+ 108\(^{\circ}\) = 180\(^{\circ}\) \(\angle\)JKN=36\(^{\circ}\) Now consider polygon JKQR Sum of all interior angle would be = (42)*180\(^{\circ}\) = 360\(^{\circ}\) \(\angle\)JKN+\(\angle\)KJN+\(\angle\)JRQ+\(\angle\)RQK= 360\(^{\circ}\) 36\(^{\circ}\) +108\(^{\circ}\) +90\(^{\circ}\) +\(\angle\)RQK= 360\(^{\circ}\) \(\angle\)RQK = 126\(^{\circ}\) Consider Line Segment LR \(\angle\)RQK+\(\angle\)KQL = 180\(^{\circ}\) \(\angle\)KQL = 180\(^{\circ}\) 126\(^{\circ}\) \(\angle\)KQL = 54\(^{\circ}\)
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In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 10:11
niks18So you mean to say that: For intersecting line segments LR and NK, \(\angle\) KQR will be equal to \(\angle\) LQN since both line segments intersect only at point Q.I do not need any other conditions, correct?
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 10:45
adkikani wrote: Attachment: g1.JPG In the figure above, JKLMN is a regular pentagon. Find the measure of ∠KQL A. 45 B. 48 C. 51 D. 54 E. 60 The sum of all the angles in pentagon = (52) * 180 = 540 since the pentagon is regular, all the sides and angles are equal, therefore the measurement of <KJN = 540/5 = 108 Now, <KJN + <JKN +<JNK = 180 [<JKN = <JNK, since triangle JKN is isosceles triangle] ==> <JKN = <JNK =(180108)/2 = 36 unit Next in triangle QRN, <QRN = 90 <RNQ = 36 so, the angle(RQN) = 180  (90+36) = 54 angle(RQN) = <KQL [corresponding angles] Therefore Our answer will be (D) NOTE> Help by hitting KUDOS



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In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 11:02
adkikani wrote: Attachment: The attachment g1.JPG is no longer available In the figure above, JKLMN is a regular pentagon. Find the measure of ∠KQL A. 45 B. 48 C. 51 D. 54 E. 60 Attachment:
g1edit.png [ 69.27 KiB  Viewed 1224 times ]
I found the third angle measure of ∆ RQN ( EDIT: as did Hero8888 and zishu912 , I see  dense text and different names made me ) That third angle, ∠RQN = ∠KQL (vertical angles) ∆ RQN, three angles One angle = 90° Second angle easily derived: ∠JNK = 36° Subtract those two from 180° Third angle ∠RQN = ∠KQL • Derive measure of ∠JNK (second figure) A regular pentagon: each vertex* = 108° A regular pentagon: can be divided into three isosceles triangles ∆ JKN is one of those three. It is isosceles Its angles: (180  vertex 108) = 72 left. Two small angles. Each = 36 ∠JNK = 36 • So the second angle of ∆ RQN = 36° • Third angle of ∆ RQN? (third figure) ∠RQN + 36 + 90 = 180 ∠RQN = 54, and by vertical angle property ∠KQL = 54
Answer D
*108° at each vertex A regular pentagon has 108° at each vertex Total of interior angle measures, regular polygon, n = number of sides (n2)*180 = 3*180 = 540° TOTAL 5 vertices. 5 interior angles. \(\frac{540}{5}\) = 108° at each vertex Vertical angles: lines that intersect create congruent opposite angles
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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26 Apr 2018, 11:18
adkikani wrote: niks18So you mean to say that: For intersecting line segments LR and NK, \(\angle\) KQR will be equal to \(\angle\) LQN since both line segments intersect only at point Q.I do not need any other conditions, correct? Hi adkikaniYes \(\angle\) KQR and \(\angle\) LQN are the other pair of opposite angles for the intersecting line LR & NK. And as vertically opposite angles are equal so you have  \(\angle\) KQR \(=\angle\) LQN \(\angle\) KQL \(=\angle\) RQN



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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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27 Apr 2018, 04:06
Hii adkikaniVertically opposite angle property is valid for any 2 intersecting lines, parallel lines are not required.The vertically opposite angles opposite each other when two lines cross. They are always equal. In this example a° and b° are vertically opposite angles. Attachment:
images (2).png [ 1.72 KiB  Viewed 1076 times ]
"Vertical" refers to the vertex (where they cross), NOT up/down. adkikani wrote: Hero8888Quote: Sum of angles in any polygon = 180*(n2)=> Sum of angles of pentagon =180*(52)= 540. Each angle = 540/5=108 => J = 108. Since all sides of pentagon are equal then other 2 angles in equilateral triangle JKN = (180  108)/2 = 36. Triangle QRN is right triangle => Angle RQN = 180  90 36 = 54, henceAngle KQN = 54. I could not understand if you applied concept of vertically opposite angles? If yes, what are the parallel lines you considered?
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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28 Nov 2018, 10:32
adkikani wrote: Attachment: g1.JPG In the figure above, JKLMN is a regular pentagon. Find the measure of ∠KQL A. 45 B. 48 C. 51 D. 54 E. 60 Since JKLMN is a regular pentagon ... Each interior angle = (52)/5 * 180 deg. = 108 ∠KJN =108 In triangle KJN, KJ = JN ... So, ∠JKN = ∠KNJ = (180108)/2 = 36 ∠QRN = 90 ∠RQN = 9036 = 54 ∠KQL = ∠RQN = 54 Answer D
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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21 Mar 2019, 17:56
adkikani wrote: Attachment: g1.JPG In the figure above, JKLMN is a regular pentagon. Find the measure of ∠KQL A. 45 B. 48 C. 51 D. 54 E. 60 Since JKLMN is a regular pentagon, each exterior angle is 360/5 = 72, so each interior angle is 180  72 = 108. Triangle JKN is an isosceles triangle with vertex angle J = 108, so angle QNR = (180  108)/2 = 72/2 = 36. Since angle QRN = 90, angle RQN = 180  90  36 = 54. Since angle KQL and angle RQN are vertex angles, angle KQL = 54. Answer: D
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of
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