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

Question

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

Hero8888 · Accepted Answer

Sum of angles in any polygon = 180*(n-2)=> Sum of angles of pentagon =180*(5-2)= 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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Answer (D)

adkikani · Answer

Hero8888

I could not understand if you applied concept of vertically opposite angles?
If yes, what are the parallel lines you considered?

Princ · Answer

adkikani 
 https://gmatclub.com/forum/download/file.php?id=41044

Consider 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

adkikani · Answer

Thanks  Princ

bumping  VeritasPrepKarishma   Bunuel   chetan2u   niks18   gmatbusters  for an 
alternative solution.

niks18 · Answer

HI  adkikani

I 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

Princ · Answer

adkikani 
OA: D
 https://gmatclub.com/forum/download/file.php?id=41044

Let me try another way
Sum of internal angle of a polygon is given (n-2)* 180 ^{\circ}  
for Pentagon, n = 5 so sum of internal angle would be =(5-2)*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 = (4-2)*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}

adkikani · Answer

niks18

So 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?

zishu912 · Answer

The sum of all the angles in pentagon = (5-2) * 180 = 540 since the pentagon is regular, all the sides and angles are equal, therefore the measurement of Help by hitting KUDOS

generis · Answer

g1edit.png 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 (n-2)*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

niks18 · Answer

Hi  adkikani

Yes  \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

GMATBusters · Answer

Hii adkikani Vertically 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. images (2).png "Vertical" refers to the vertex (where they cross), NOT up/down.

shashankism · Answer

Since JKLMN is a regular pentagon ... Each interior angle = (5-2)/5 * 180 deg. = 108

∠KJN =108 
In triangle KJN, KJ = JN ... So, ∠JKN = ∠KNJ = (180-108)/2 = 36
∠QRN = 90
∠RQN = 90-36 = 54
∠KQL = ∠RQN = 54

Answer D

ScottTargetTestPrep · Answer

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

Hayward1gmat · Answer

Just to clarify, as we draw a straight line from L to R and it splits the angle LRJ and angle LRN into both 90. Can we conclude that angle KLR = angle MLR? (i.e. angle bisects each other)

bumpbot · Answer

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

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