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

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

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New post 26 Apr 2018, 06:42
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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  [#permalink]

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New post Updated on: 26 Apr 2018, 11:21
2
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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Originally posted by Hero8888 on 26 Apr 2018, 06:55.
Last edited by Hero8888 on 26 Apr 2018, 11:21, edited 1 time in total.
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of  [#permalink]

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New post 26 Apr 2018, 07:02
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Hero8888


Quote:
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, 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  [#permalink]

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New post 26 Apr 2018, 07:25
adkikani wrote:
Hero8888


Quote:
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, henceAngle KQN = 54.


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


adkikani
Image

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

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New post 26 Apr 2018, 07:38
Thanks Princ

bumping 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  [#permalink]

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New post 26 Apr 2018, 08:46
adkikani wrote:
Hero8888


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

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New post 26 Apr 2018, 08:46
adkikani wrote:
Thanks Princ

bumping VeritasPrepKarishma Bunuel chetan2u niks18 gmatbusters for an
alternative solution.


adkikani
OA: D
Image

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\(^{\circ}\)
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In the figure above, JKLMN is a regular pentagon. Find the measure of  [#permalink]

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New post 26 Apr 2018, 09:11
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?
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of  [#permalink]

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New post 26 Apr 2018, 09:45
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 = (5-2) * 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 =(180-108)/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)

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

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New post 26 Apr 2018, 10:02
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
g1edit.png [ 69.27 KiB | Viewed 814 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
(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
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of  [#permalink]

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New post 26 Apr 2018, 10:18
adkikani wrote:
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?


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

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New post 27 Apr 2018, 03:06
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.
Attachment:
images (2).png
images (2).png [ 1.72 KiB | Viewed 665 times ]


"Vertical" refers to the vertex (where they cross), NOT up/down. :angel:

adkikani wrote:
Hero8888


Quote:
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, 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  [#permalink]

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New post 28 Nov 2018, 09:32
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 = (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
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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of &nbs [#permalink] 28 Nov 2018, 09:32
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