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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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26 Apr 2018, 07:42
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65% (02:27) correct 35% (02:40) wrong based on 77 sessions

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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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Updated on: 26 Apr 2018, 12:21
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1
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.

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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  [#permalink]

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26 Apr 2018, 08:02
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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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26 Apr 2018, 08:25
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?

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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26 Apr 2018, 08:38
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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26 Apr 2018, 09:46
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?

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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26 Apr 2018, 09:46
Thanks Princ

bumping VeritasPrepKarishma Bunuel chetan2u niks18 gmatbusters for an
alternative solution.

OA: D

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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26 Apr 2018, 10:11
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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26 Apr 2018, 10:45
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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26 Apr 2018, 11:02
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

*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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26 Apr 2018, 11:18
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?

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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27 Apr 2018, 04:06
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 [ 1.72 KiB | Viewed 1076 times ]

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

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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28 Nov 2018, 10:32
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

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

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21 Mar 2019, 17:56
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.

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Re: In the figure above, JKLMN is a regular pentagon. Find the measure of   [#permalink] 21 Mar 2019, 17:56
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