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In the figure, JKLMNP is a regular hexagon .

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In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post 06 Nov 2015, 09:06
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In the figure, JKLMNP is a regular hexagon .Find the measure of \(\angle\) MQN.

A. \(30^{\circ}\)
B. \(45^{\circ}\)
C. \(50^{\circ}\)
D. \(60^{\circ}\)
E. \(75^{\circ}\)

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Geometry_Hexagon.PNG [ 461.86 KiB | Viewed 2739 times ]


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In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post 11 Dec 2015, 08:43
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Skywalker18 wrote:
In the figure, JKLMNP is a regular hexagon .Find the measure of \(\angle\) MQN.

A. \(30^{\circ}\)
B. \(45^{\circ}\)
C. \(50^{\circ}\)
D. \(60^{\circ}\)
E. \(75^{\circ}\)



vinnisatija
Theres a very simple explanation for this question.

Remember the formula for the sum of internal angles of a polygon.

Sum of angles= 180 X (n-2) ; where n= number of sides

In our case ; 180 ( 6-2) = 720.


Now this is a regular hexagon. All the angles are equal and all the sides are also equal.


therefore Each angle = 120

Now consider triangle LMN

we know angle M=120.
Since all the sides are equal, we know this is an isosceles triangle.

2x+120=180
x=30.

So angle L= N= 30.

Similarly, in triangle KLM,

angle k and M = 30.

From what we have proved above,

Angle QMN 120-30 = 90

Now we have angle QMN+ Angle MQN + angle QNM = 180

90+MQN+30=180
MQN=60




Hope its clear. Let me know through the kudos button :)
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Re: In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post 05 Dec 2015, 07:04
can anyone provide an alternate explanation on this ?
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Re: In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post 05 Dec 2015, 08:05
Skywalker18 wrote:
In the figure, JKLMNP is a regular hexagon .Find the measure of \(\angle\) MQN.

A. \(30^{\circ}\)
B. \(45^{\circ}\)
C. \(50^{\circ}\)
D. \(60^{\circ}\)
E. \(75^{\circ}\)


An interesting question that requires you to be a bit creative with the information that you have been given.

The given polygon is a REGULAR hexagon --> all sides and all angles are equal. Every interior angle of a hexagon = 120 degrees.

As per the attached image, draw a circle around the given hexagon. Thus , now you get equal arcs LM=MN=NP=JP=JK=KL . As there are 6 arcs and all arcs must subtend a total of 360 degrees at the center of the circle ---> each arc subtends 60 degrees at the center.

Additionally, an angle subtended by any arc of a circle at the center = 2* angle subtended by the same arc on the circumference.

Consider triangle QNM,\(\angle{QNM}\) = angle subtended by LM = 30 degrees (as it is the angle subtended by arc LM on the circumference.)

Similarly, Arcs KJ+JP+PN subtend \(\angle{QMN}\) on the circumference. ---> \(\angle{QMN} = 90 degrees\).

Finally, in triangle QMN, \(\angle{QMN} + \angle {QNM} + \angle {MQN} = 180 degrees\) --->\(\angle {MQN}\) = 180-30-90 = 60 degrees.

D is the correct answer.

Hope this helps.
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In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post Updated on: 20 May 2016, 07:29
I think there's an easier way to solve this problem.

As we were saying a regular hexagon has each angle measuring \(120°\).

Now let's look at the image posted above,
the theorem of vertical and adjacent angle pairs says really simply that opposite angles born from the intersection of two lines must be equal.
Therefore if the angle of the hexagon is \(120°\) the opposite angle will be \(120°\) as well,
but we also know that the round angle is 360°, therefore, the sum of the remaining opposite pair must be \(360° - 240° = 120°\),
and since they are opposite as well they have to be equal -> \(\frac{120°}{2}= 60°\)

:)

Originally posted by DensetsuNo on 20 May 2016, 04:57.
Last edited by DensetsuNo on 20 May 2016, 07:29, edited 1 time in total.
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Re: In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post 20 May 2016, 05:58
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Since it is a regular hexagon, angles made at dissection point by all diagonals will be equal.

Total number of angles made by all diagonals= 6
Total angle made at dissection point = 360

Any single angle measurement will b = 360/6= 60

D is the answer
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Re: In the figure, JKLMNP is a regular hexagon .  [#permalink]

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New post 09 Jan 2019, 00:33
Skywalker18 wrote:
In the figure, JKLMNP is a regular hexagon .Find the measure of \(\angle\) MQN.

A. \(30^{\circ}\)
B. \(45^{\circ}\)
C. \(50^{\circ}\)
D. \(60^{\circ}\)
E. \(75^{\circ}\)


do you have easy solution?
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Re: In the figure, JKLMNP is a regular hexagon . &nbs [#permalink] 09 Jan 2019, 00:33
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In the figure, JKLMNP is a regular hexagon .

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