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# The measures of the interior angles in a polygon are

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The measures of the interior angles in a polygon are  [#permalink]

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03 Feb 2011, 14:45
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The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

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03 Feb 2011, 15:02
4
10
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, $$n_{th}$$ angle, will be $$136+(n-1)$$ degrees. The sum of the $$n$$ consecutive integers (the sum of $$n$$ angles) is given by $$\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n$$;

So we have that $$180(n-2)=\frac{271+n}{2}*n$$ --> $$360(n-2)=(271+n)*n$$, now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits.

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04 Feb 2011, 10:20
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Sum of interior angles for n sides: (n-2)*180
And each interior angle are increasing by 1.
This can be written as : 136+137+138+139+... = 136+(136+1)+(136+2)+(136+3)+...
For n sides

136n+(1+2+3+4....+n-1)
136n+(n-1)n/2 --- Sum of n-1 natural number

Equate:
136n+(n-1)n/2 = (n-2)*180
n^2-89n+720=0
n^2-80n-9n+720=0
n(n-80)-9(n-80)=0
(n-9)(n-80)=0

n=9
n=80

Since 9 is one of the options.

Ans: "B"
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03 Feb 2011, 20:05
11
1
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

When I look at this question, I say, "I know how to find the Interior angle of a regular polygon. But this is not a regular polygon since it has angles 136, 137, 138, 139, 140, 141, 142 ....etc."

Also,
Interior angle of 8 sided regular polygon = 180*6/8 = 135
Interior angle of 9 sided regular polygon = 180*7/9 = 140
Interior angle of 10 sided regular polygon = 180*8/10 = 144
etc

The average of the given angles can only match 140 (such that effectively, all the angles are 140) Hence, the polygon must have 9 sides.

Remember, capitalize on what you know.
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04 Feb 2011, 09:47
Karishma,
Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks
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04 Feb 2011, 11:56
3
ajit257 wrote:
Karishma,
Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks

Sum of interior angles of a polygon = (n-2)*180 (not necessarily regular polygon)

Why? See the figure below:

Attachment:

Ques2.jpg [ 5.14 KiB | Viewed 36505 times ]

A 6 sided polygon can be split into 4 triangles each of which has a sum of interior angles 180 degrees.
An n sided polygon can be split into n - 2 triangles.

When the polygon is regular, each angle is same so the sum is divided by the number of sides to get the measure of each angle e.g. in a regular hexagon, each interior angle = 4*180/6 = 120 degrees.

Now if I have a hexagon whose angles are 115, 117, 119, 121, 123 and x, what will be the angle x?
We can see it in two ways -
1. The sum of all angles should be 4*180 = 720
So 115 + 117 + 119 + 121 + 123 + x = 720
or x = 125

2. The average of the angles should be 120. (Since the sum of the angles is 720 and there are 6 sides)
119 and 121 average out as 120. (119 is 1 less than 120 and 121 is 1 more than 120)
117 and 123 average out as 120.
So 115 and x should average out as 120 too. Therefore, x should be 125.

In the question, the average of the given angles of the polygon can only be 140. So it must have 9 sides. To confirm,
136, 137, 138, 139, 140, 141, 142, 143, 144 - 9 angles with average 140. So the polygon must have 9 sides.

(It cannot be 144 or anything else because 10 angles (136, 137, 138, 139, 140, 141, 142, 143, 144, 145) will not give an average of 144)
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05 Feb 2011, 08:01
tough one. simple solution, but its not an ez one.
am i right?
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06 Feb 2011, 06:25
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144144 wrote:
tough one. simple solution, but its not an ez one.
am i right?

Actually, I wouldn't say it is very tough if we know that the sum of interior angles of a polygon is 180(n -2) (I explained above why it is so). We can also use a very straight forward but long approach.

Interior angles of the given polygon: 136, 137, 138, 139, 140, 141, 142, 143....

Using options:

If the polygon had 8 sides, it would have had 8 interior angles. Their sum: (8-2)180 = 1080
Sum of 8 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 is more than 1080 hence this polygon does not have 8 sides.

If the polygon had 9 sides, it would have had 9 interior angles. Their sum: (9-2)180 = 1260
Sum of 9 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 + 144 is 1260. Hence this polygon does have 9 sides.
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06 Feb 2011, 06:45
144144 wrote:
tough one. simple solution, but its not an ez one.
am i right?

I wouldn't say it's very hard either, though I wouldn't recommend trial and error in this cases:

...
Calculating the sum of the interior angles of a polygon with 7 sides then checking whether 7 consecutive integers starting from 136 add up to that value;
Calculating the sum of the interior angles of a polygon with 8 sides then checking whether 8 consecutive integers starting from 136 add up to that value;
Calculating the sum of the interior angles of a polygon with 9 sides then checking whether 9 consecutive integers starting from 136 add up to that value;

Whereas equating two formulas and working on answer choices should give an answer in less time: $$180(n-2)=\frac{271+n}{2}*n$$ --> $$360(n-2)=(271+n)*n$$ --> $$n=9$$.
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06 Feb 2011, 13:03
1
ye... well i guess i need to work on my geometry more so the way i process these things will be better...

thanks both as usual.
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09 Aug 2012, 09:08
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Hey guys...please let me know whether this method is correct or not.
the stem says that the angles are consecutive integers, 136 being the smallest one. Also going by the formula of sum of interior angles, we know atleast this much that sum of angles cant be number like 729, 847, 653, 542 but it can be a number of the form xx0. So going by this knowledge, if the smallest angle is 136 then the sum of angles can only be of the form xx0 when the largest angle is 144. That is 9 sides.
Yeah the largest angle could be 154, 164 but in that case the number of sides must be 19 and 29 respectively.
Since the largest option is 13, hence the answer is 9.
Please correct me if I am wrong.
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Re: The measures of the interior angles in a polygon are  [#permalink]

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05 Apr 2014, 15:32
1

But a polygon has sum of sides equal to (n-2)*180, thus sum will always have 0 as units digit.

Therefore if we begin with 136+137 etc... only 9 fits the bill since 4 is not a valid answer choice

Thanks!
Cheers
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Re: The measures of the interior angles in a polygon are  [#permalink]

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03 Nov 2014, 07:36
Bunuel wrote:
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, $$n_{th}$$ angle, will be $$136+(n-1)$$ degrees. The sum of the $$n$$ consecutive integers (the sum of $$n$$ angles) is given by $$\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n$$;

So we have that $$180(n-2)=\frac{271+n}{2}*n$$ --> $$360(n-2)=(271+n)*n$$, now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits.

Hi Bunnel,

In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits'.
Why are you saying that LHS is 0.

Regards,
Ammu
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Re: The measures of the interior angles in a polygon are  [#permalink]

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03 Nov 2014, 07:39
ammuseeru wrote:
Bunuel wrote:
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, $$n_{th}$$ angle, will be $$136+(n-1)$$ degrees. The sum of the $$n$$ consecutive integers (the sum of $$n$$ angles) is given by $$\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n$$;

So we have that $$180(n-2)=\frac{271+n}{2}*n$$ --> $$360(n-2)=(271+n)*n$$, now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits.

Hi Bunnel,

In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits'.
Why are you saying that LHS is 0.

Regards,
Ammu

Because the left hand side (LHS) is 360(n-2) = 10*(36(n-2)) = 10*integer = something with the units digit of 0.

Does this make sense?
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Re: The measures of the interior angles in a polygon are  [#permalink]

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03 Nov 2014, 08:22
Hi Bunnel,

In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits'.
Why are you saying that LHS is 0.

Regards,
Ammu[/quote]

Because the left hand side (LHS) is 360(n-2) = 10*(36(n-2)) = 10*integer = something with the units digit of 0.

Does this make sense?[/quote]

I understood it now. Thank you
Actually I had misinterpreted it . I thought you were saying "RHS to end with zero -->(271+n)*n= 0" . Its my Bad.
Thanks for clarification.

Regards,
Ammu
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The measures of the interior angles in a polygon are  [#permalink]

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07 Oct 2016, 06:10
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

oh..nice question..
i haven't reviewed the geometry and i took some additional time to solve it...
total sum of the interior angle of a polygon is 180(n-2)
i started by trial...
suppose we have 10 sides, then 180(10-2) = 180*8 = 1440
sum of the angles in this case...
i solved this way - 10*136 = 1360
then, since we have consecutive numbers, let's find the sum of the numbers from 1 to 9. best way, to use the formula n(n+1)/2 -> 9*10/2 = 45.
1360+45 = 1405.
the sum doesn't match with what we need.
in this case, we know for sure that C, D, and E are not correct.
we are left with 2 answers, and we need to test one more time...

9 sides -> 180(9-2) = 1260
now..don't need to go all the way through with what we did earlier...
sum of the consecutive numbers is 1405, and the greatest number is 136+9 or 145.
1405 - 145 = 1260.

we have a match!

woah, so many more ways to solve it here! awesome!
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The measures of the interior angles in a polygon are  [#permalink]

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30 Mar 2017, 17:37
1
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

OFFICIAL SOLUTION

We are told that the smallest angle measures 136 degrees--this is the first term in the consecutive set. If the polygon has S sides, then the largest angle--the last term in the consecutive set--will be (S - 1) more than 136 degrees.

The sum of consecutive integers = (Average Term) * (# of Term)= $$\frac{First + Last}{2}$$ * (# of Terms).

Given that there are S terms in the set, we can plug in for the first and last term as follows:

$$\frac{136 + 136 + (S-1)}{2} * S$$ = sum of the angles in the polygon.

We also know that the sum of the angles in a polygon = 180 (S-2) where S represents the number of sides.

Therefore: $$180(S-2) = \frac{136+136+(S-1)}{2} * S$$. We can solve

for S by cross-multiplying and simplifying as follows:

$$2(180)(S-2) = [272 + (S-1)] S$$
$$360S - 720 = (271 + S)S$$
$$360S - 720 = 271S + S^2$$
$$S^2 - 89S + 720 = 0$$

A look at the answer choices tells you to try (S - 8), (S - 9), or (S - 10) in factoring.

As it turns out (S-9)(S-80)=0, which means S can be 9 or 80. However S cannot be 80 because this creates a polygon with angles greater than 180.

Therefore S equals 9; there are 9 sides in the polygon.

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Re: The measures of the interior angles in a polygon are  [#permalink]

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31 Mar 2017, 06:50
Bunuel wrote:
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, $$n_{th}$$ angle, will be $$136+(n-1)$$ degrees. The sum of the $$n$$ consecutive integers (the sum of $$n$$ angles) is given by $$\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n$$;

So we have that $$180(n-2)=\frac{271+n}{2}*n$$ --> $$360(n-2)=(271+n)*n$$, now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits.

This is the best and the most proper way of doing this task but may I suggest another way:
the sum of all angles must be divisible by 10.
Let's sum last figures of consecutive angles until we'll get 0 (but the number of angles should be more then 7):
we have angles #1 - 136, #2 - 137, #3 - 138, 139, 140, #6 - 141, #7 - 142.
So 6+7=13, 8 (next angle) + 3 (last figure in 13)=11, 1>0.
9+1=0 great! but this is only angle #4, and we don't have this in our answers so we move on.
0+0=0, but see above.
0+1=1, 1+2(angle #7)=3.
Now we are in the area of answer therefore we should be more accurate. So 3 (angle #8 - 143) +3=6.
4+6=0! great. This correspondes to the angle #9.
Just to check if we don't have other answers for angles 10-13:
0+5=5>0
5+6=11,1>0
1+7=8>0,
8+8=16, 6>0.
Here we go, the only possible answer is B. This is pretty long description but this is definitely is not time consuming way.
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Re: The measures of the interior angles in a polygon are  [#permalink]

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24 Apr 2017, 05:20
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Reverse Approach using external angle

We are given : smallest internal angle = 136 deg ;

to find : number of sides 'n'

Solution : We know that, internal angle = (n - 2)*180;

But, we dont know value of n. However, we know one thing for sure. Irrespective of the value of n, Sum of all the external angles will be 360 deg.

So, Corresponding external angle for internal angle of 136 deg = 180 - 136 = 44 deg. (Since, Sum of internal + external angle = 180 deg)

As internal angle increases by 1 external angle decreases by 1.

So, now 2nd external angle will be 43 deg, 3rd external angle will be 42 deg, 4th will be 41 deg and so on. We keep doing this till the point our sum of all external angles turns out to be 360 deg.

So, 44 + 43 + 42 + 41 + 40 + 39 + 38 + 37 + 36 = 360

total number of terms in above equation is 9. So the number of sides of polygon = 9

( We can also use concept of AP. All the terms are in AP. We know S = 360, t1 = 44, d = -1, n=?
360 = n/2* (2*44 + (n-1)*-1) => n^2 - 89n + 720 = 0 => n = 80 or n = 9)
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The measures of the interior angles in a polygon are  [#permalink]

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24 Apr 2017, 12:05
rxs0005 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

a plugin approach:
(n-2*180°)/n=average internal angle
assume n is odd; an even number of n consecutive integers won't sum to n-2*180°
average internal angle will be middle term of sequence
starting with B,
7*180°=1260°; 1260°/9=140°, the 5th, or middle, of 9 terms
9
B
The measures of the interior angles in a polygon are &nbs [#permalink] 24 Apr 2017, 12:05

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