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# If the sum of the interior angles of a regular polygon measures up to

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Manager
Joined: 22 Sep 2010
Posts: 71
If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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05 Oct 2010, 05:40
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Difficulty:

15% (low)

Question Stats:

74% (00:55) correct 26% (01:14) wrong based on 121 sessions

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If the sum of the interior angles of a regular polygon measures up to 1440 degrees, how many sides does the polygon have?

1. 10 sides
2. 8 sides
3. 12 sides
4. 9 sides
5. None of these
Math Expert
Joined: 02 Sep 2009
Posts: 58445
Re: If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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05 Oct 2010, 06:18
pzazz12 wrote:
If the sum of the interior angles of a regular polygon measures up to 1440 degrees, how many sides does the polygon have?

1. 10 sides
2. 8 sides
3. 12 sides
4. 9 sides
5. None of these

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

So $$180(n-2)=1440$$ --> $$n=10$$.

For more on this check Polygons chapter of Math Book: math-polygons-87336.html

Hope it helps.
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Joined: 30 Sep 2010
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Re: If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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05 Oct 2010, 06:19
Hi

The formula for interior angle is (n-2)*180

(10-2)*180 = 1440.

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Re: If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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05 Oct 2010, 06:56
If you don't know the formula you can get to the answer this way. A little more involved but probably still doable in the crunch:

Start with the second to highest number in the answers (i.e. 10) so you can go higher or lower depending on what you find.

Draw or visualize a decagon, and run the five diagonals that go through the center, so that you end up with 10 identical 'wedges' or 'pizza slices' which will be isosceles triangles.

All 10 triangles will have the center, where all diagonals cross, in common. You know the sum of all those angles is 360, so since you have 10 of them, that angle is 360/10 = 36 for each triangle. Now you can see that for each slice, the sum of the other triangles must be 180 - 36 = 144. Therefore, since you have 10 triangles, the total sum of the angles that make the decagon is 144 *10 = 1440. Voila.

If this had been too low, you know the polygon would have to have more sides, if it had been too big, you would need less. so you could work that way.

Still, formula is MUCH simpler so this tells me I need to brush up!!
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Joined: 15 Apr 2010
Posts: 105
Re: If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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05 Oct 2010, 12:15
1
Sum of interior angles of a regular polygon = 180 * (n-2), where n is the number of sides.

180 * (n-2) = 1440.

Solve for n.

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Joined: 12 Sep 2015
Posts: 4009
Re: If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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19 Jan 2018, 14:01
Top Contributor
pzazz12 wrote:
If the sum of the interior angles of a regular polygon measures up to 1440 degrees, how many sides does the polygon have?

1. 10 sides
2. 8 sides
3. 12 sides
4. 9 sides
5. None of these

Nice rule: The sum of the interior angles in an N-sided polygon is equal to 180(N - 2) degrees

So, we can write: 180(N - 2) = 1440
Divide both sides by N-2 to get: N-2 = 8
Solve: N = 10

Cheers,
Brent
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Joined: 24 Aug 2017
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Re: If the sum of the interior angles of a regular polygon measures up to  [#permalink]

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08 May 2019, 12:25
we know that the sum of the angles is 1440
and we should use the formula that the sum of the angles = 180*(n-2)
1440=180*(N-2)=> 180N=1800 N= 10.
so the answer is A (10)
Re: If the sum of the interior angles of a regular polygon measures up to   [#permalink] 08 May 2019, 12:25
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