The measures of the interior angles in a polygon

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The measures of the interior angles in a polygon are consecutive integers. The largest angle measures 110 degrees. How many sides does this polygon have?

A) 5
B) 6
C) 7
D) 9
E) 11

OA A

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by rintoo22 » Tue Apr 16, 2013 1:06 pm
The measures of the interior angles in a polygon are consecutive integers. The largest angle measures 110 degrees. How many sides does this polygon have?
110 + 110-x + 110-2x + 110-3x + 110-4x = 540
550 - 10x = 540
x = 1 ...(1)
Now put the value of x

110 + 109 + 108 + 107 + 106 = 540.

Option (A) is the correct answer.

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by guerrero » Tue Apr 16, 2013 1:10 pm
rintoo22 wrote:
The measures of the interior angles in a polygon are consecutive integers. The largest angle measures 110 degrees. How many sides does this polygon have?
110 + 110-x + 110-2x + 110-3x + 110-4x = 540
how did you derive the above equation . Could you please elaborate ?

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by srcc25anu » Tue Apr 16, 2013 2:05 pm
sum of interior angles = n-2 * 180
lets start with options given.

Lets start with Option A. If polygon has 5 sides, sum of interior angles = 3 * 180 = 540
now given that all angles are consequitive integers, Let those angles be x-2, x-1, x, x+1 and x+2
sum of above (x - 2 + x - 1 + x + x + 1 + x + 2)= 5x = 540
x = 108 degrees
If x = 108, x-2 = 106, x-1 = 107, x+1 = 109 and x+2 (largest) = 110

hence A is our solution.

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by rintoo22 » Tue Apr 16, 2013 8:57 pm
how did you derive the above equation . Could you please elaborate ?
Important points in the question
1. Largest angle measures 110
2. Interior angles in a polygon are consecutive integers.

Let x be the difference in consequitive integers (angles)

110 + (110-x) + (110-2x) + (110-3x) + (110-4x).....

Lets plug this equation in the (A) option, 5 sides.
If polygon has 5 sides, sum of interior angles = 3 * 180 = 540.

So 5 angles
110 + (110-x) + (110-2x) + (110-3x) + (110-4x) = 540
we get x = 1
Now put the value of x

110 + 109 + 108 + 107 + 106 = 540.

Hope this helps

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by Anju@Gurome » Tue Apr 16, 2013 9:14 pm
guerrero wrote:The measures of the interior angles in a polygon are consecutive integers. The largest angle measures 110 degrees. How many sides does this polygon have?
Plugging answer choices is a good strategy for this problem.
As that has been discussed, I'll post another two methods to solve this problem.

Tricky Approach:
If a polygon has n sides, the sum of the interior angles of the polygon = (n - 2)*180°
Hence, the sum must be a multiple of 10, i.e. units digit of the sum must be 0.

As the largest angle measures 110° and the angles are consecutive integers, the next angles will be 109°, 108°, ... etc

The 1st time we get zero as units digit when we start adding the units place of these measures is (0 + 9 + 8 + 7 + 6) ---> Hence, 5 angles --> n = 5

The 2nd time we get zero as units digit when we start adding the units place of these measures is (0 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 + ...) ---> Hence, more than 11 angles angles --> n > 11

Hence, the polygon has 5 sides.

The correct answer is A.
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by Anju@Gurome » Tue Apr 16, 2013 9:54 pm
Algebraic Approach:
Say, the number of sides of the polygon is n.
Hence, sum of the interior angles of the polygon = (n - 2)*180°

Largest angle is 110° and all the interior angles are consecutive integers.
Hence, the measures of the interior angles of the polygon in degrees are 110, (110 - 1), (110 - 2), ..., and (110 - (n - 1)).

So, sum of the angles = [110 + (110 - 1) + (110 - 2) + ... + (110 - (n - 1))]
= (110 + 110 + ... n times) - (1 + 2 + ... + (n - 1))
= 110n - sum of first (n - 1) integers
= 110n - n(n - 1)/2

So, 110n - n(n - 1)/2 = (n - 2)*180
--> 220n - n² + n = 360n - 720
--> n² + 139n - 720 = 0
--> n² - 5n + 144n - 720 = 0
--> (n - 5)(n + 144) = 0

As n cannot be negative, only possible value of n is 5.

The correct answer is A.
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