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by sanju09 » Sat Mar 31, 2012 5:08 am
Md.Nazrul Islam wrote:A polygon has 44 diagonals .Number of sides are .

If you are talking about an n-sided convex polygon, then its number of diagonals are given by nC2 - n; and if

nC2 - n = 44 (given), then

½ n (n - 1) - n = 44

n^2 - n - 2 n = 88

n^2 - 3 n - 88 = 0

Solving the quadratic, we have

n^2 - 11 n + 8 n - 88 = 0

(n - 11) (n + 8) = 0

Since, n = -8 is not applicable, hence the polygon in question must have [spoiler]11[/spoiler] number of sides to it.
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by Brent@GMATPrepNow » Sat Mar 31, 2012 6:42 am
Md.Nazrul Islam wrote:A polygon has 44 diagonals .Number of sides are .
We can also solve this without using combinations.

If there are n sides, there are n vertices.

Let's examine 1 vertex. From this vertex, how many other vertices can this vertex connect to to create a diagonal line? Well, we can't draw a diagonal line to the 2 adjacent vertices, and we can't draw a line to the vertex we are starting from.
So, from one particular vertex, there are n-3 candidates to create a diagonal line to.

So, we have n vertices altogether, and from each vertex we can create n-3 diagonals.
This means that the total number of diagonals = n(n-3)
Now keep in mind that this happens to count each diagonal twice.
So, to account for this duplication, we can say that:

In any n-sided convex polygon, the total number of diagonals = n(n-3)/2

Now onto the question. We're told that there are 44 diagonals.

So, n(n-3)/2 = 44
Multiply both sides by 2: n(n-3) = 88
Expand: n^2 - 3n = 88
Set quadratic equal to zero: n^2 - 3n - 88 = 0
Factor: (n+8)(n-11)=0
So, n=-8 or n=11
n can't be negative, so the answer is [spoiler]n=11[/spoiler]

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by GMATGuruNY » Sat Mar 31, 2012 2:26 pm
If a polygon has 44 diagonals, then the number of sides is?

7,8,9,10,11
We can plug in the answers, which represent the number of sides.
The number of vertices is equal to the number of sides.
A diagonal can be drawn between any combination of 2 non-adjacent vertices.
(Adjacent vertices form the sides of the polygon and thus cannot serve as endpoints of a diagonal.)

Answer choice C: 9 vertices
Number of combinations of 2 that can be formed from 9 choices = 9C2 = 36.
Too small.
Eliminate A, B and C.

Answer choice D: 10 vertices
Number of combinations of 2 that can be formed from 10 choices = 10C2 = 45.
Since these 45 combinations include the 10 sides of the polygon, we lose 10 options:
45-10 = 35.
Too small.

The correct answer is E.

Answer choice E: 11 vertices
Number of combinations of 2 that can be formed from 11 choices = 11C2 = 55.
Since these 55 combinations include the 11 sides of the polygon, we lose 11 options:
55-11 = 44.
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