How many sides does a polygon have, if it has 44 diagonals?

There is a rule for this:
For any geometric figure, the number of sides is equal to the number of vertexes. Let "n" be that number.

At first, you may think that n.n is the number of diagonals but that is not the case for two reasons.

First, with such procedure, you are counting each line joining two vertexes, including the edges of the figure, which are not diagonals. So, for each vertex, you need to discard two lateral lines. That is n-2. In addition, you need to discard the same vertex you are focusing on. That is n-2-1.

So far, we have n vertexes, each with n-1-2 diagonals coming from it, but that is not the end for a second reason. With n(n-2-1) you are counting each diagonal twice; so divide the number by 2.

The formula is then: n(n-3)/2.

For this question, we have that n(n-3)/2=44. Let's operate.

n(n-3)=88, you can solve the quadratic equation n^2-3n-88 and find that n is equal to 11 (remember that no negative number applies here); or you can notice that 88 must be the result of multiplying 8 x 11. So 11 is the answer.

Hi prab Bunuel
Please provide options to the question.

Asked: A polygon has 44 diagonals, and then numbers of sides are?

If a polygon has n sides
the number of diagonals = n(n-3)/2 = 44;
n^ - 3n = 88 ;
n^2 - 3n - 88 = 0
n^2 - 11n + 8n - 88 = 0
(n-11)(n+8) = 0
n = 11
numbers of sides = n = 11

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