Question

What is the sum of the interior angles of a polygon that has a total of 3740 diagonals?

Answer 1

`15480°`

Explanation:

I guess I will just answer my own question after two months.

The equation for diagonals:

`D=(n(n-3))/2`
`n=` number of sides the polgyon has

Now let’s plug in `3740` as `D` in the equation.

`3740=(n(n-3))/2`

To cancel the denominator `2` out, multiply both sides by `2`:

`7480= n^2 -3n`

Let’s have both the variables and the constant on one side set to `=0`. To do this, you subtract `7480` from both sides.

`n^2 -3n -7480 =0`

Factor the equation out:

`(n+85)(n-88)`

Set `n+85=0` and `n-88=0`

`n=-85` or `n=88`

We cannot use `n=-85` in our search for the sum of interior angles because a polygon cannot have negative sides. So let’s use
`n=88`.

This is the equation to find the sum of interior angles:

`180(n-2)`

Now that we have the number of sides of the polygon, we can plug it into the equation:

`180(88-2)`

`15480°`

Hope this helps someone out!