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

Jun 11, 2018

15480°

#### Explanation:

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

The equation for diagonals:

$D = \frac{n \left(n - 3\right)}{2}$
$n =$ number of sides the polgyon has

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

$3740 = \frac{n \left(n - 3\right)}{2}$

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

$7480 = {n}^{2} - 3 n$

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

${n}^{2} - 3 n - 7480 = 0$

Factor the equation out:

$\left(n + 85\right) \left(n - 88\right)$

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:

$180 \left(n - 2\right)$

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

$180 \left(88 - 2\right)$

15480°

Hope this helps someone out!