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

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Answer 1 · 2018-06-11T19:37:41

#15480°#

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!