If the number of diagonal lines of a polygon is exactly 3 times of the number of sides of the polygon, then what is the sum of the inner angles of the polygon?

If the number of diagonal lines of a polygon is exactly 3 times of the number of sides of the polygon, then what is the sum of the inner angles of the polygon?

If the diagonal of n-polygon is n (n-3) / 2, then:
n(n－3)/2=3n
n=9
This is a 9-sided shape, then the sum of internal angles (n-2) × 180 ° = 1260 °

What is the formula for the number of diagonal lines of a polygon
How to find the number of diagonal lines of polygon?
What is the formula
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The number of diagonals of each vertex is n-3 if n-polygon is set
There are n * (n-3) vertices
So each vertex is evaluated twice
So divide by two
The result is n * (n-3) / 2

If one side of the rectangle is xcm, its area is SCM;
(1) Please use the formula with X
Constant and variable are pointed out
I looked at the reference answer, s = x (5-x). How did I get it?

Let the other side of the rectangle be y
Then the perimeter of the rectangle is 2 × (x + y) = 10 (1)
The area of a rectangle is xy = s (2)
It can be obtained from formula (1) and (2) by eliminating y
s=xy=x（5-x）