+312 The number of diagonals in a certain regular polygon is equal to $2$ times the number of sides. How many sides does this polygon have?
+2159
The number of diagonals in a certain regular polygon is equal to $2$ times the number of sides. How many sides does this polygon have?
d = s(s – 3) / 2
d = s(s – 3) / 2
2s = s(s – 3) / 2
2s = (s2 – 3s) / 2
4s = (s2 – 3s)
4s = s2 – 3s
s2 – 3s – 4s = 0
s2 – 7s = 0
s • (s – 7) = 0
s = 0 discard
s = 7
.
+2 Hey CInsbstnkng!
The number of diagonals of a polygon can be defined as \(d = \frac{n(n-3)}{2}\) where n is the number of sides (for any amount of sides over 3)
so then you can form an equation like so:
\(2n = \frac{n(n-3)}{2}\)
this sets 2 times the sides (n) against the number of diagonals.
Hope this helps!
(If you are interested in how to find the diagonal formula you can look here https://byjus.com/diagonal-of-polygon-formula)
+2159
The number of diagonals in a certain regular polygon is equal to $2$ times the number of sides. How many sides does this polygon have?
The number of diagonals in a polygon with s sides is d = s(s – 3) / 2.
d = s(s – 3) / 2
2s = s(s – 3) / 2
2s = (s2 – 3s) / 2
4s = (s2 – 3s)
4s = s2 – 3s
s2 – 3s – 4s = 0
s2 – 7s = 0
s • (s – 7) = 0
s = 0 discard
s = 7
.
+2159
The number of diagonals in a certain regular polygon is equal to $2$ times the number of sides. How many sides does this polygon have?
d = s(s – 3) / 2
d = s(s – 3) / 2
2s = s(s – 3) / 2
2s = (s2 – 3s) / 2
4s = (s2 – 3s)
4s = s2 – 3s
s2 – 3s – 4s = 0
s2 – 7s = 0
s • (s – 7) = 0
s = 0 discard
s = 7
.
