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?
We can write an equation to solve this problem.
First, the equation for the number of diagonals in a polygon is \(n(n-3)/2\) where n is the number of sides or vertices.
Thus, we can write the equation
\(n(n-3)/2=2n\)
Now, we simply solve for n.
\(n^2-3n=4n\\ n^2-7n=0\)
Using the quadratic equation, we get
\(n=\frac{7\pm \sqrt{(-7)^{2}-4\cdot 1\cdot 0}}{2\cdot 1}\\ \)
From this, we get
\(n=7\\ n=0\)
So we have 7 sides.
So 7 is our answer.
Thanks! :)
We can write an equation to solve this problem.
First, the equation for the number of diagonals in a polygon is \(n(n-3)/2\) where n is the number of sides or vertices.
Thus, we can write the equation
\(n(n-3)/2=2n\)
Now, we simply solve for n.
\(n^2-3n=4n\\ n^2-7n=0\)
Using the quadratic equation, we get
\(n=\frac{7\pm \sqrt{(-7)^{2}-4\cdot 1\cdot 0}}{2\cdot 1}\\ \)
From this, we get
\(n=7\\ n=0\)
So we have 7 sides.
So 7 is our answer.
Thanks! :)