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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?

 Jun 20, 2024

Best Answer 

 #1
avatar+1252 
+1

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! :)

 Jun 20, 2024
 #1
avatar+1252 
+1
Best Answer

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! :)

NotThatSmart Jun 20, 2024

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