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# Geometry

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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

#1
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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.

Thanks! :)

Jun 20, 2024

#1
+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.