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The number of diagonals in a certain regular polygon is equal to $1/2$ times the number of sides.  How many sides does this polygon have?

 Jan 7, 2024

Best Answer 

 #1
avatar+36923 
+1

Recall the formula for the number of diagonals of a polygon

D= n * (n-3) /2       where n is the number of sides and D is the number of diagonals

So you have this: 

1/2 n = n * (n-3)/2        <==== solve for 'n'

1/2n = ( n^2 - 3n )/2      multiply through by 2

n = n^2 - 3n 

n^2 - 4n = 0 

n(n-4) = 0     shows n = 0  (throw out) or n = 4 sides     ( a quadrilateral.....square, rectangle etc) 

 Jan 7, 2024
 #1
avatar+36923 
+1
Best Answer

Recall the formula for the number of diagonals of a polygon

D= n * (n-3) /2       where n is the number of sides and D is the number of diagonals

So you have this: 

1/2 n = n * (n-3)/2        <==== solve for 'n'

1/2n = ( n^2 - 3n )/2      multiply through by 2

n = n^2 - 3n 

n^2 - 4n = 0 

n(n-4) = 0     shows n = 0  (throw out) or n = 4 sides     ( a quadrilateral.....square, rectangle etc) 

ElectricPavlov Jan 7, 2024

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