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# Diagonals

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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

#1
+36923
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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
+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)

ElectricPavlov Jan 7, 2024