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A polygon with n sides has n*(n-3)/2 diagonals.

So you need to solve n*(n-3)/2 = 54

n2 - 3n - 108 = 0 which has the solutions n = 12 or n = -9.

Since a polygon cannot have a negative number of sides, the answer is 12.

A polygon with n sides has n*(n-3)/2 diagonals.

So you need to solve n*(n-3)/2 = 54

n2 - 3n - 108 = 0 which has the solutions n = 12 or n = -9.

Since a polygon cannot have a negative number of sides, the answer is 12.

A polygon with n sides has n*(n-3)/2 diagonals.

So you need to solve n*(n-3)/2 = 54

n2 - 3n - 108 = 0 which has the solutions n = 12 or n = -9.

Since a polygon cannot have a negative number of sides, the answer is 12.

A polygon with n sides has n*(n-3)/2 diagonals.

So you need to solve n*(n-3)/2 = 54

n2 - 3n - 108 = 0 which has the solutions n = 12 or n = -9.

Since a polygon cannot have a negative number of sides, the answer is 12.

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12y ago

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More answers

A polygon with n sides has n*(n-3)/2 diagonals.

So you need to solve n*(n-3)/2 = 54

n2 - 3n - 108 = 0 which has the solutions n = 12 or n = -9.

Since a polygon cannot have a negative number of sides, the answer is 12.

User Avatar

Wiki User

12y ago
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Q: What kind of polygon has 54 diagonals?
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