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No not normally.

The formula of the number of diagonals is (n/2)(n - 3).

(n/2)(n - 3) = 10

n(n - 3) = 20

n2 - 3n - 20 = 0 we cannot factor it as (n - r1)(n - r2), where r1 and r2 are natural numbers, so there is any polygon with 10 diagonals.

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