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I've used this formula to calculate the solution.

The number of diagonals of a polygon is $(n(n-3))/2$, where $n$ is the number of sides of a polygon.

But I'm getting answer as decimal point. Is the question wrong or am I doing any calculation mistake?

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  • $\begingroup$ Sorry about that. The question is wrong. $\endgroup$
    – soupless
    Commented Mar 28, 2022 at 13:16

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Obviously the question is worded wrongly. In any polygon, the number of diagonals is one less than a triangular number, and 60+1=61 is not triangular.

In three dimensions, a regular dodecahedron has 60 face diagonals, plus 100 internal (body) diagonals.

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    $\begingroup$ Another way to see it can't be 60 is that the number of diagonals can't be divisible by 3 without also being divisible by 9. $\endgroup$
    – eyeballfrog
    Commented Mar 28, 2022 at 13:43

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