The number of diagonals in an octagon is:

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(B) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.For an octagon,the number of sides $n = 8$.Substitu

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$20$

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(B) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.For an octagon,the number of sides $n = 8$.Substituting $n = 8$ into the formula:Number of diagonals $= \frac{8(8-3)}{2} = \frac{8 	imes 5}{2} = \frac{40}{2} = 20$.Alternatively,using combinations: The total number of ways to choose $2$ vertices out of $8$ is $^8C_2 = \frac{8 	imes 7}{2} = 28$. Subtracting the $8$ sides of the octagon,we get $28 - 8 = 20$ diagonals.

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