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# Question and Answer

The number of diagonals in a polygon of $$n$$ sides is
（ ）
A. $$\frac {n(n-1)}{2}$$.
B. $$\frac {n(n-2)}{2}$$.
C. $$\frac {n(n-3)}{2}$$.
D. $$n(n-3)$$.
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## QuestionMathsClass 8

The number of diagonals in a polygon of $$n$$ sides is
（ ）
A. $$\frac {n(n-1)}{2}$$.
B. $$\frac {n(n-2)}{2}$$.
C. $$\frac {n(n-3)}{2}$$.
D. $$n(n-3)$$.

## Answer

C
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## Solution

The number of diagonals in a polygon of $$n$$ sides is $$\frac {n(n-3)}{2}$$. For example: In quadrilateral we know that there are two diagonals.
Now, put $$n=4$$ in the formula $$\frac {n(n-3)}{2}$$,
$$=\frac {4(4-3)}{2}$$
$$=\frac {4\times1}{2}$$
$$=2$$
Take another example: we know that in a pentagon number of diagonals are $$5$$.Now, put $$n=5$$ in the formula $$\frac {n(n-3)}{2}$$,
$$=\frac {5(5-3)}{2}$$
$$=\frac {5\times 2}{2}$$
$$=5$$.
So, the number of diagonals in a polygon of $$n$$ sides is $$\frac {n(n-3)}{2}$$.
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