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The no. of diagonals of $n$ sided polygon =?

$\begin{array}{1 1}(A) \;\large\frac{n(n-1)}{2} \\ (B) \large\frac{n(n-1)(n-2)}{6} \\ (C) \;\large\frac{n(n-3)}{2} \\(D) \;n(n-1) \end{array}$

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Total number of lines that can be drawn using the $n$ vertices of the polygon
$=^nC_2=\large\frac{n.(n-1)}{2}$.
Out of these lines $n$ lines are the $n$ sides of the polygon.
The remaining lines are diagonals of the poygon.
$\therefore$ The no. of diagonals $=^nC_2-n=\large\frac{n.(n-1)}{2}$$-n$
$=\large\frac{n.(n-3)}{2}$
answered Aug 13, 2013 by rvidyagovindarajan_1
 

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