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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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A convex polygon has 44 diagonals.Find the number of its sides

$\begin{array}{1 1}(A)\;9\\(B)\;10\\(C)\;11\\(D)\;12\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Let the number of sides of polygon =n
$\therefore$ Number of angular points =n
$ \therefore$ Number of straight lines joining any two of these n points =$nC_2$
Now the number of sides of the polygon =n
$\therefore$ Number of diagonals =$nC_2-n$
But it is given the number of diagonals =44
$nC_2-n=44$
$\large\frac{n(n-1)}{2}$$-n=44$
$n^2-n-2n=88$
$n^2-3n-88=0$
$(n-11)(n+8)=0$
$n=11$ or $n=-8$
Rejecting negative quantity,
$n=11$
Hence the required number of sides =11
Hence (C) is the correct answer.
answered May 15, 2014 by sreemathi.v
 

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