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A polygon has $44$diagonals, then the number of its sides are
$A.$ $11$
$B.$ $7$
$C.$ $8$
$D.$ $None{\text{ }}of{\text{ }}these$

Answer Verified Verified
Hint: Here we have to select two such points on joining them we create a diagonal .So for selection We Use combination to find no. of diagonals.
Let the polygon have n sides.
This means that the polygon has n vertices.
No. of diagonals of the polygon $ = $ No. of ways two vertices could be connected by a line segment $ - $
                                                              No. of sides of the polygon
                                                          $ = c\left( {n,2} \right) - n{\text{ - }}\left( 1 \right)$

We know that,
\[
  c\left( {n,r} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} \\
   \Rightarrow c\left( {n,2} \right) = \dfrac{{n!}}{{\left( {n - 2} \right)!2!}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!2!}} = \dfrac{{n\left( {n - 1} \right)}}{2} \\
 \]
Using the above obtained formula in equation $\left( 1 \right)$
No. of diagonals of the polygon $ = c\left( {n,2} \right) - n = \dfrac{{n\left( {n - 1} \right)}}{2} - n = \dfrac{{n\left( {n - 3} \right)}}{2}$
Given in the problem the polygon has $44$diagonals.
Using the above formula, we get,
$
  44 = \dfrac{{n\left( {n - 3} \right)}}{2} \\
   \Rightarrow 88 = {n^2} - 3n \\
   \Rightarrow {n^2} - 3n - 88 = 0 \\
$
Factoring the above quadratic equation, we get,
$
  {n^2} - 11n + 8n - 88 = 0 \\
   \Rightarrow n\left( {n - 11} \right) + 8\left( {n - 11} \right) = 0 \\
   \Rightarrow \left( {n + 8} \right)\left( {n - 11} \right) = 0 \\
$
From the above equation either $n = - 8$ or $n = 11$.
Since the no. of sides of a polygon is a positive real integer, we neglect the value $n = - 8$
$ \Rightarrow n = 11$
Therefore, the number of sides of the polygon with $44$diagonals is $11$.
Hence option A is correct.

Note: Try to analyse the problems of above type using permutations and combinations. Sides of the polygon are to be excluded from the combination as mentioned in the above solution.



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