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Question: A polygon has \(44\)diagonals, then the number of its sides are \(A.\) \(11\) \(B.\) \(7\) \(C...

A polygon has 4444diagonals, then the number of its sides are
A.A. 1111
B.B. 77
C.C. 88
D.D. None of theseNone{\text{ }}of{\text{ }}these

Explanation

Solution

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(n,2)−n - (1)= c\left( {n,2} \right) - n{\text{ - }}\left( 1 \right)

We know that,

c(n,r)=n!(n−r)!r! ⇒c(n,2)=n!(n−2)!2!=n(n−1)(n−2)!(n−2)!2!=n(n−1)2  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 (1)\left( 1 \right)
No. of diagonals of the polygon =c(n,2)−n=n(n−1)2−n=n(n−3)2 = 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 4444diagonals.
Using the above formula, we get,
44=n(n−3)2 ⇒88=n2−3n ⇒n2−3n−88=0  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,
n2−11n+8n−88=0 ⇒n(n−11)+8(n−11)=0 ⇒(n+8)(n−11)=0  {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=−8n = - 8 or n=11n = 11.
Since the no. of sides of a polygon is a positive real integer, we neglect the value n=−8n = - 8
⇒n=11\Rightarrow n = 11
Therefore, the number of sides of the polygon with 4444diagonals is 1111.
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.