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Question

Answers

$A.$ $11$

$B.$ $7$

$C.$ $8$

$D.$ $None{\text{ }}of{\text{ }}these$

Answer
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.

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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