
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
535.2k+ views
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.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
What is the difference between superposition and e class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
