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Find the number of sides in a polygon if it has 65 diagonals.

 

Option: 1

11 
 


Option: 2

15 


Option: 3

12 

 


Option: 4

13


Answers (1)

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In this case, let's say that the polygon has n sides. Therefore, there will also be 'n' vertices.

If we connect any of these n points to form a line, the result will either be a diagonal or a side of the polygon.

Thus, there are \mathrm{^{n}C_{2}}methods in total to choose two points from a total of n points.

The total number of lines formed using n vertices = Number of diagonal + Number of sides.

\mathrm{\begin{aligned} &{ }^n C_2=65+n\\ &{ }^n C_2-65=n\\ &\frac{n !}{2 !(n-2) !}-65=n \end{aligned}}

We can write \mathrm{\mathrm{n!}\: \: as\: \: n !=n(n-1)(n-2) !}
\mathrm{ \begin{aligned} & \frac{n(n-1)(n-2) !}{2 !(n-2) !}-n=65 \\ & \frac{n(n-1)}{2}-n=65 \\ & \frac{n^2-3 n}{2}=65 \end{aligned} }
Rearranging the equation to get,

\mathrm{ n^2-3 n-130=0 }

Solving the quadratic equation to get,

n = 13 and n = -10 

The value of n cannot be negative.

Therefore, the number of sides in a polygon is 13.

 

 

 

Posted by

manish painkra

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