Question

A polygon has 54 diagonals. Number of sides of the polygon is
A. 9
B. 12
C. 15
D. 16

Answer 1

Hint: We first establish the fact that the joining of any two points of the vertices of the polygon gives either diagonal or the sides of the polygon. We use the combinatorial formula of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to find the number of ways we can choose 2 points out of $n$ points.

Complete step-by-step answer:
We have to find the number of diagonals that can be formed by joining vertices of a polygon.
As the points are on a polygon, it is clear that joining any two points of the vertices will give either diagonal or the sides of the polygon.
We are going to use the concept of permutation to find the number of triangles.
We assume the number of sides of the polygon is $n$.
We know that the number of ways we can choose r things out of n things where they are not alike is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Here we have to choose 2 points out of $n$ points which can be done in ${}^{n}{{C}_{2}}=\dfrac{n!}{2!\times \left( n-2 \right)!}=\dfrac{n\left( n-1 \right)}{2}$ ways. But this number includes the number of sides of the polygon. Therefore, number of diagonals will be $\dfrac{n\left( n-1 \right)}{2}-n=\dfrac{n\left( n-3 \right)}{2}$
So, $\dfrac{n\left( n-3 \right)}{2}=54$ gives ${{n}^{2}}-3n-108=0$.
The roots are \[n=\dfrac{3\pm \sqrt{441}}{2}=\dfrac{3\pm 21}{2}=-9,12\]. The number of sides will eb 12 as it can’t be negative. The correct option is B.
So, the correct answer is “Option B”.

Note: The number of vertices, sides and angles are the same for a triangle. Therefore, the number of vertices decides the number of sides which help in combinational evaluation. If we had to find the diagonals in that case, we had to subtract the number of sides.