Question

A convex polygon has 44 diagonals, find the number of sides of this polygon.

Answer 1

Hint: A diagonal is drawn by taking two points of a polygon at a time. By joining two points of a polygon, we can get either a side of the polygon or a diagonal of the polygon. Assume a variable ‘n’ for the number of sides of the polygon, then the number of diagonals will be ${}^{n}{{C}_{2}}-n$. Equate ${}^{n}{{C}_{2}}-n$ with 44 and solve the obtained equation to get the value of n.

Complete step-by-step answer:
We have to find the number of sides of a polygon which has 44 diagonals.
Let us assume the number of sides of this polygon to be n. So, the number of vertices will also be ‘n’.
If we draw a line by joining any of these n points, the lines will either be a diagonal or a side of the polygon.
Total number of lines that can be drawn using these points = Total number of ways of selecting two points out of ‘n’ points.
We know that total number of ways of selecting r objects out of n objects ${}^{n}{{C}_{r}}$.
So, the total number of ways of selecting two points out of n points ${}^{n}{{C}_{2}}$.
And thus the total number of lines that can be made using n points (i.e. these n vertices of the polygon) $={}^{n}{{C}_{2}}$.
These lines will either be a diagonal of the polygon or a side of the polygon.
i.e. total number of line formed using n vertices = Number of diagonal + Number of sides
On putting total number of line formed using n vertices $={}^{n}{{C}_{2}}$ as calculated above and number of diagonals = 44 as given in question and number of sides = n as our assumption, we will get,
$\Rightarrow {}^{n}{{C}_{2}}=44+n$
Taking terms containing ‘n’ to LHS, we will get,
$\Rightarrow {}^{n}{{C}_{2}}-n=44$
We know,
$\begin{align}
  & {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\
 & \Rightarrow \dfrac{n!}{2!\left( n-2 \right)!}-n=44 \\
 & \Rightarrow \dfrac{n\left( n-1 \right) {\left( n-2 \right)!}}{2! {\left( n-2 \right)!}}-n=44\ \ \ \ \left[ We\ can\ write\ n!=n\left( n-1 \right)\left( n-2 \right)! \right] \\
 & \Rightarrow \dfrac{n\left( n-1 \right)}{2}-n=44 \\
\end{align}$
Taking LCM in LHS, we will get,
$\begin{align}
  & \dfrac{n\left( n-1 \right)-2n}{2}=44 \\
 & \dfrac{{{n}^{2}}-n-2n}{2}=44 \\
 & \dfrac{{{n}^{2}}-3n}{2}=44 \\
\end{align}$
Multiplying both sides of equation by ‘2’, we will get,
$\begin{align}
  & \Rightarrow {{n}^{2}}-3n=2\times 44 \\
 & \Rightarrow {{n}^{2}}-3n=88 \\
\end{align}$
Taking all the terms to one side of equation, we will get,
$\Rightarrow {{n}^{2}}-3n-88=0$
Now, we have got a quadratic equation, let us factorise this equation by splitting its middle term,
$\Rightarrow {{n}^{2}}-11n+8n-88=0$
Taking ‘n’ common from first two terms and ‘8’ common from last two terms, we will get,
$\Rightarrow n\left( n-11 \right)+8\left( n-11 \right)=0$
Taking $\left( n-11 \right)$ common, we will get,
$\begin{align}
  & \Rightarrow \left( n-11 \right)\left( n+8 \right)=0 \\
 & n=11,\ n=-8 \\
\end{align}$
‘n’ can’t be negative as ‘n’ is the number of sides of a polygon and number of sides can’t be negative.
Hence, the number of sides in a polygon which has 44 diagonals is 11.

Note: Shortcut: The number of diagonals in a convex polygon having ‘n’ sides is $\dfrac{n\left( n-3 \right)}{2}$.
Here, the number of diagonals is 44.
$\begin{align}
  & \Rightarrow \dfrac{n\left( n-3 \right)}{2}=44 \\
 & \Rightarrow \dfrac{{{n}^{2}}-3n}{2}=44 \\
 & \Rightarrow {{n}^{2}}-3n=88 \\
 & \Rightarrow {{n}^{2}}-3n-88=0 \\
 & \Rightarrow {{n}^{2}}-11n+8n-88=0 \\
 & \Rightarrow n\left( n-11 \right)+8\left( n-11 \right)=0 \\
 & \Rightarrow \left( n-11 \right)\left( n+8 \right)=0 \\
 & n=11\ ,n=-8 \\
\end{align}$
‘n’ can’t be negative. So, required n = 11.