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If a convex polygon has 44 diagonals, then find the number of its sides.
Solution:
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Verified Answer
Let the convex polygon has $n$ sides
$\therefore$ Number of diagonals $={ }^n C_2-n$
According to the question,
$$
\begin{aligned}
&{ }^n C_2-n=44 \\
&\Rightarrow \frac{n !}{2 !(n-2) !}-n=44 \Rightarrow \frac{n(n-1)}{2}-n=44 \\
&\Rightarrow n\left[\frac{(n-1)}{2}-1\right]=44 \Rightarrow n\left(\frac{n-1-2}{2}\right)=44 \\
&\Rightarrow n(n-3)=44 \times 2 \Rightarrow n(n-3)=88 \\
&\Rightarrow n^2-3 n-88=0 \quad[\because n \neq-8] \\
&\Rightarrow n=11,-8 \quad[(n-11)(n+8)=0 \\
&\therefore n=11
\end{aligned}
$$
$\therefore$ Number of diagonals $={ }^n C_2-n$
According to the question,
$$
\begin{aligned}
&{ }^n C_2-n=44 \\
&\Rightarrow \frac{n !}{2 !(n-2) !}-n=44 \Rightarrow \frac{n(n-1)}{2}-n=44 \\
&\Rightarrow n\left[\frac{(n-1)}{2}-1\right]=44 \Rightarrow n\left(\frac{n-1-2}{2}\right)=44 \\
&\Rightarrow n(n-3)=44 \times 2 \Rightarrow n(n-3)=88 \\
&\Rightarrow n^2-3 n-88=0 \quad[\because n \neq-8] \\
&\Rightarrow n=11,-8 \quad[(n-11)(n+8)=0 \\
&\therefore n=11
\end{aligned}
$$
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