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A polygon has 44 diagonals. The number of its sides is
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The correct answer is:
11
No. of diagonals in a polygen $={ }^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$
$\Rightarrow 44={ }^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$
$\Rightarrow 44=\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}-\mathrm{n}$
$\Rightarrow \quad 44=\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}$
$\Rightarrow \quad 44=\frac{\mathrm{n}(\mathrm{n}-3)}{2}$
$\Rightarrow \mathrm{n}^{2}-3 \mathrm{n}-88=0$
$\Rightarrow(\mathrm{n}-11)(\mathrm{n}+8)=0$
$\mathrm{n} \neq-8$
$\mathrm{n}=11$
$\Rightarrow 44={ }^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$
$\Rightarrow 44=\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}-\mathrm{n}$
$\Rightarrow \quad 44=\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}$
$\Rightarrow \quad 44=\frac{\mathrm{n}(\mathrm{n}-3)}{2}$
$\Rightarrow \mathrm{n}^{2}-3 \mathrm{n}-88=0$
$\Rightarrow(\mathrm{n}-11)(\mathrm{n}+8)=0$
$\mathrm{n} \neq-8$
$\mathrm{n}=11$
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