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એક બહુકોણમાં બે ક્રમિક અંતઃકોણોનો તફાવત $5^{\circ}$ છે. જો સૌથી નાનો ખૂણો $120^{\circ}$ નો હોય, તો તે બહુકોણની બાજુઓની સંખ્યા શોધો.
Solution
The angles of the polygon will form an $A.P.$ with common difference $d$ as $5^{\circ}$ and first term $a$ as $120^{\circ}$
It is known that the sum of all angles of a polygon with $n$ sides is $180(n-2)$
$\therefore S_{n}=180^{\circ}(n-2)$
$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=180^{\circ}(n-2)$
$\Rightarrow \frac{n}{2}\left[240^{\circ}+(n-1) 5^{\circ}\right]=180^{\circ}(n-2)$
$\Rightarrow n[240+(n-1) 5]=360(n-2)$
$\Rightarrow 240 n+5 n^{2}-5 n=360 n-720$
$\Rightarrow 5 n^{2}-125 n+720=0$
$\Rightarrow n^{2}-25 n+144=0$
$\Rightarrow n^{2}-16 n-9 n+144=0$
$\Rightarrow n(n-16)-9(n-16)=0$
$\Rightarrow(n-9)(n-16)=0$
$\Rightarrow n=9$ or $16$
