A) 8
B) 9
C) 10
D) 7
Correct Answer: B
Solution :
[b] Let 'a' and "d" be the 1st term & common difference in A.P. respectively. Given, \[a={{120}^{{}^\circ }}\And d={{5}^{{}^\circ }}\] \[S=\frac{n}{2}[2a+(n-1).d]\] By Geometry, Sum of angles of the polygon be \[\left( 2n-4 \right)\times {{90}^{{}^\circ }}\] \[\therefore (2n-4)\times {{90}^{{}^\circ }}=\frac{n}{2}[2\times {{120}^{{}^\circ }}+(n-1)5]\] \[\Rightarrow {{180}^{{}^\circ }}(2n-4)=240n+5{{n}^{2}}-5n\] 5n2 + 240n - 5n - 360n + 720 = 0 \[\Rightarrow 5{{n}^{2}}-125n+720=0\Rightarrow {{n}^{2}}-25n+144=0\]\[\Rightarrow {{n}^{2}}-16n-9n+144=0\Rightarrow (n-16)(n-9)=0\] \[n=9,16\] \[\because n=16\] (It is not possible) \[\therefore n=9\] Hence, the no. of sides of the polygone, \[n=9.\] Hence, option [b] is correct
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