Nine-gon

Calculate the perimeter of a regular nonagon (9-gon) inscribed in a circle with a radius 13 cm.

Result

x =  80.03 cm

Solution:

θ=1809=π9=20 a=2rsin(π9)=8.89 cm x=9a=80.03 cm\theta = \dfrac{ 180 ^\circ }{ 9} = \dfrac{ \pi}{ 9} = 20 ^\circ \ \\ a = 2r \sin( \dfrac{ \pi}{ 9} ) = 8.89 \ cm \ \\ x = 9a = 80.03 \ \text{cm}



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