<PRE><B>Question 1154051</B>
{{{A=(1/2)(r^2)(sin(360/n))(n)}}}

{{{A=(1/2)(9^2)(sin(360/15))(15)}}}

{{{A=(1/2)(81)(sin(24))(15)}}}

{{{A}}}≈ {{{247mm^2}}}


or this way:

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Observe that m &lt; {{{AOB=360/n=360/15=24}}}, as there are {{{n}}} congruent triangle around {{{O}}}
and m &lt; {{{MOB=24/2=12}}}

if {{{n=15}}}, m &lt; {{{theta=180/15=12}}}


The sides of that triangle {{{MBO}}} are: 
 {{{BO=r=9}}},  
 {{{MB=r*sin(12)=9*sin(12)}}} 
 {{{OM=r*cos(12)=9*cos(12)}}} 



and area of the triangle {{{MBO}}} is:

 {{{(1/2)(9*sin(12)*9*cos(12))=8.24}}}

then double it to get area of the isosceles triangle {{{ABO}}}: 
{{{2*8.24=16.48}}}


the area of the regular 15-gon is: 

{{{A=15*16.48=247.2}}}


Therefore area of the regular 15-gon with {{{9mm}}} radius is ≈ {{{247mm^2}}}


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