SOLUTION: Find the area of the regular 15-gon with radius 9 mm

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Question 1154051: Find the area of the regular 15-gon with radius 9 mm
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of the regular 15-gon with radius 9 mm
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It's 15 isosceles triangles, each with a central angle of 360/15 = 24 degs.
---> 30 right triangles with a hypotenuse of 9 mms and a smallest angle of 12 degs.
base b = 9*cos(12)
Height h = 9*sin(12)
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The area of each triangle is b*h/2

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
A=%281%2F2%29%28r%5E2%29%28sin%28360%2Fn%29%29%28n%29
A=%281%2F2%29%289%5E2%29%28sin%28360%2F15%29%29%2815%29
A=%281%2F2%29%2881%29%28sin%2824%29%29%2815%29
A≈ 247mm%5E2

or this way:
View post on imgur.com


Observe that m < AOB=360%2Fn=360%2F15=24, as there are n congruent triangle around O
and m < MOB=24%2F2=12
if n=15, m < theta=180%2F15=12

The sides of that triangle MBO are:
BO=r=9,
MB=r%2Asin%2812%29=9%2Asin%2812%29
OM=r%2Acos%2812%29=9%2Acos%2812%29


and area of the triangle MBO is:
%281%2F2%29%289%2Asin%2812%29%2A9%2Acos%2812%29%29=8.24
then double it to get area of the isosceles triangle ABO:
2%2A8.24=16.48

the area of the regular 15-gon is:
A=15%2A16.48=247.2

Therefore area of the regular 15-gon with 9mm radius is ≈ 247mm%5E2