anonymous
  • anonymous
how do I figure out area of a polygon with perimeter of 108
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
It depends. Is the polygon regular (that is, does it have equal internal angles and equal sides)?
xkehaulanix
  • xkehaulanix
Is it a regular polygon?
anonymous
  • anonymous
It's a 9 sided polygon

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anonymous
  • anonymous
Okay, it sounds like it's regular, since 9 divides 108. If it's regular, you can look at the shape and break it up into 9 isosceles triangles. You can find the area using a combination of cosine rule (to find the other sides of the isosceles triangle) and then formula for area of a triangle (1/2 ab sinC). The angle C will be 360degrees/(number of sides). The formula for the area will be \[A=\frac{na^2\sin (360^o/n)}{4(1-\cos(360^o/n))}\]where n is the number of sides, a is the side length. Here, \[a=\frac{108}{9}=12\] and n = 9, so\[A=\frac{9 \times 12^2.\sin(40^o)}{4 \times (1-\cos(40^o))} \approx 890 \]square units. If you want a proper derivation of the formula, let me know. I have to rush right now. Also, the formula only holds for regular polygons; that is, polygons that have the same side length and equal internal angles.

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