What is the area of a regular decagon with a radius of 8 centimeters?

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What is the area of a regular decagon with a radius of 8 centimeters?

Mathematics
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did you get this yet?
No, I can't find a formula that I understand.
do you know trig or are you using something else?

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Other answers:

Do not know trig! Trying my hardest, I have difficulties with even simple math.
i am not sure i can do it without trig. maybe Anwar?
oops he just left.
:/ Can you explain it to me? If you can explain the formula, I can do that.
ok you need to think of splitting it up in to right triangles. you will have twenty of them.
wait i have a better idea, because i see this one involves trig as well. so we might as well do this. the formula for the area of a triangle is \[\frac{1}{2}bcsin(A)\] where A is the angle between the sides. in this case the sides are both of length 8 and the angle is 360/10=36
so unless i miss my guess each of these 10 triangles (not 20, just ten) has area \[32sin(36)=18.809\]
multiply by 10 to get the total area. now i am going to check to see if this is correct. but i cannot see a way to do it without using some trig.
Thank you for trying so hard, though, and this I can understand fairly easy.
welcome, but don't quote me!
yea but i think i am wrong. i will keep trying.
ok let me try again. is this for a trig class?
in any case break this thing up into 20 right triangles, each with hypotenuse 8. the small angle between the long sides of those triangle is 18 degrees because it is 360/20
call the long side of those triangles b \[cos(18)=\frac{b}{10}\] so the long side is \[b=10cos(18)=9.51\] rounded. the last side we find by pythagoras since we know the hypotenuse and the long side. the short side is \[\sqrt{10^2-9.51^2}=3.09\] rounded.
the area of each of those twenty triangles is therefore \[\frac{1}{2} 9.21 \times3.09\]
since there are twenty of them just compute \[10\times 9.21 \times 3.09\]
i know this answer is correct (i mean i know the area is) because i did checked it here: http://www.cleavebooks.co.uk/scol/calpolyg.htm
i see that i used 10 instead of 8. you should use 8. but i also got the same answer using the sine method so forget it.

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