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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    did you get this yet?

  2. anonymous
    • 5 years ago
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    No, I can't find a formula that I understand.

  3. anonymous
    • 5 years ago
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    do you know trig or are you using something else?

  4. anonymous
    • 5 years ago
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    Do not know trig! Trying my hardest, I have difficulties with even simple math.

  5. anonymous
    • 5 years ago
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    i am not sure i can do it without trig. maybe Anwar?

  6. anonymous
    • 5 years ago
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    oops he just left.

  7. anonymous
    • 5 years ago
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    :/ Can you explain it to me? If you can explain the formula, I can do that.

  8. anonymous
    • 5 years ago
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    ok you need to think of splitting it up in to right triangles. you will have twenty of them.

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    so unless i miss my guess each of these 10 triangles (not 20, just ten) has area \[32sin(36)=18.809\]

  11. anonymous
    • 5 years ago
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    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.

  12. anonymous
    • 5 years ago
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    Thank you for trying so hard, though, and this I can understand fairly easy.

  13. anonymous
    • 5 years ago
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    welcome, but don't quote me!

  14. anonymous
    • 5 years ago
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    yea but i think i am wrong. i will keep trying.

  15. anonymous
    • 5 years ago
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    ok let me try again. is this for a trig class?

  16. anonymous
    • 5 years ago
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    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

  17. anonymous
    • 5 years ago
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    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.

  18. anonymous
    • 5 years ago
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    the area of each of those twenty triangles is therefore \[\frac{1}{2} 9.21 \times3.09\]

  19. anonymous
    • 5 years ago
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    since there are twenty of them just compute \[10\times 9.21 \times 3.09\]

  20. anonymous
    • 5 years ago
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    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

  21. anonymous
    • 5 years ago
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    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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