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AdhaZeera

  • 2 years ago

What area can the goat roam?

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  1. AdhaZeera
    • 2 years ago
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    |dw:1350884315409:dw|

  2. AdhaZeera
    • 2 years ago
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    Fence SHED 4feet 5feet 8feet

  3. UnkleRhaukus
    • 2 years ago
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    |dw:1350884748553:dw|

  4. UnkleRhaukus
    • 2 years ago
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    ops i dident see the fence

  5. Fellowroot
    • 2 years ago
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    Can he go inside the shed? And is he/she connected by a chain that is connected to the corner of the shed?

  6. UnkleRhaukus
    • 2 years ago
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    |dw:1350884878590:dw|

  7. AdhaZeera
    • 2 years ago
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    No he's not connected. I'm not sure. That's all my teacher wrote

  8. Fellowroot
    • 2 years ago
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    he has to be connected by a chain to the shed right?

  9. AdhaZeera
    • 2 years ago
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    Wait... yes..

  10. UnkleRhaukus
    • 2 years ago
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    |dw:1350885122452:dw|

  11. UnkleRhaukus
    • 2 years ago
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    |dw:1350885496696:dw|

  12. AdhaZeera
    • 2 years ago
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    How do I calculate it?

  13. UnkleRhaukus
    • 2 years ago
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    break the shape in to smaller shapes

  14. Fellowroot
    • 2 years ago
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    Unkle Rhaukus I like your drawings, but what is the actual value?

  15. AdhaZeera
    • 2 years ago
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    81 sq ft??

  16. Fellowroot
    • 2 years ago
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    I'm very sure I know how to do it. Just give me some time to work it out.

  17. Fellowroot
    • 2 years ago
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    its not easy, but i'm still working.

  18. Fellowroot
    • 2 years ago
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    Ok, here it comes. Writing it up. :) !!!!!

  19. AdhaZeera
    • 2 years ago
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    You there?

  20. Fellowroot
    • 2 years ago
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    yeah hang on almost done

  21. Fellowroot
    • 2 years ago
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    you might be surprised but i got it.

  22. Fellowroot
    • 2 years ago
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    GOT IT!!!!

  23. UnkleRhaukus
    • 2 years ago
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    im not sure why you had to integrate

  24. AdhaZeera
    • 2 years ago
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    Thank you!

  25. UnkleRhaukus
    • 2 years ago
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    we know the are of a circle is πr^2, so a semicircle has area (πr^2)/2 and a quater circle has area (πr^2)/4

  26. Fellowroot
    • 2 years ago
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    I integrated because I wanted to take the whole circle and then subtract the parts where the goat can't go. The area of the top of the circle is curved and the goat can't go there. So I created a function and integrated it so that I could find that exact area that is fenced off.

  27. UnkleRhaukus
    • 2 years ago
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    |dw:1350888457508:dw|

  28. Fellowroot
    • 2 years ago
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    @ AdhaZeera Do you understand what I did when I integrated. I will explain it to you if you want. I used calculus to arrive at that answer.

  29. UnkleRhaukus
    • 2 years ago
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    |dw:1350888515709:dw|

  30. UnkleRhaukus
    • 2 years ago
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    the sector is 1/12 of a circle

  31. Fellowroot
    • 2 years ago
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    @ UnkleRhaukus I know that there is a formula that is A= (theta)r^2/2 but I don't how it works here.

  32. UnkleRhaukus
    • 2 years ago
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    \[A=\frac{\pi\times 3^2}4+\frac{\pi \times8^2}{2}+\frac{\pi \times8^2}{12}+\frac{4\times 8}{2}\]\[=\pi\times\frac{465}{12}+16\]\[\approx 140\]

  33. UnkleRhaukus
    • 2 years ago
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    well theta was 30° right? angles of segment / angle in circle \[30°/360°=1/12\] area of segment / area of circle \[\frac{\pi r^2}{12}\qquad/\qquad\pi r^2\]

  34. UnkleRhaukus
    • 2 years ago
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    i had to use a trig ratio to to get 30° btw

  35. Fellowroot
    • 2 years ago
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    Your value comes to 140.35 mine is 141.7, its off only by a hair and is a very good approximation, but mine is exact.

  36. UnkleRhaukus
    • 2 years ago
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    where did i error?

  37. Fellowroot
    • 2 years ago
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    not sure, but its pretty darn good.

  38. UnkleRhaukus
    • 2 years ago
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    \[A=\frac{\pi\times 3^2}4+\frac{\pi \times8^2}{2}+\frac{\pi \times8^2}{12}+\frac{4\times 8}{2}=\pi\times\frac{475}{12}+16\approx140.35\]

  39. UnkleRhaukus
    • 2 years ago
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    @Fellowroot i think we got different answers because we were solving different regions our pictures are different

  40. Fellowroot
    • 2 years ago
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    |dw:1350891993500:dw| The only thing I don't know if that if the 5 feet fence touches the edge of the circle or if there is a gab. For the solution i provided it doesn't matter though. On my pic, the 'x', I really don't know that distance, it may make a difference for your answer, i don't know.