anonymous
  • anonymous
What area can the goat roam?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1350884315409:dw|
anonymous
  • anonymous
Fence SHED 4feet 5feet 8feet
UnkleRhaukus
  • UnkleRhaukus
|dw:1350884748553:dw|

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More answers

UnkleRhaukus
  • UnkleRhaukus
ops i dident see the fence
Fellowroot
  • Fellowroot
Can he go inside the shed? And is he/she connected by a chain that is connected to the corner of the shed?
UnkleRhaukus
  • UnkleRhaukus
|dw:1350884878590:dw|
anonymous
  • anonymous
No he's not connected. I'm not sure. That's all my teacher wrote
Fellowroot
  • Fellowroot
he has to be connected by a chain to the shed right?
anonymous
  • anonymous
Wait... yes..
UnkleRhaukus
  • UnkleRhaukus
|dw:1350885122452:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1350885496696:dw|
anonymous
  • anonymous
How do I calculate it?
UnkleRhaukus
  • UnkleRhaukus
break the shape in to smaller shapes
Fellowroot
  • Fellowroot
Unkle Rhaukus I like your drawings, but what is the actual value?
anonymous
  • anonymous
81 sq ft??
Fellowroot
  • Fellowroot
I'm very sure I know how to do it. Just give me some time to work it out.
Fellowroot
  • Fellowroot
its not easy, but i'm still working.
Fellowroot
  • Fellowroot
Ok, here it comes. Writing it up. :) !!!!!
anonymous
  • anonymous
You there?
Fellowroot
  • Fellowroot
yeah hang on almost done
Fellowroot
  • Fellowroot
you might be surprised but i got it.
Fellowroot
  • Fellowroot
UnkleRhaukus
  • UnkleRhaukus
im not sure why you had to integrate
anonymous
  • anonymous
Thank you!
UnkleRhaukus
  • UnkleRhaukus
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
Fellowroot
  • Fellowroot
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.
UnkleRhaukus
  • UnkleRhaukus
|dw:1350888457508:dw|
Fellowroot
  • Fellowroot
@ 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.
UnkleRhaukus
  • UnkleRhaukus
|dw:1350888515709:dw|
UnkleRhaukus
  • UnkleRhaukus
the sector is 1/12 of a circle
Fellowroot
  • Fellowroot
@ UnkleRhaukus I know that there is a formula that is A= (theta)r^2/2 but I don't how it works here.
UnkleRhaukus
  • UnkleRhaukus
\[A=\frac{\pi\times 3^2}4+\frac{\pi \times8^2}{2}+\frac{\pi \times8^2}{12}+\frac{4\times 8}{2}\]\[=\pi\times\frac{465}{12}+16\]\[\approx 140\]
UnkleRhaukus
  • UnkleRhaukus
well theta was 30° right? angles of segment / angle in circle \[30°/360°=1/12\] area of segment / area of circle \[\frac{\pi r^2}{12}\qquad/\qquad\pi r^2\]
UnkleRhaukus
  • UnkleRhaukus
i had to use a trig ratio to to get 30° btw
Fellowroot
  • Fellowroot
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.
UnkleRhaukus
  • UnkleRhaukus
where did i error?
Fellowroot
  • Fellowroot
not sure, but its pretty darn good.
UnkleRhaukus
  • UnkleRhaukus
\[A=\frac{\pi\times 3^2}4+\frac{\pi \times8^2}{2}+\frac{\pi \times8^2}{12}+\frac{4\times 8}{2}=\pi\times\frac{475}{12}+16\approx140.35\]
UnkleRhaukus
  • UnkleRhaukus
@Fellowroot i think we got different answers because we were solving different regions our pictures are different
Fellowroot
  • Fellowroot
|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.

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