What area can the goat roam?

- anonymous

What area can the goat roam?

- schrodinger

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- anonymous

|dw:1350884315409:dw|

- anonymous

Fence
SHED 4feet
5feet 8feet

- UnkleRhaukus

|dw:1350884748553:dw|

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## More answers

- UnkleRhaukus

ops i dident see the fence

- 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

|dw:1350884878590:dw|

- anonymous

No he's not connected.
I'm not sure. That's all my teacher wrote

- Fellowroot

he has to be connected by a chain to the shed right?

- anonymous

Wait... yes..

- UnkleRhaukus

|dw:1350885122452:dw|

- UnkleRhaukus

|dw:1350885496696:dw|

- anonymous

How do I calculate it?

- UnkleRhaukus

break the shape in to smaller shapes

- Fellowroot

Unkle Rhaukus I like your drawings, but what is the actual value?

- anonymous

81 sq ft??

- Fellowroot

I'm very sure I know how to do it. Just give me some time to work it out.

- Fellowroot

its not easy, but i'm still working.

- Fellowroot

Ok, here it comes. Writing it up. :) !!!!!

- anonymous

You there?

- Fellowroot

yeah hang on almost done

- Fellowroot

you might be surprised but i got it.

- Fellowroot

GOT IT!!!!

##### 3 Attachments

- UnkleRhaukus

im not sure why you had to integrate

- anonymous

Thank you!

- UnkleRhaukus

we know the are of a circle is Ï€r^2,
so a semicircle has area (Ï€r^2)/2
and a quater circle has area (Ï€r^2)/4

- 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

|dw:1350888457508:dw|

- 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

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- UnkleRhaukus

the sector is 1/12 of a circle

- 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

\[A=\frac{\pi\times 3^2}4+\frac{\pi \times8^2}{2}+\frac{\pi \times8^2}{12}+\frac{4\times 8}{2}\]\[=\pi\times\frac{465}{12}+16\]\[\approx 140\]

- UnkleRhaukus

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

- UnkleRhaukus

i had to use a trig ratio to to get 30Â° btw

- 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

where did i error?

- Fellowroot

not sure, but its pretty darn good.

- 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

@Fellowroot
i think we got different answers because we were solving different regions
our pictures are different

- 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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