math puzzle: a rectangular field, a rope and a cow

[SphinxnihpS]

Originally posted by: joesmoke
turns out that Mitt Romney has been living in the hollowed out ribcage of your cow. the answer is zero. forever.

Originally posted by: SphinxnihpS
Originally posted by: SphinxnihpS
So now you hate robots? Fucking fundamentalist lunatic Luddite girl subverter!!!

I can just picture the Pedobear stickers all over your van, the lilting music...

ahhh the summertime...

Fuck yeah! Sig quoted! SCHWING!

 

[child of wonder]

R = sq( A^2/Pi )

 

[eits]

Originally posted by: SlitheryDee
Wouldn't it be like:

Find area of rectangle:

A x 2A = 2A^2

Find half area of rectangle:

(2A^2)/2 = A^2

State that the area of the circle = half area of the rectangle:

- R^2 = A^2

Square root both sides to remove the squares: (Not sure about this one)

-^(1/2) x R = A

Divide both sides by the square root of Pi to leave R on one side and the answer on the other side in terms of A as requested.

R = A/-^(1/2)

I've never been good at math. If I'm wrong I'd like to know where though.

that's exactly what i did

 

[SunnyD]

Originally posted by: eits

Originally posted by: tmc
i guess too many ppl asking for HW solutions in ATOT.

anyways, it is not that hard to see i wasn't one of them (hint: profile, name, google).

owned.

oooh, I have your phone number now... You never know if it's going to be me or Ed McMahon!

 

[SphinxnihpS]

Originally posted by: tmc
There is a rectangular grass field of size 'a' by '2a'.

A cow is tied by a rope of length 'r' at the center of the field.

The cow is able to graze exactly half the area of the field.

What is the length of the rope (in terms of 'a')?

edit - this is not my HW and this is not that straight-forward (easy) as it looks.

Would I be wrong in assuming this is impossible?

Since the cow can not graze outside the rectangle, r can not exceed .5a. Since the rectangle is 2a^2, no value for r will equal (2a^2)/2.

Am I right?

 

[SlitheryDee]

Originally posted by: SphinxnihpS

Originally posted by: tmc
There is a rectangular grass field of size 'a' by '2a'.

A cow is tied by a rope of length 'r' at the center of the field.

The cow is able to graze exactly half the area of the field.

What is the length of the rope (in terms of 'a')?

edit - this is not my HW and this is not that straight-forward (easy) as it looks.

Would I be wrong in assuming this is impossible?

Since the cow can not graze outside the rectangle, r can not exceed .5a. Since the rectangle is 2a^2, no value for r will equal (2a^2)/2.

Am I right?

That occurred to me after reading dullard's post. The problem doesn't say that the cow can't walk off the field, just that he only has access to half of it That means that the area of the whole circle is actually larger than half the area of the rectangle, but the area of the circle minus the parts that hang over the edges is half the area of the rectangle. That is harder and makes my above post incorrect.

 

[tmc]

Originally posted by: eits

Originally posted by: SlitheryDee
Wouldn't it be like:

Find area of rectangle:

A x 2A = 2A^2

Find half area of rectangle:

(2A^2)/2 = A^2

State that the area of the circle = half area of the rectangle:

- R^2 = A^2

Square root both sides to remove the squares: (Not sure about this one)

-^(1/2) x R = A

Divide both sides by the square root of Pi to leave R on one side and the answer on the other side in terms of A as requested.

R = A/-^(1/2)

I've never been good at math. If I'm wrong I'd like to know where though.

that's exactly what i did

thats the trick.

in this case R > 0.5 A. then the cow goes outside the field. thus, the equation of - R^2 = A^2 is invalid.

see the earlier posts for correct solution.

 

[SunnyD]

Originally posted by: SphinxnihpS

Originally posted by: tmc
There is a rectangular grass field of size 'a' by '2a'.

A cow is tied by a rope of length 'r' at the center of the field.

The cow is able to graze exactly half the area of the field.

What is the length of the rope (in terms of 'a')?

edit - this is not my HW and this is not that straight-forward (easy) as it looks.

Would I be wrong in assuming this is impossible?

Since the cow can not graze outside the rectangle, r can not exceed .5a. Since the rectangle is 2a^2, no value for r will equal (2a^2)/2.

Am I right?

Yes and no. The rope can slack at .5a, but you need to compensate along the 2a length of the pasture for the slacked sides. See dullard's post.

 

[SphinxnihpS]

Well I ain't getting out no trig book, not that I'd remember a damn thing about it anyway. GL fellas.