[Prof. Templeton]

After winning a goat in a recent game show, Farmer Templeton put the goat out to graze. The goat was put out in a square field at 7:00 am in the morning. The goat immediately started eating grass at a constant rate of 122 square meters per hour. When the goat had finished grazing he stopped at a point inside the pasture that was 4 meters from the pasture's SE corner, 28 meters from the NW corner, and 20 meters from the NE corner. When the goat had finished grazing he was immediately removed from the pasture. At what time was he removed?

[bonanova]

After winning a goat in a recent game show ...

Farmer T apparently didn't switch doors

bn is pondering the goat's munchings as he chides...

[Prof. Templeton]

Farmer T apparently didn't switch doors

bn is pondering the goat's munchings as he chides...

Sometimes switching still nets you a goat.

[bonanova]

Approx 11:12 a.m.

Spoiler for The field is approx:

22.6 m on a side, with area approx 512 m².

512/122 is about 4.2 hours = 4 hr 12 min + 7:00 a.m. = 11:12 a.m.

[Guest]

After winning a goat in a recent game show...

Thanks Prof., that was a solid 10 seconds of very loud LOL'ing.

Did the goat cover the entire area. That part isn't clear in the OP.

I am assuming he/she did, until you say otherwise.

[Prof. Templeton]

Thanks Prof., that was a solid 10 seconds of very loud LOL'ing.

Did the goat cover the entire area. That part isn't clear in the OP.

I am assuming he/she did, until you say otherwise.

Yeah, sorry. The goat ate the whole pasture.

[bonanova]

Suppose the goat ended up x m to the west of and y m to the north of the SE corner

The field is a m on a side. Then,

x² + y² = 4²

[a-y]² + x² = 20²

[a-y]² + [a-x]² = 28²

Solve these for a, x and y.

Area is a².

Divide by 122

Add that time to 7:00a.m.

[Prof. Templeton]

Approx 11:12 a.m.

Spoiler for The field is approx:

22.6 m on a side, with area approx 512 m².

512/122 is about 4.2 hours = 4 hr 12 min + 7:00 a.m. = 11:12 a.m.

You must have a lot of written on envelopes at your house.

[Guest]

I'm thinking goat for goat - your one is E-type, GTE, or basically as good as a heard of cows (which have as many as half a length of string) - you can rent the goat out and use the money to lease the car you cld have won.

[Guest]

The above picture represents the square field where the variable X represents a single side of the field, the Northwestern corner has an unknown angle a, and the Southeastern corner has an unknown angle b.

Solving for x:

sin a = (20 / 28); cos a = (x / 28)

a = arcsin (20 / 28)

x / 28 = cos (arcsin (20 / 28))

x = 28 * cos (arcsin (20 / 28))

x = ~19.595918

-or-

cos b = (4 / 20); sin b = (x / 20)

b = arccos (4 / 20)

x / 20 = sin (arccos (4 / 20))

x = 20 * sin (arccos (4 / 20))

x = ~19.595918

Now that we have the length of the side, we can square it to get the size of the field in total square meters which just happens to be exactly 384 m².

x² = ~19.595918² = 384 m²

Ok, now the goat eats the grass at 122 m²/h which means it takes 3 hours 8minutes and approximately 51 seconds for the goat to eat all the grass.

384 / 122 = 3 9/61 = 3h 8m ~51s

Now if he started at 7:00 am then he would finish just before 10:09 am.

Did I do something wrong? I'm a little rusty...

Unless of course the goat you're refering to happens to be the Generalized Occupational Aptitude Test...

Edited December 9, 2008 by August717

[bonanova]

The goat ended up on a diagonal of the field [x=y].

[1] x² + y² = 4² = 16

[2] [a-y]² + x² = 20² = 400

[3] [a-y]² + [a-x]² = 28² = 784

[2'] a² - 2ay + [x² + y²] = 400 => a[a-2y] = 384

[3'] 2a² - 2a[x+y] + [x² + y²] = 784 => a[a-(x+y)] = 384

thus 2y = x+y or x=y

The field diagonal is thus 28+4 = 32 and a = 32/sqrt(2) or a² = 512

512/122 = 4.1967213114754098360655737704918 or approx 4 hours and 12 minutes.

Here's a sketch

[bonanova]

Did I do something wrong? I'm a little rusty...

Unless of course the goat you're refering to happens to be the Generalized Occupational Aptitude Test...

You're applying sin and cos formulas to triangles that are almost right triangles, giving you a result that is almost correct.

If you look at the EW and NS coordinates of where the goat ended up, and drop perpendiculars to the sides of the field,

then you have right triangles you can use.

angles a and b are both 45 degrees

[Guest]

LOL, wow it has been a while since I’ve used them, and so I completely misused them... wow, so this is what happens when good formulas go bad!

Or should I say, when carelessness is carefully applied... *sigh* I guess this just means I need to reeducate myself! Thanks for the explanation, it really helped out...

[Izzy]

The above picture represents the square field where the variable X represents a single side of the field, the Northwestern corner has an unknown angle a, and the Southeastern corner has an unknown angle b.

Solving for x:

sin a = (20 / 28); cos a = (x / 28)

a = arcsin (20 / 28)

x / 28 = cos (arcsin (20 / 28))

x = 28 * cos (arcsin (20 / 28))

x = ~19.595918

-or-

cos b = (4 / 20); sin b = (x / 20)

b = arccos (4 / 20)

x / 20 = sin (arccos (4 / 20))

x = 20 * sin (arccos (4 / 20))

x = ~19.595918

Now that we have the length of the side, we can square it to get the size of the field in total square meters which just happens to be exactly 384 m².

x² = ~19.595918² = 384 m²

Ok, now the goat eats the grass at 122 m²/h which means it takes 3 hours 8minutes and approximately 51 seconds for the goat to eat all the grass.

384 / 122 = 3 9/61 = 3h 8m ~51s

Now if he started at 7:00 am then he would finish just before 10:09 am.

I tried doing that during maths today. Different method.

I had the SW corner as the origin and drew a 4m (to scale) circle around it. The goat has to be somewhere on the circle, and the NE corner has to line up with the origin. So, did the distance formla a few times, found the NE corner. Then did it again to find the NE corner. Don't think I did it right though, 'cos none of the numbers I came up with made sense..

I'll try again during lunch and upload when I get home.

[Guest]

After winning a goat in a recent game show, Farmer Templeton put the goat out to graze. The goat was put out in a square field at 7:00 am in the morning. The goat immediately started eating grass at a constant rate of 122 square meters per hour. When the goat had finished grazing he stopped at a point inside the pasture that was 4 meters from the pasture's SE corner, 28 meters from the NW corner, and 20 meters from the NE corner. When the goat had finished grazing he was immediately removed from the pasture. At what time was he removed?

There isn't enough information to solve this. Nothing is said about the path that the goat takes in its grazing, is it well known that unconstrained goats graze in a systematic fashion? If not, the coordinates of where it finished are immaterial.

Since a clarification to the OP stated that the goat 'Ate the entire pasture'. If so:

X = size of pasture in m2

T = time of removal

T = 7:00 + (X/122)

[bonanova]

There isn't enough information to solve this.

Nothing is said about the path that the goat takes in its grazing, is it well known that unconstrained goats graze in a systematic fashion?

If not, the coordinates of where it finished are immaterial.

Since a clarification to the OP stated that the goat 'Ate the entire pasture'. If so:

X = size of pasture in m2

T = time of removal

T = 7:00 + (X/122)

Sure there is.

The coordinates of where the goat finished are material. But it's not because he finished grazing there.

They establish that that point exists somewhere in the field..

That fact lets you calculate X [in your terminology] and then your analysis gives the finish time.

[Guest]

Sure there is.

The coordinates of where the goat finished are material. But it's not because he finished grazing there.

They establish that that point exists somewhere in the field..

That fact lets you calculate X [in your terminology] and then your analysis gives the finish time.

Hiding head in shame...