Remember the goat in the 40 foot square pen?
The goat is on a 40 foot rope, tethered
to one of the corners. There a square obstruction
in the middle of the grassed 40 x 40 pen.
Each side of the obstruction is 16.5 feet from
the nearest fence line. Neither the goat nor the rope
can pass thru the obstruction.
How many square feet can the goat graze.
(Figure to the nearest 1/100 of a sq ft)
to one of the corners. There a square obstruction
in the middle of the grassed 40 x 40 pen.
Each side of the obstruction is 16.5 feet from
the nearest fence line. Neither the goat nor the rope
can pass thru the obstruction.
How many square feet can the goat graze.
(Figure to the nearest 1/100 of a sq ft)
posted by Ragknot at
1:53 PM
20 Comments:
Ragknot ...that's my goat ...did he get loose again? ...
Zaux,
In the previous post, your goat wasn't tied up. He was eating our fruit tree in the middle of our pen.
LOL, Thanks for the idea of this post. I wanted to put a circular fence around the tree, but I couldn't find any circular planks.
The rope is long enough for the goat to get round the side of the 7 x 7 obstruction and past the opposite corner.
So the goat's tether can extend in a straight line in an arc extending from each adjacent side to the point where the tether touches the corner of the obstruction. The angle of this arc is tan(-1) 16.5/23.5 = 35.07 degrees and the length of the tether to this corner is 28.71.
So the grass available which is thus unobstructed by the obstruction has area 2 * 35.07 / 360 * pi * 40^2 = 979.34 sq ft.
At the point where the tether touches the corner of the obstruction there is 11.29 feet left of the rope. This allows for another arc extending from the interior side of the previous arc to the side of the obstruction. The angle here is 54.93 degrees, so the area of the two smaller arcs on each side of the obstruction is 2 * 54.93 / 360 * pi * 11.29^2 = 122.20 square feet.
These two smaller arcs overlap each other past the far corner of the obstruction, forming a square of side 4.29, area 18.40. This has to be subtracted from the areas calculated so far.
Answer: 979.34 + 122.20 - 18.40 = 1083.14 square feet.
Actually, that last little bit where the two smaller arcs overlap isn't a square as I thought but two overlapping arcs.
It looks rather hard to work out, so how about we agree on, say, 16.54 for this area, making the total answer a round 1085 square feet. Near enough, eh?
I'll post this, then see if I can work out this area when I have more time.
try again
660 Square Feet.
Well Wiz, you got a picture of it in your mind. That's good, I was afraid that few would get that far.
Check out a diagram at
http://ragknot.blogspot.com/
Will do. Good question, Lata
I don't know how to determind a small sickle shaped area (probably need calculus)... therefore I can't determine the exact area accesible by MY GOAT. Bit I am going to say it is very close to 1200 sq. ft. I got 1207.64 less the small area I can't determine.
yep ...looked at your drawing Ragknot ... its the same one I determined
1200 ft is very close.
I doubt anyone will get much closer.
I'll post another picture with dimensions and see if anyone can get closer.
MY GOAT could roam a quarter circle if the center obstruction was no there.
area of the quarter circle:
3.14(40)^2= 5026.56 sq ft
area of center square:
7^2 = 49sq ft
actual goat roaming area is 1256.64 sq ft - 49 sq ft = 1207.64 sq ft. less the small sickle shaped area which must be about 7 - sq ft ... so I am guessing (heh heh) approx. 1200 sq ft.
Trying to work out the overlap area that the goat can graze coming round from either side of the obstruction . . . here goes:
Draw a line from the far corner of the obstruction to the far corner of the paddock from where the goat is tethered. This line bisects the overlap area. The angle of this line is 135 degrees to the two sides of the obstruction.
This gives us a wedge shaped triangle formed by this line and the side of the obstruction and the 11.29 length of the tether remnant coming round the obstruction corner. The area of this triangle is 12.88 and the small angle alongside the obstruction is 19 degrees.
The area of a 19 degree arc whose sides are formed by the 11.29 tether remnant is 19/360 * pi * 11.29^2 = 21.13 sq ft.
So the area of overlap is twice (21.13 - 12.88) = 16.50.
So, my revised answer for the total grazing area is 979.34 + 122.20 - 16.50 = 1085.04 sq ft.
This is just 0.04 sq ft, or just one small tuft of grass away from my inspired guess above.
That is, if I haven't made any mistakes along the way. I probably have.
Wiz,
You have an error somewhere.
Without the obstruction, there would
be a 1/4 of a circle.
The complete 1/4 circle is 1256.6371 sq ft
Subtract the 49 sqft obstruction leaves 1207.6371 sqft.
Lastly two smaller arcs need to be subtracted from the 40 ft arc.
Since this area is symmetrical you can figure the difference in the 40 ft arc and smaller arc.
A picture show lots of dimensions.
http://ragknot.blogspot.com/
(3 pictures now)
This problem is not like most. It depends on actual abilities or special skills, not a trick.
I am very proud of Zaux and Wiz for getting this close with a real problem.
Goats pen
A=A1+A2+A3+A4
A=2*40 deg arcs+2 triangles up to obstruction+ 2 arcs after obstruction+ 2 triangles after obstruction to arc intesection
2 arcs
To bottom right corner of box (16.5,40-16.5)
arctan(16.5/23.5)=35.07375 deg
A1=Pi*10^2*(2*35.07375/360)=979.443
Rope left after hit corner of box =
40-sqrt(16.5^2+23.5^2)=40-28.7141=11.2859
Triangles from (0,0) to (16.5,16.5) to(23.5,16.5)
A=1/2*b*h=0.5*(2 3.5-16.5)*16.5=57.75
two triangles
A2=2*57.75=115.5
2 arcs after obstruction
By symmetery arcs intesect on line from from 0,0 to 40,40
(x-23.5)^2+(y-16.5)^2=11.2859^2
x=y
x^2-47x+23.5^2+x^2-33x+16.5^2-11.2859^2=0
2x^2-80x+697.1285=0
x=(-b+-sqrt(b^2-4ac))/(2a)
x=(20+-7.17187)
x=27.17187 or 12.8281
we need x on right sight of box so x=27.17187
need to find degrees of arc between (27.17187,27.17187) and (23.5,16.5)
arctan((27.17187-16.5)/(27.17187-23.5))=71.013198 deg
71.013198- 35.07375=35.9394 deg
2 arcs
A3=Pi*11.2859^2*(2*35.9394/360)=79.8953
Now the triangles from the obstruction to arc intersection are:
(16.5,23.5) to (23.5,23.5) to (27.17187,27.17187)
A=1/2*b*h
A=1/2*(23.5-16.5)*(27.17187-23.5)=12.8515
A4=2*12.8515=25.703
Atotal=A1+A2+A3+A4
Atotal=979.443+115.5+79.8953+25.703=1200.5413
Answer:
The goat can cover 1200.54
The above was from me
Cam
Ragnot -- what program did you use to create those diagrams? I spent a few minutes with Google Sketchup getting quite frustrated (mostly because I don't know it well)
Just to address some typos to make solution easier to follow:
"A=2*40 deg arcs+2 triangles up to obstruction+ 2 arcs after obstruction+ 2 triangles after obstruction to arc intesection"
should be
"A=2*40' radius arcs+2 triangles up to obstruction+ 2 arcs after obstruction+ 2 triangles after obstruction to arc intesection"
and
"A1=Pi*10^2*(2*35.07375/360)=979.443"
should be
"A1=Pi*40^2*(2*35.07375/360)=979.443"
Cam
Congrats Cam,
There are at least three ways to get this.
I didn't see where Wiz had an error, but he was headed down the right path. Zaux was very very close, but hit a wall figuring out the area between three arcs.
Cam did the very detailed method that I think Wiz was working on got it exactly right.
Wow Cam, you really amazed me again.
See my summary
http://ragknot.blogspot.com/
my 1200 sq ft. partial guess wasn't too bad ... heh heh ... if Cam is right ... and I'd bet on that
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