School-Safe Puzzle Games

The Goat Problem

A goat is tied on the corner post of a 20×20 ft barn in a flat grass field.

The goat can reach a length of 45 ft at maximum from the corner post

How many square feet can the goat eat exluding the area of the barn?

Thanks to MichaelC for submitting!

Please refer to the diagram below to better understand the challenge:

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If you can figure this out, enter your answer below in the comment section.  Will reveal submissions (as well as the correct answer) in several days.

33 Comments to “The Goat Problem”


  1. NitzOO | Profile

    I think the total area is approximately 6130 square ft (6129.50543 to be more precise).


  2. suineg | PUZZLE MASTER | Profile

    Very nice puzzle MichaelC, ok let me try it:
    For me you got to get the area of the circle of 45 feet of radio and substract the area of the square, so I think is:
    1)Area of the circle=pi*square(45)=6361.74 (pi taken as 3.1416)
    2)Area of the square= 20*20=400 square feet
    3)Dogfood= 1)-2)=6361.74-400= 5961.74
    Ok hope I got it right cool.


  3. aaronlau | Profile

    did a ton of calculations, so would be difficult to share.
    hope i did not make any careless mistakes.


    ANS = 5729.254943 sq ft


  4. suineg | PUZZLE MASTER | Profile

    I was wrong jajajaja, my last comment is wrong, because the can move the whole circle jajaja, MichaelC this was tricky jajaja
    so you definitely has:
    1) 3/4 Area of the circle of 45 feet
    2) 2/4 Area of the circle of 25 feet (when it get to the opossite corner, both directions)
    3)-1/4 Area of the circle of 5 feet (common area between the two 25 feet circles)


    so: pi*( 3/4*2025+ 2/4*625-1/4*25)–>pi*(1518.75+312.5-6.25)–>
    pi*(1837.5)= 3,1416*1837.5= 5772.69 square feet—> GoatFood


    Cool MichaelC, very nice, I am 99 percent sure now


  5. KMESON | Profile

    3/4 pi 45^2 gets the North and East yards.
    2 * 1/2 pi 25^2 gets the South and West yards.
    What’s that you say ? South and West overlap? Then yes, let’s subtract that out.
    A = SW corner of barn
    B = SE corner of barn
    C = intersection of the South and west yards.
    then sin of angle ACB is asin( 4/(5 sqrt(2)) ) :: law of sines
    and angle ABC is pi/4 – angle ACB.
    The overlap is two regions that are wedge – triangle
    triangle = 25 20 sin(ABC) / 2
    wedge = pi 25^2 ABC/(2pi)


    numerically this is ~ 5660.09412 sqft


  6. hex | PUZZLE MASTER | Profile

    Area =
    3/4 of a circle with radius=45 (1)
    + 2 circles with radius=25 (2)
    - intersection of the last 2 circles (3)


    The last possible circle with radius=5 would already be covered by the above.


    (1) = 3/4 x Pi x 45^2 / 4
    (2) = 2 x Pi x 25^2 / 4
    (3) which is the tricky part = 4.578^2 + 2 x 1.292
    Total area=2151.035 sq. ft.


  7. diegote | Profile

    The goat can eat 5729,5 square feet.


    I added the area of
    1 x 270°sector r=45
    2 x 79,4° sector r=25
    2 x triangles base=20 heigh=4,57


  8. hex | PUZZLE MASTER | Profile

    Minor glitch above.
    Area =
    3/4 of a circle with radius=45 (1)
    + 2 x 1/4th circles with radius=25 (2)
    - intersection of the last 2 circles (3)


    The last possible circle with radius=5 would already be covered by the above.


    (1) = 3/4 x Pi x 45^2 / 4
    (2) = 2 / 4 x Pi x 25^2 / 4
    (3) which is the tricky part = 4.578^2 + 2 x 1.292
    Total area=1414.724 sq. ft.


  9. joe | Profile

    My answer is only an approximation, there should be some integration involved but my brain went mushy when i got deep and failed so I am posting what I think is a reasonable approximation.. (assume pi = 3.1416)


    The goat can go round the barn in what is effectively a circle of radius 45ft. It will cover 3/4 of the area of this circle before the barn interrupts the rope and it then starts forming the area of a different circle of radius= rope length minus barn side length,(45ft-20ft =25ft. Going round the other side it does the same but if we draw it there is an overlap at the bottom left of the barn (in relation to the drawing given).
    Area of circle is (pi) r^2.
    The first larger area then will be 3/4 of (pi) r^2
    = 0.75 x 3.1416 x 45^2 = 4,771.305 sq ft
    The next circle segments are both 1/4 of a circle of radius 25 (as explained above)
    1/4 x 3.1416 x 25^2 = 490.875 each ( x 2 = 981.75)
    The bit that overlaps is 5 ft of radius in each circle, so a bit less than a perfect circle of side, 5. OR a bit more of a 1/4 of a circle of radius 5
    If it were a square then 5×5 = 25 sq ft
    If it were a circle r=5 then 1/4 pi r^2 = 19.635 sq ft
    So I am going to approximately go for 22.3175 (half of it), which we subtract from the total area.
    So 4,771.305 + 981.75 – 22.3175


    = 5,730.75 sq feet !


  10. NzNerd | Profile

    we have 3/4 of a circle of radius 45ft plus 2 lots of 1/4 of a circle radius 25ft


    Total area = 7325*pi /4 square feet


    roughly = 5753.042 square feet


  11. Shawn | PUZZLE GRANDMASTER | Profile

    Using 3.14 for pi and rounding, I get 5,235 sqft


    I assumed that the barn had 4 rigid sides and no doors that the goat could access. He could not walk straight through and come out the opposite corner with 16+ feet of rope to work with!


  12. bizarette18 | PUZZLE MASTER | Profile

    5729.5 sq ft?
    3/4 a 45 radius circle + 2/4 a 25 radius circle, take away the not quite square where the two quarter circles overlap. I got that bit to be 23.54


  13. Hendy | Profile

    3/4 Pi 45^2 + 1/2 Pi 25^2 – 3/4 Pi 5^2


  14. Falwan | Profile

    0.75 of a 45 ft. radius circle + 0.25 of a 25 ft. radius circle + 0.25 of a 5 ft. radius circle:


    4768.875+490.625+19.625 = 5279.125 sq. ft.


  15. LadyInsomnia | Profile

    The goat can graze on about 4769 square feet of grass.


    Consider his tether to be the radius of a circle of accessible grass. Since the corner of the barn is a right angle, there’s 3/4 of a circle for which we need the area: 3/4*(pi)*(45^2)


  16. Falwan | Profile

    oops!


    I change my mind :


    4768.875 + 2( 490.625 ) = 5750.125 sq. ft.


    I took the union of the 2 clockwise & anticlockwise areas.


  17. bilbao | Profile

    I get 5724.62 square feet.
    I will calculate half area and double it afterwards. The square diagonal will be the simmetry axis. Starting where the picture shows and going unclockwise:
    1st-the goat covers an area of 3/8*pi*45^2 till the rope gets parallel to the wall of the barn (2385.64 sq.feet)
    2nd-the rope bends at the corner and the goat can eat an extra 1/4*pi*25^2 (490.87 sq.feet)
    3rd-I need to substract the area of the portion of circle that has crossed the diagonal (14.20 sq.feet)
    Total area=2x(2385.64 + 490.87 – 14.20)=5724.62 sq.feet
    3rd step was tricky!Nice puzzle, MichaelC!


  18. Hendy | Profile

    I looked again at my drawing of the circles created by the rope and I don’t know how to remove the parts that the regions cross over one another. My previous guess is probably close, but definitely not right. 3/4 Pi 45^2 + 1/2 Pi 25^2 is the area that can be reached stretching the rope clockwise and counter clockwise together about the barn, but there is a region that is shared by the two and must be halved to show actual area. Unfortunately, I don’t know how to express the full region, Part of it is 3/8 pi 5^2, but there is another small region that does not fit into a circular section.


  19. Bobo The Bear | PUZZLE MASTER | Profile

    The total grazing area is (approximately) 5729.5 sq ft.


    The area is made up of the following:


    1) a 3/4-circle – a sector with angle 270 degrees and a radius of 45 feet (area is 4771.29)
    2) two sectors, each with an angle of 79.45 degrees and a radius of 25 feet (area is 433.33*2 = 866.6)
    3) two thin triangles, each with sides of 20 and 25 feet with an angle of 10.55 degrees between these sides (area is 45.77*2 = 91.54)


    The the 79.45 dergee angle can be found by solving for the base angles of an isosceles triangle with legs of 25 feet, a base of 20sqrt(2) feet, and a vertex point where the two smaller sectors meet. The angle of the sectors will be 135 degrees minus the base angles. The 10.55 degree angle is the complement of the 79.45 degree angle.


    That and some of the high school trig will get you to the final area.


  20. Emily | Profile

    ~5238.63 square feet.


  21. RK | Founder | Profile

    I’m going to defer to MichaelC on this one, but it looks like Aaronlau, Diegote, Joe, Bizarette18, BobotheBear, Bilbao (slight error in calculation?) and ?possibly Hendy got the right answer.


    You may view MichaelC’s full solution here:http://www.smart-kit.com/wp-co.....lution.bmp


  22. hex | PUZZLE MASTER | Profile

    I have goofed using area of circle = Pi x R^2 / 4 instead of Pi x R^2 (or Pi x D^2 / 4), and amazingly integrated the intersection area correctly (I am more calculus oriented than geometry)
    Correcting:
    Area =
    3/4 of a circle with radius=45 (1)
    + 2 x 1/4th circles with radius=25 (2)
    - intersection of the last 2 circles (3)


    The last possible circle with radius=5 would already be covered by the above.


    (1) = 3/4 x Pi x 45^2 = 4771.2938426395
    (2) = 2 / 4 x Pi x 25^2 = 981.74770424681
    (3) which is the tricky part = 4.57737973711325^2 + 2 x 1.29185834947498 = 23.536121956685
    Thus area = 5729.50542492962


    Why does Bilbao get special treatment for making a slight mistake while I do not? :P


    Nice puzzle!


  23. RK | Founder | Profile

    hahahha that’s kind of funny cause I did feel a twinge of guilt for being too lenient with Bilbao above.


    And after all the other impossibly hard/tricky ones you’ve gotten right, figured you probably made a silly mistake in your calculations


  24. michaelc | Profile

    Wow!


    I’m really impressed with the calculus answers to the problem. Calculus isn’t my strong subject, but I’m trying to get more familiar with the techniques used there.


    Geometry is just a natural kind of math (for me anyway). Easy to see and “know” what you’re doing is accurate by looking at the drawing. With calculus, I can look at the drawing and the calculations 15 times, and still not know if it is accurate!


    Thanks for the nice comments. Hopefully this thing got you guys thinking as much as it did me! I’ll probably re-visit this problem when my calculus gets stronger.


  25. alexc | Profile

    i think through all the calculus people have missed the obvious. the goat can’t eat in the barn!
    Anyway, here are my workings
    (1) 3/4 x Pi x 45^2 = 4771.29
    (2) 20 * 20 = 400
    (3) work out section of r25 circle
    sqrt(20^2 + 20^2) = 25^2 + 25^2 – 2* 25 * 25 cos (A)
    a bit of jiggling the angles around gets the r25 circle area to be
    79.4499/360 * 25^2 * Pi
    (4) area of remaining triagle
    Rejigged the angle to work out height or triagle to be 3.66
    1/2 * 25 * 3.66 = 45.77


    total area = (1) – (2) + 2(3) + 2(4)


    So edible area = 5329.505


  26. bilbao | Profile

    I deserve to be dropped off the solvers list ;-)
    When dealing with this type of mathematical puzzles, if you don´t get the exact solution something is wrong.
    At some point in my calculations I dragged on a value of 11 degrees for an angle instead of 10.55 degrees.
    This changes my 3rd step from 14.20 sq.feet to 11.768058, so the total area would be:
    2 x (2385.646921 + 490.873852 – 11.768058) = 5729.50543 sq.feet


  27. hex | PUZZLE MASTER | Profile

    Michaelc,
    Calculus/algebraic solutions tend normally to be straight forward to start with, but difficult to compute. The common area between the R=25 circles was not easy to compute (the result involves inverse trigonometrical functions). On the other hand, geometry provided a simple answer. In general, the former is more powerful.


    One thing I have noticed though is that nobody mentioned the partial circle with R=5, when the goat reaches point C on your drawing. Luckily, it turns out that this circle is included in the R=25 circles.


    RK,
    Now Bilbao will have a twinge of guilt because he’s caused you inadvertently some suffering :D


    Alexc,
    At least most of the posters, if not all, excluded the barn area.


  28. alexc | Profile

    hex, yeah i realised i stuffed that up when lying in bed last night. Sorry guys


  29. Excitol | Profile

    I started doing this thinking it would take five minutes, damn. Decided in the end to resort to calculus so my answer should be pretty accurate. Anyway the equation that gave me the answer was:


    3?/4 x 45^2 + ?/2 x 25^2 – 2 x integral of [(625 - X^2)^(1/2) - (X + 20)] between the limits of 0 and -10 + root(850)/2


    The answer this gives is 5729.50542 although this relies on my calculater button pressing skills being spot on.


    Sorry not much explanation of where that comes from but i felt that might get tedious. Nice Puzzle btw!


  30. Excitol | Profile

    Sorry in my answer the “?” were meant to be pi symbols


  31. Neon Maze | Smart-Kit Puzzles and Games | Guest

    [...] next week, someone has come up with a sequel to MichaelC’s Goat Problem! [...]


  32. tomlaidlaw | Profile

    hex has it right.
    3/4 of a 45 ft. radius circle
    +2*79.45/360 of 25 ft. radius circle
    +Area of triangle of sides 25,25,28.28(diagonal of barn)
    -1/2 barn area
    Approx. 5729.5
    See my complete discussion of this puzzle at http://tomlaidlaw.com/goatandsilo


  33. Guggs | Profile

    The area of grass the goat can eat is 5,729.52 sq ft and is made up of three separate sections:
    1. 3/4 of a circle produced by the goats 45 ft tether = 4,771.30 sq ft
    2. 2 segments of a circle of radius 25 ft centred at nearest corners
    = 866.66 sq ft
    3. 2 triangles to complete the common area eaten from clockwise and anticlockwise grazing by the goat = 91.55 sq ft


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