April 27, 2009

Maths Challenge #2 – Pig in a Field

A pig tied to the corner of a 100m-sided equilateral triangle field can cover half its area. How long is the rope?


- 4 comments by 3 or more people Not publicly viewable

  1. I got 109m, which suggests I’m rather wrong!

    27 Apr 2009, 20:13

  2. Eleanor Lovell

    The answer is approximately 64.30m and here is how…(courtesy of @matmannion)

    It would appear that the area that the pig can cover whilst on the rope is half the area of the field, since it can eat half the grass. Given the situation below (with the pig at the bottom), we know that the area of the arc is half that of the area of the entire equilateral triangle.

    Not to scale (and probably not an equilateral triangle either…)

    The area of the arc, is a=\frac{\pi r^2 d}{360} where d is the angle of the arc. Since this is an equilateral triangle, d is 60 degrees. This gives the area of the arc as \frac{\pi r^2}{6}

    The area of the equilateral triangle is given as a=\frac{s^2 \sqrt{3}}{4} where s is the length of one side; in this case, 100m.

    Since the area of the arc is half the area of the triangle, we can solve for r:

    \frac{\pi r^2}{6}=\frac12 \frac{{100}^2 \sqrt{3}}{4} r=\sqrt{\frac{7500 \sqrt3}\pi}

    thus r is approximately 64.30m.

    28 Apr 2009, 13:19

  3. Mathew Mannion

    Since the a=\frac{s^2 \sqrt{3}}{4} is a bit of a leap for the area of a triangle (I had to look it up myself), you can also use the general equation to get the area of a triangle, a=\frac12bh where h is the perpendicular height.

    To get the perpendicular height, we can bissect the triangle in two to create a triangle with half the base, and use the Pythagorean theorem (the square of the base plus the square the height is equal to the square of the hypoteneuse) to get:

    50^2+h^2=100^2, so h=\sqrt{100^2-50^2}=\sqrt{7500}

    We can then get the area of the whole equilateral triangle:

    a=\frac12bh=\frac12*100\sqrt{7500}=50\sqrt{7500}

    The cow can cover half of this, so the area of the arc needs to be equal to half of this, so:

    25\sqrt{7500}=\frac{\pi r^2}6, therefore r=\sqrt{\frac{150\sqrt{7500}}{\pi}}

    which is approximately 64.30.

    28 Apr 2009, 14:23

  4. Steven Jones

    You can make those surds a little nicer:
    r = \sqrt{\frac{7500\sqrt{3}}{\pi}}=50\sqrt{\frac{3\sqrt{3}}{\pi}}

    01 May 2009, 12:13


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