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A horse's rope (Posted on 2003-12-05) Difficulty: 2 of 5
Two ranchers shared a circular corral of 10-meter radius. One rancher wanted to tie his horse to one point on the fence and allow him to graze on just half the area of the corral. How long should the horse's rope be to make the portion available to this horse be equal to half the area of the corral?

See The Solution Submitted by Antonio    
Rating: 3.3333 (6 votes)

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finding a solution the easy way. | Comment 5 of 8 |
Penny's method was slick, but I managed to find an even easier way to solve the problem than searching the internet (or clicking 'favorites' ->'Dr Math) finding the relevant problem - copy-pasting into flooble and tacking on some snide barb - 'cos that would've taken ages!
I just copy-pasted Penny's.

The horse's rope should be 11.5872847 meters long.

Anyone requiring an explanation, is probably someone who, before he starts his car in the morning, first finds a large rock and then re-invents the wheel. But here is the explanation from the Dr. Math website:

Draw a circle with radius r (the corral).

Now take a point C on the circumference and with a slightly larger radius R draw an arc of a circle to cut the first circle in points A and B. Join AC and BC.

Let O be the centre of the first circle of radius r. Let angle OCA = x (radians). This will also be equal to angle OCB.

The area we require is made up of a sector of a circle radius R with angle 2x at the centre, C, of this circle, plus two small segments of the first circle of radius r cut off by the chords AC and BC.

The area of the sector of circle R is (1/2)R^2*2x = R^2*x

The area of the two segments

= 2[(1/2)r^2(pi-2x) - (1/2)r^2sin(pi-2x)]
= r^2[pi - 2x - sin(2x)]

We also have R = 2rcos(x) so R^2*x = 4r^2*x*cos^2(x)
We add the two elements of area and equate to (1/2)pi*r^2

4r^2*x*cos^2(x) + r^2[pi-2x-sin(2x)] = (1/2)pi*r^2 divide out r^2
4x*cos^2(x) + pi - 2x - sin(2x) = (1/2)pi

4x*cos^2(x) + (1/2)pi - 2x - sin(2x) = 0

We must solve this for x and we can then find R/r from R/r = 2cos(x)

Newton-Raphson is a suitable method for solving this equation, using a starting value for x at about 0.7 radians.

The solution I get is x = 0.95284786466 and from this cos(x) = 0.579364236509

and so finally R/r = 2cos(x) = 1.15872847

  Posted by Lee on 2003-12-05 12:52:18
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