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Answer to "Old MacDonald's Farm"


The contended goat could happily graze over  59. 7 square meters  of lush grass.
Restricted  by the two ropes r1 and r2, old MacDonald's goat could graze only over the shaded area in fig 1. (All figures are illustrative and not necessarily drawn to scale).
The shaded area is the sum of the areas of segment-a and segment-b.
The area of segment-a is found by deducting the area of triangle-a from the area of sector-a (fig 2).
The area of segment-b is found by deducting the area of triangle-b from the area of sector-b (fig 2).
A simplified solution follows.
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The 9, 12 and 15 meter lengths of ropes (r1 and r2) and peg distance (d) are in the well-known ratios of  3, 4, 5 which of course make a right-angled triangle (fig 3). This simplifies the solution.
Angle C is 90 and the 15 meter distance between the pegs at A and B is the hypotenuse.
Therefore the angles of triangle-c are,
Angle A = arccos (12 / 15) = 36.9
Angle B = 90 - A = 53.1   

Note that the kite-figure in fig 1 and fig 4 is actually the triangle-c plus its inverted shape. The figure can be redrawn as a rectangle in fig 5 (not drawn to scale). 
Area of the rectangle = r1 * r2 = 108 square meters
This is the area of the shape in fig 1 (also the sum of the areas of triangle-a + triangle-b in fig 2.)
The angles of sector's (a) and (b) are double the angles of A and B respectively.

Area of sector-a = (2 * A) / 360 * Pi * r1 * r1 = 92.6 square meters
Area of sector-b = (2 * B) / 360 * Pi * r2 * r2 = 75.1 square meters
Grazing area = (sum of areas of sector-a and sector-b) - area of rectangle in fig 5.
= 92.6 + 75.1 - 108 = 59.7 square meters

Copyright   1996, 2000   S. Zervos

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