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Answer to "Old MacDonald's Farm"
**The contended goat could happily graze over

Restricted by the two ropes r1 and r2, old MacDonald's goat could graze only over the shaded area in fig 1.

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 (

The area of segment-b is found by deducting the area of triangle-b from the area of sector-b (

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

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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