Determining Areas of Irregular Shaped Fields Using Standard Geometric Shapes

Very few agricultural fields or lawns are standard geometric shapes. They may not be as complicated as the field illustrated in Figure 13.15, but the principles

Geometrical Shapes Agriculture Tools
FIGURE 13.15. Example of determining the area of an irregular shape.

Determining Areas of Irregular-Shaped Fields Using Standard Geometric Shapes

for determining areas are the same. The total area is the sum of each individual area. The key to efficient calculations is dividing the irregular shape into the fewest possible number of standard shapes.

Shape A is a sector of a circle with a known radius and angle. The area of this shape can be found by using the sector equation.

360 360

12004778 2

Shape B is a triangle with the length of the sides known.

a + b + c 68.6 ft + 79.2 ft + 115.4 ft 263.2 S =---=---= —-— = 131.6

= J 131.6(131.6 - 68.6)(131.6 - 79.2)(131.6 - 115.4) = y/131.6 (63) (52.4) (16.2) = V7037894.304 = 2652.90... or 2,650 ft2

Shape C is another triangle with the lengths of all three sides known.

= s/127.25(127.25 - 68.4)(127.25 - 70.7)(127.25 - 115.4)

Area D is a triangle with one angle known and the length of the adjacent sides. The area (AD) is:

A = — x a x b x sine 6 = - x 70.7 ft x 161.2 ft x sine 59.0 22

11396.84 x 0.8571... 9768.99... 2 =-=-= 4884.49... or 4,880 ft2

Shape E is the same as shape D. The area of shape E is:

A = -x a x b x sine 0 = - x 161.2 ft x 150.0 ft x sine 38.3

Now that the area for each shape is known, the total area (AT) is calculated by combining the areas for the individual shapes:

= 3330 ft2 + 2650 ft2 + 2240 ft2 + 4880 ft2 + 7490 ft2 = 20590 or 20,600 ft2

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