Occasionally it becomes necessary to determine the area of a field or other property that has one irregular-shaped boundary. This is a very common problem when one of the boundaries is formed by surface water. In this situation, the area can be determined by dividing the shape into a series of trapezoids and summing the individual areas. In situations where the trapezoids can be laid out so the heights are the same, the trapezoidal summation equation can be used.
Problem: What is the area (acres) for the field illustrated in Figure 13.16?
400i ft
250I ft
375 ft
400i ft
250I ft
375 ft
400 ft
200 ft
300 ft
FIGURE 13.16. Example of using trapezoids for irregular shape.
Determining Areas of Irregular-Shaped Fields Using Trapezoidal Equations
Solution: The total area (At) is the sum of each trapezoidal shape (dashed lines).
Í 500 ft + 400 ft\ / 400 ft + 250 ft \ Í 400ft x -2-j + Í 200 ft x -2-j
= (400 x 450) + (200 x 325) + (300 x 312.5) = 180000 + 65000 + 93750 = 338,750 ft2
1 ac
43560 ft2
The field illustrated in Figure 13.16 contains approximately 7.78 acres. Why is it approximate? The dashed lines used to define the trapezoidal shapes only approximate the actual boundary lines of the field, Figure 13.17.
When the irregular boundary is uniform enough to allow a series of equal distances (d) along the base line to be established, the summation trapezoidal equation can be used. Study Figure 13.16 and the following equation for an illustration of this method.
+ E h + hf where A = Area; d = Distance between offsets (must be equal); ho and hn = End offsets measured from base line AB; ^ h = Summation of all interior offsets (excluding the two end offsets).
500 ft
400 ft
450 ft
L 200 ft
450 ft
400 ft
200 ft
FIGURE 13.18. Summation trapezoidal equation applied to a field.
200 ft
FIGURE 13.18. Summation trapezoidal equation applied to a field.
To use the summation equation the field length, AB, is divided into a number of equal distances (d) and the offsets (perpendicular distances from line AB to the curved edge of the field), h, are measured and recorded.
Problem: Determine the area (ac) of the irregular-shaped field illustrated in Figure 13.18.
Solution:
500 ft 400 ft = 200 ft x I-+ (450 ft + 390 ft + 275 ft + 370 ft + 390 ft) + -
= 200 ft x (250 ft + 450 ft + 390 ft + 275 ft + 370 ft + 390 ft + 200 ft) = 200 ft x 2325 ft = 465,000 ft2
1 ac
43,560 ft2
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