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Problem (6)
To calibrate her pacing, a surveyor paced a known 176.5 ft ling base line four times: 93.5, 94.5, 95.0, 93.0. She then preoceded to pace the perimeter of a warehouse in order to determine its area. Her three pacing trips around the perimter were:
- 144.5, 110.0, 144.0, 111.5
- 145.5, 110.5, 143.5, 110.5
- 146.0, 109.5, 144.0, 110.5
What is the area of the warehouse to the nearest 10 sq ft?
Solution
Her pace conversion is (176.5 ft) / [(93.5+94.5+95.0+93.0)/4] = 176.5 ft / 94.225 p
Average length is (144.5+145.5+146.0+144.0+143.5+144.0) / 6 = 144.5833 p
Average width is (110.0+110.5+109.5+111.5+110.5+110.5) / 6 = 110.4167 p
Area = (144.5833 p x 176.5 ft / 94.225 p) x (110.4167 x 176.5 / 94.225) = 56,015.733 = 56,020 sq ft