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Question 4 options: It is desired to lay off certain horizontal distances for building layouts. The length of the tapes are not 100.00 ft. Determine the field dimensions that should be used with the incorrect tape the lengths so that the correct dimensions are obtained. Desired dimension: 725.00 by 180.00 ft Correct tape length: 100.06 ft

Respuesta :

Answer:

true length = 730.225 ft

true width = 181.297 ft

Explanation:

given data

length of the tapes L1= 100.00 ft

Correct tape length L = 100.06 ft

Desired dimension D = 725.00 × 180.00 ft

solution

first we get here true dimension that is

true dimension = [tex](\frac{L1}{L})^2 *D[/tex]    ..........1

true dimension = [tex](\frac{100.06}{100} )^2[/tex] × 725.00 × 180.00

true dimension = 725.87 × 180.216 ft

and

field dimension = width × length

so field dimension = 725.87 × 180.216

so here true length and true width will be

true length = [tex]\frac{L1}{L}[/tex] × desired dimension length

true length = [tex]\frac{100.06}{100}[/tex] × 725.87

true length = 730.225 ft

and

true width = [tex]\frac{L1}{L}[/tex] × desired dimension width

true width = 1.006 × 180.216

true width = 181.297 ft

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