# Answer to Question #155316 in Civil and Environmental Engineering for Ian

Question #155316

A rectangular lot was measured using a 50 m metallic tape which was 1.02 cm too short. The recorded distances were 170.55 m long by 137.05 m wide. What is the error in area introduced due to the erroneous length of tape?

1
2021-01-18T01:17:25-0500

Length of tape = 50m

Convert to Centimetres where 1m = 100cm

50m = 50 x 100 = 5000cm

Measured Length = 170.55m

Measured Length in cm = 170.55 x 100 = 17055cm

Measured Width = 137.05m

Measured Length in cm = 137.05 x 100 = 13705cm

Area is given by Length x Width.

Measured area = 17055 x 13705 = 233,738,775 cm2

Actual length.

If 5000cm generate an error of 1.02cm, what about 17055 cm

(17055 x 1.02)/5000 = 3.47 cm

Actual length = 17055 + 3.47 = 17058.47cm

Actual Width.

If 5000cm generate an error of 1.02cm, what about 13705 cm

(13705 x 1.02)/5000 = 2.79 cm

Actual length = 13705 + 2.79 = 13707.79 cm

Actual area = 17058.47 x 13707.79 = 233,833,924.48 cm2

Error in Area = 233,833,924.48 - 233,738,775 = 95,149.48 cm2

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