The length of a line originally 100 mm long on a map plotted to a scale of 1/1000, was found to be 96 mm due to shrinkage of the map. The map prepared using a tape of length 20 m was later found to be actually 20.03 m. If a certain area on the map, measured using a planimeter, is 282 mm², what is the correct area on the ground?

Concept:

Shrinkage factor = Shrunk length/ Actual length

True area\(=S.F.^2\times\)measured area

Explanation:

Given

Actual length = 100 mm

Shrunk length = 96 mm

Original Scale = 1/1000

Actual length of tape  (L) = 20 m

Incorrect length of tape (L') = 20.03 m

Shrinkage factor =\(\frac{96}{100}=0.96\)

So, measured area of the ground = \(\frac{C.A.}{S.F.^2}=\frac{282}{0.96^2}=305.98958mm^2\)

\(S.F. = \frac{L'}{L}=\frac{20.03}{20}=1.0015\)

True Area = \(S.F.^2\times M.A=1.0015^2\times305.98958=306.91mm^2=307mm^2\) 

The correct area on the ground is 307 mm²