Question

A plan of an old survey plotted to a scale of 10 m to 1 cm carried on a note stating that the chain having nominal length 30 m was 20 cm too short. It was also found that plan has shrunk so that a line originally 10 cm long was 9.69 cm. The area of the plot on the available plan was found to be $60c{m}^2$. The correct area of the plot is _______ ha.

• A. 0.63

Answer 1

The correct option is A 0.63
Correct area on map = ${\left[\frac{\text{Original length of tape}}{\text{Nominal length of tape}}\right]}^2\times {\left[\frac{1}{\text{Shrinkage factor}}\right]}^2\times \text{area measured}$

Nominal length of tape = 30 m

$\because$ Tape is 20 cm too short

$\therefore$ Originally length of tape = 30 - 0.2 = 29.80 m

$\text{Shrinkage factor}=\frac{\text{Shrunk length}}{\text{Original length}}=\frac{9.69}{10}=0.969$

$\therefore$ Correct area on map = ${\left(\frac{29.80}{30}\right)}^2\times {\left[\frac{1}{0.969}\right]}^2\times 60$

$=63.05c{m}^2$

correct area on ground = $63.05\times 100\left[\begin{array}{c}\because 1cm=10m\\ \therefore 1c{m}^2=100m^2\end{array}\right]$

$=6305m^2=0.63ha$