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Here, we are given a rectangular field that has some error of margin in the length and the breadth of the field, so the area will also have some error margin which we have to calculate in the solution. The field has the following dimensions

Length $ = \left( {120 \pm 2} \right)m$

Breadth $ = \left( {100 \pm 5} \right)m$

We know that the area for the field is

$A = l \times b$

$ \Rightarrow A = 120 \times 100$

$ \Rightarrow A = 12000{m^2}$

Here we are given the dimensions (length and breadth) of a rectangular field. It is also mentioned that there is an $ \pm 2m$ error margin in the length of the field. It means that the length of the field may be $2m$ longer or $2m$ shorter than the measured length. It is also mentioned that there is an $ \pm 5m$ error margin in the breadth of the field. It means that the breadth of the field may be $5m$ longer or $5m$ shorter than the measured breadth.

So, the margin of error in the area of the field can be calculated by using the equation

$\dfrac{{\Delta A}}{A} = \left( {\dfrac{{\Delta l}}{l} + \dfrac{{\Delta b}}{b}} \right)$

$ \Rightarrow \Delta A = \left( {\dfrac{{\Delta l}}{l} + \dfrac{{\Delta b}}{b}} \right) \times A$

$ \Rightarrow \Delta A = \left( {\dfrac{2}{{120}} + \dfrac{5}{{100}}} \right) \times 12000$

$ \Rightarrow \Delta A = 800.4{m^2}$