Answer
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Hint: You can start by finding the area of the field without the error, i.e. $A = l \times b$ . Then calculate the margin of error in the area of the field by using the equation $\dfrac{{\Delta A}}{A} = \left( {\dfrac{{\Delta l}}{l} + \dfrac{{\Delta b}}{b}} \right)$ . Then express the area of the field with error in the form $A + \Delta A$
Complete step-by-step answer:
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}$
So, the area of the field $ = A + \Delta A = \left( {12000 \pm 800.4} \right){m^2} = \left( {1.2 \pm 0.08} \right) \times {10^4}{m^2}$
Note: The concept of errors teaches us a very important thing. You may have conducted experiments in the lab, and as you may know, we take multiple observations for the same experiments and then take an average for the final result, because if we only take one observation it may be errors that can be human error or any other error.
Complete step-by-step answer:
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}$
So, the area of the field $ = A + \Delta A = \left( {12000 \pm 800.4} \right){m^2} = \left( {1.2 \pm 0.08} \right) \times {10^4}{m^2}$
Note: The concept of errors teaches us a very important thing. You may have conducted experiments in the lab, and as you may know, we take multiple observations for the same experiments and then take an average for the final result, because if we only take one observation it may be errors that can be human error or any other error.
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