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The length and breadth of a field are 22.4 cm and 15.8 cm respectively and have been measured to an accuracy of 0.2 cm .find the percentage error in the area of this?
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The length and breadth of a field are 22.4 cm and 15.8 cm respectively...
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The length and breadth of a field are 22.4 cm and 15.8 cm respectively...
Introduction:
In this problem, we are given the length and breadth of a field along with their respective accuracies. We need to calculate the percentage error in the area of the field.

Given:
Length of the field = 22.4 cm
Breadth of the field = 15.8 cm
Accuracy = 0.2 cm

Calculating the area:
The area of a rectangle is given by the formula: Area = Length * Breadth

Substituting the given values:
Area = 22.4 cm * 15.8 cm
Area = 353.92 cm^2

Calculating the maximum and minimum possible values for the length and breadth:
Since the measurements have been taken with an accuracy of 0.2 cm, the actual values of the length and breadth can vary within ±0.2 cm.

Maximum possible length = 22.4 cm + 0.2 cm = 22.6 cm
Minimum possible length = 22.4 cm - 0.2 cm = 22.2 cm

Maximum possible breadth = 15.8 cm + 0.2 cm = 16 cm
Minimum possible breadth = 15.8 cm - 0.2 cm = 15.6 cm

Calculating the maximum and minimum possible areas:
Using the maximum and minimum possible values for the length and breadth, we can calculate the maximum and minimum possible areas.

Maximum possible area = Maximum possible length * Maximum possible breadth
Maximum possible area = 22.6 cm * 16 cm
Maximum possible area = 361.6 cm^2

Minimum possible area = Minimum possible length * Minimum possible breadth
Minimum possible area = 22.2 cm * 15.6 cm
Minimum possible area = 346.32 cm^2

Calculating the percentage error in the area:
Percentage error in the area can be calculated using the formula:

Percentage error in the area = (|Maximum possible area - Minimum possible area| / Actual area) * 100

Substituting the values:
Percentage error in the area = (|361.6 cm^2 - 346.32 cm^2| / 353.92 cm^2) * 100
Percentage error in the area = (15.28 cm^2 / 353.92 cm^2) * 100
Percentage error in the area ≈ 4.32%

Conclusion:
The percentage error in the area of the field is approximately 4.32%. This means that the actual area of the field can vary by around 4.32% due to the measurement inaccuracies in the length and breadth.
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The length and breadth of a field are 22.4 cm and 15.8 cm respectively and have been measured to an accuracy of 0.2 cm .find the percentage error in the area of this?
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The length and breadth of a field are 22.4 cm and 15.8 cm respectively and have been measured to an accuracy of 0.2 cm .find the percentage error in the area of this? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The length and breadth of a field are 22.4 cm and 15.8 cm respectively and have been measured to an accuracy of 0.2 cm .find the percentage error in the area of this? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The length and breadth of a field are 22.4 cm and 15.8 cm respectively and have been measured to an accuracy of 0.2 cm .find the percentage error in the area of this?.
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