Alex measured the field to the nearest meter and got a width
of 7m and a length of 9m calculate the length absolute relative and percentage error
Answers
Area
When working out areas you need to think about both the width and length ... they could possibly both be the smallest measure or both the largest.
Example: Alex measured the field to the nearest meter, and got a width of 6 m and a length of 8 m.
Measuring to the nearest meter means the true value could be up to half a meter smaller or larger.
area 6x8 measure error 41.25,48,55.25
The width (w) could be from 5.5m to 6.5m:
5.5 ≤ w < 6.5
The length (l) could be from 7.5m to 8.5m:
7.5 ≤ l < 8.5
The area is width × length:
A = w × l
The smallest possible area is: 5.5m × 7.5m = 41.25 m2
The measured area is: 6m × 8m = 48 m2
And the largest possible area is: 6.5m × 8.5m = 55.25 m2
41.25 ≤ A < 55.25
Absolute, Relative and Percentage Error
The only tricky thing here is ... which is the absolute error?
From 41.25 to 48 = 6.75
From 48 to 55.25 = 7.25
Answer: pick the biggest one! So:
Absolute Error = 7.25 m2
Relative Error = 7.25 m248 m2 = 0.151...
Percentage Error = 15.1%
(Which is not very accurate, is it?)