If L = 20.04 m `pm`0.01 m , B= 2.52 m `pm ` 0.02 m . What are the values of `(L xx B) and (L+B)` ?

Explanation

Given errors are absolute errors, while the rule says that percentage errors are to be added. (you can use relative or fractional errors also in place of percentage errors, but you will find it less confusing while dealing with percentage errors ). Hence , the first step will be to convert the given absolute errors into pecentage errors.
`therefore L = 20.04 m pm (0.01)/(20.04) xx 100% = 20.04 m pm 0.049 %`
B = 2.52`m pm (0.02)/(2.52)xx 100% = 2.52 m pm 0.79%`
`therefore L xx B = (20.04xx2.52)m^2 pm (0.049+ 0.79)% = 50.50 m^2 pm 0.839 %`
This is the result. However , since the data given in the question was in terms of absolute errors , so we should give our result also in absolute errors.
`L xx B = 50.50 m^2 pm (0.839)/(100)xx50.50 m^2 = 50.50 m^2 pm 0.423 m^2 pm 0.423 m^2 = 50.50 m^2 pm 0.42 m^2 `
Similarly, `L/B = (20.04 cancelm)/(2.52 cancelm) pm (0.049 + 0.79)% = 7.95 pm 0.839% = 7.95 pm (0.839)/(100)xx 7.95 = 7.95 xx 0.066`