# Math Help - Significant figures problem...

1. ## Significant figures problem...

Hi, I put it in the physics forum, but it's applicable to this thread too, and I'm really confused

A rectangular plate has a length of (17.4 ± 0.2) cm and a width of (9.4 ± 0.1) cm. Calculate the area of the plate, including its uncertainty.

I proceed as follows:

the % of error for Length = 0.2/17.4*100% = 1.15%
the % of error for Width = 0.1/9.4*100% = 1.06%

the area of the plate = W*L = 17.4*9.4 = 163.56cm^2

the total uncertainty = [163.56cm^2 *(1.15%+1.06%)]/100% = 3.6 cm^2

since the length has 2 significant figures, I have to write the result as
1.6*10^2 ±4 cm^2 right,

I should write the area in only 2 significant figures which is 1.6*10^2 , right, but still it gives me wrong...

your input would be greatly appreciated,
dokrbb

2. ## Re: Significant figures problem...

The most direct way to do this is to note that the largest the length and width could be is 17.6 cm and 9.5 cm, respectively. That means the largest possible area is (17.6)(9.5)= 167.2 sq. cm. The smallest possible length and width will be 17.2 cm and 9.3 cm. That means the smallest possible area is (17.2)(9.3)= 159.96 sq. cm. (17.4)(9.4)= 163.56 which is 167.2- 163.56= 3.65 smaller than the largest possible value and 163.56- 159.96= 3.6 sq. cm. larger than the smallest value. So we are certain the true area is $163.56\pm 3.6$ sq. cm. Rounding that "error" to one decimal place would give your "4" but I don't see why you would do that.

There is an engineer's "rule of thumb" that "when quantities are added, errors are add; when quantities are multiplied, relative errors are added". So you are correct that since the "relative errors" are $\frac{0.2}{17.4}= 0.0115$ (or 1.15%) and $\frac{0.1}{9.4}= 0.0106$ (or 1.06%), the relative error in there product is 1.15+ 1.06= 2.21% so the error would be (.0221)(163.56)= 3.62. I can see rounding that to 3.6 (the error I have, above) but not to "4".