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