The length and breadth of a rectangle are `( 5.7 +- 0.1 ) cm and ( 3.4 +- 0.2 ) cm`, respectively calculate the area of rectangle with error limits.

Explanation

Here `l = (5.7+-0.1)cm, b = (3.4+-0.2) cm`
Area: `A = l xx b = 5.7xx3.4 = 19.38 cm^(2) = 19cm^(2)`
(rounding off to two significant figures)
`:. (Delta A)/(A) = +- ((Delta l)/(l) + (Delta b)/(b)) = +- ((0.1)/(5.7) + (0.2)/(3.4))`
`= ((0.34 + 1.14)/(5.7xx3.4)) = +- (1.48)/(19.38)`
`implies Delta A = +- (1.48)/(19.38) xx A = +- (1.48)/(19.38) xx 19.38 = +- 1.48 = +- 1.5`
(rounding off to two significant figures)
So, Area `= (19.0 +- 1.5) cm^(2)`