A sidewalk is to be constructed around a pool that measures #(10.0+-0.1) m# by #(17.0+-0.1) m#. If the sidewalk is to measure #(1.00+-0.01) m# wide by #(9.0+-0.1) cm# thick, what volume of concrete is needed and what is its approximate uncertainty?

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Answer 1 · 2017-07-26T00:38:33

Area of the pool #=lxxb=17xx10=170m^2#
Area of pool #+# sidewalk#=(l+2)xx(b+2)#
(sidewalk is #1m# wide on all sides).
#=(17+2)xx(10+2)=19xx12=228m^2#

Area of sidewalk#=228-170=58m^2#

Volume of concrete needed for the sidewalk #="Area"xx"thickness"#
#=58xx0.9=5.22m^3#

We know that

Volume #V=lxxbxxt#

Taking #log# of both sides we get

#logV=log(lxxbxxt)#
#=>logV=logl+logb+logt#

From theory of errors

#(DeltaV)/V=(Deltal)/l+(Deltab)/b+(Deltat)/t#
where #DeltaV# stands for error or uncertainty in volume and similarly for other terms.

Inserting given values we get

#(DeltaV)/V=(+-0.1)/10.0+(+-0.1)/17.0+(+-0.1)/9.0#

Errors/uncertainties are always taken as additive
#(DeltaV)/V=+-[(0.1)/10.0+(0.1)/17.0+(0.1)/9.0]#

#=>(DeltaV)/V=+-(0.01+0.006+0.011)#
#=>(DeltaV)/V=+-0.027#, rounded to three decimal places.

In terms of % uncertainty#=+-0.027xx100=+-2.7%#