Question Number 20847 by j.masanja06@gmail.com last updated on 04/Sep/17
$${the}\:{rectangle}\:{is}\:{known}\:{to}\:{be}\:{twice} \\ $$$${as}\:{long}\:{as}\:{its}\:{wide}.{if}\:{the}\:{width}\:{is} \\ $$$${measured}\:{as}\:\mathrm{20}\:\pm\mathrm{0}.\mathrm{2}{cm}. \\ $$$${find}\:{the}\:{area}\:{in}\:{the}\:{form}\:{of}\:\left({A}\pm{b}\right) \\ $$$$ \\ $$
Answered by ajfour last updated on 04/Sep/17
$${A}={lw}\:\:\:\:{and}\:{as}\:{l}=\mathrm{2}{w} \\ $$$$\:\:\:{A}=\mathrm{2}{w}^{\mathrm{2}} =\mathrm{2}×\left(\mathrm{20}\right)^{\mathrm{2}} \:=\:\mathrm{800}{cm}^{\mathrm{2}} \\ $$$$\Rightarrow\:\Delta{A}=\mathrm{4}{w}\Delta{w}\:=\mathrm{4}×\mathrm{20}\left(\pm\mathrm{0}.\mathrm{2}\right){cm}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pm\mathrm{16}{cm}^{\mathrm{2}} \\ $$$${Area}\:=\:{A}+\Delta{A}\:=\mathrm{800}{cm}^{\mathrm{2}} \pm\mathrm{16}{cm}^{\mathrm{2}} \:. \\ $$$$ \\ $$