The length , breadth , and thickness of a metal sheet are `4.234 m , 1.005 m , and 2.01 cm`, respectively. Give the area and volume of the sheet to the correct number of significant figures.

Explanation

length (l)=4.234m breadth (b)=1.005 m
thickness `(t)=2.01 cm=2.01xx10^(-2)m`
Therefore area of the sheet `=2lxxb+bxxt+txxl)`
`=2(4.234xx1.005+1.005xx0.0201+0.0201xx4.234)m^(2)`
`=2(4.3604739)m^(2)=8.720978 m^(2)`
Since area can contain a `"max"^(m)` of 3 SF (Rule II of article 4.2) therefore, rounding off, we get
Area `=8.72 m^(2)`
Like wise volume `=lxxbxxt`
`=4.234xx1.005xx0.0201 m^(3)=0.0855289 m^(3)`
Since volume can contain 3 SF, therefore, rounding off, we get
Volume `=0.0855 m^(3)`