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

Explanation

length `(l) = 4.234 m`
breadth `(b) = 1.005 m`
thickness `(t) = 2.01 cm = 2.01 xx 10^(–2) m`
Therefore area of the sheet
`=2(lxxb+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 `= 8.72m^(2)`
Like wise volume `=lxxbxxt`
`=4.234xx1.005xx0.0201m^(3)=0.0855289m^(3)`
Since volume can contain 3 SF, therefore, rounding off, we get
Volume `=0.0855 m^(3)`