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) = 4.324 m` Breadth `(b) = 1.005 m` Thickness `(t) = 2.01 cm = 2.01 xx 10^(4) m`
Therefore are of the sheet `= 2(lxxb+bxxt+txxl)`
`=(4.25517+0.0202005+0.0851034)`
`=2(4.255+0.0202+0.0851)`
`=2(4.360)=8.7206 = 8.73)`
Since area can contain a maximum of 3SF (Rule II article 2) therefore, rounding off.
we get : Area `= 8.72 m^(2)`
Like wise volume `= lxxbxxt=2.234xx1.005xx0.0201 m^(3)=0.0855289 m^(2)`
Since volume can contan 3SF, therefore, rounding off, we get : volume `= 0.0855 m^(3)`