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

Here, length I = 4.234 m, breadth b = 1.005m, thickness, `t = 2.01 cm = 2.01xx10^(-2)m`
Area of the sheet = `2 (lxxb + bxxt + +txxl) = 2(4.234xx1.005 + 1.005 xx0.0201 + 0.0201 xx 4.234)`
`= 2 (4.3604739) = 8.7209478 m^2`
As area can contain a maximum of three significant digits, therefore, rounding off, we get
`Area = 8.72 m^2`
Also, volume = `lxxbxxt`
`V = 4.234xx1.005xx0.0201 = 0.0855289 = 0.0855 m^3` ("containing three singinficant figures")`