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

Given, length (l) = 4.234 m

Breadth (b) =1.005 m

Thickness (t) =2.01cm =0.0201m

Area of sheet (A)

$=2(\ell \times b+b\times t+t\times \ell )$

$=2[(4.234\times 1.005)+(1.005\times 0.0201)+(0.020\times 4.234\text{})]$

$=2\times 43604739$

$=8.7209478m^2$

As thickness has least number of significant figures 3, therefore rounding-off area up to three significant figures, we get

Area of sheet $\left(A\right)=8.72m^2$

Volume of sheet (V) = l $\times$ b $\times$ t

$=4.234\times 1.005\times 0.0201$

= 0.0855289

Rounding-off up to three significant figures, we get  Volume of the sheet

$=0.0855m^3$