Question

The length, breadth and thickness of a rectangular sheet of metal are 3.234 m, 2.005 m, and 1.01 cm respectively. Give the area and volume of the sheet to correct significant figures.​

Answer 1

Answer:

A=2×(L×B+B×T+T×L)

∴A=2×(4.234×1.005+1.005×0.0201+0.0201×4.234)

∴A=2×(4.2552+0.0202+0.0851)

∴A=8.721m

2

∴A=8.721m

2

to correct significant digits

V=L×B×T

∴V=4.234×1.005×0.0201

∴V=0.0855m

3

to correct significant digits

Answer 2

Given: The length, breadth and thickness of a rectangular sheet of metal are 3.234 m, 2.005 m, and 1.01 cm respectively.

To find: The area and volume of the sheet to correct significant figures.​

Solution:

• The rectangular sheet is actually a cubiod.

• To find the area of the surface of a cuboid, we use the following formula,

       $Area = 2 * (l*b + b*h + h*l)$

• Here, l is the length, b is the breadth and h is the height ot the thickness of the cuboid.

• We multiply by 2 because in a cuboid, there are two surfaces of each rectangle present in the cuboid.
• The thickness or height is given as 1.01 cm which is equal to 0.0101 m.
• Hence, the area of the cuboid is,

       $Area = 2 * ((3.234*2.005) + (2.005*0.0101) + (0.0101*3.234))$

                $= 2 * (6.48 + 0.02 + 0.03)$

                $= 13.06 m^{2}$

• To find the volume of the rectangluar sheet, we use the formula of volume of cuboid,

        $Volume = l * b * h$

                     $= 3.234 * 2.005 * 0.0101$

                     $= 0.0655 m^{3}$

Therefore, the area and volume of the sheet to correct significant figures is 13.06 m² and 0.0655 m³.