Question

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








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Answer 1

Answer:

# Answer- A=8.72 m^2 and V=0.0855 m^3. Area and volume both must have least significant figures i.e. 3. Hence, our answer is A=8.72 m^2 and V=0.0855 m

or

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=8.721m2

∴A=8.721m2  to correct significant digits

V=L×B×T

∴V=4.234×1.005×0.0201

∴V=0.0855m3 to correct significant digits

Answer 2

Explanation:

Given:-

Length = 4.234 m

Breadth = 1.005 m

Height (thickness of sheet) = 2.01 cm = 0.0201 m

Solution:-

$Area \: = \: 2(lb \: + \: bh \: + hl) \\ = 2(4.234 \: \times1.005 \: + 1.005 \: \times 0.201 \: + \: 0.201\: \times \: 4.234) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: 2(4.25517 + 0.0202005 + 0.08551034) \\ = (8.7209478) \\= 8.72 \: {m}^{2}$

Now,

$Volume \: = \: l \: \times \: b \: \times \: h \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4.234 \: \times 1.005 \: \times \: 0.0201 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0.0855 \: m ^{3}$

$Hence \: The \: area \: is \: 8.72 \: m ^{2} \: \: and \: \\ Volume \: is \: 0.0855 {m}^{3}$