Question

the length, breadth and thinness of a metal sheet are 4.234m, 1.005m and 2.01cm respectively. give the area and volume of the sheet to correct number of significant figures.

Answer 1

Answer:

Explanation:hope it helps

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,

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

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

Hope it helps you.