Question

A cubical block of metal was melted and recast into 20 cubes of edge 4.5 cm. what was the volume of the cubical block ?​

Answer 1

Given:

• Edge of one cube = 4.5 cm
• Number of cubes recasted from a cubical block of metal = 20

To find:

• Volume of the cubical block

$\:$

Solution:

We know that,

$\boxed {\mathfrak {\pink {Volume\ of\ cube = s^{3}}}}$

Where,

• s denotes the edge or side of the cube

Substitute the values,

Volume of 1 cube = (4.5)³

Volume of 1 cube = 4.5×4.5×4.5

$\bigstar {\sf {\orange {Volume\ of\ 1\ cube = 91.125\ cm^{3}}}}$

$\:$

Now,

Volume of 20 cubes = 20 × 91.125cm³

$\bigstar {\sf {\red {Volume\ of\ 20\ cubes = 1,822.5\ cm^{3}}}}$

$\:$

As, the cubical block of metal is recast into 20 cubes, therefore volume of the 20 cubes is equal to volume of the cubical block of metal

Volume of the cubical block of metal = Volume of the 20 cubes

$\boxed {\bf {\purple {Volume\ of\ cubical\ block\ of\ metal = 1,822.5\ cm^{3}}}}$

$\:$

Volume of cubical block:

Volume of cubical block of metal is 1,822.5 cm³

Answer 2

Given:-

• The edge of a cube is 4.5 cm, where 20 such cubes were recasted from a cubical block.

To find:-

• The volume of the cubical block.

Formula applied:-

• Volume of cube = (side)³ cu.units

Solution:-

It is given that 20 cubes of side 4.5 cm were recasted from a cuboidal block and we have to find the volume of the cuboidal block. From the analysis, we can assert that,

$\boxed{ \sf{Volume_{(20 \: cubes) }= Volume_{(cubical \: block)}}}$

So, the volume of 20 cubes equals the required answer, i.e, the volume of the cubical block. Let's find the volume of 20 cubes by multiplying 20 with the volume of a cube!

$\\ \longmapsto{ \sf{ \pmb{Volume_{(20\:cubes)} = 20 \times {(side)}^{3} }}} \\ \\ \longmapsto{ \sf{Volume_{(20\:cubes)} = 20 \times {(4.5)}^{3} }} \\ \\ \longmapsto{ \sf{Volume_{(20\:cubes)} = 20 \times 91.125}} \\ \\ \longmapsto{ \sf{Volume_{(20\:cubes)} = \boxed{ \tt{ \pmb{ \purple{ 1822.5 \: {cm}^{3} }}}}}}$

$\\ \therefore \underline{ \sf{ \pmb{The \: volume \: of \: the \: cubical \: block \: is \: \red{1822.5 \: {cm}^{3}} . }}}$

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Some formulae related to the concept:-

• Volume of cuboid = Length × Breadth × Height sq.units
• Volume of sphere = 4/3 πr³ cu.units
• Volume of cone = 1/3 πr²h cu.units
• Volume of cylinder = πr²h
• TSA of cube = 6a² sq.units
• TSA of cuboid = 2(lb + bh + lh) sq.units
• TSA/LSA of sphere = 4πr²
• TSA of cone = πr (l + r) sq.units
• TSA of cylinder = 2πr (h + r) sq.units
• LSA of cube = 4a² sq.units
• LSA of cuboid = 2(l + b)h sq.units
• CSA of cone = πrl sq.units
• CSA of cylinder = 2πrh sq.units

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