Question

12. Three solid metal cubes of edges 6 cm, 8 cm and 10 cm are melted and recasted into a
single solid cube. Find the length of the edge of the cube so obtained
​

Answer 1

$\underline{\boxed{\bold{Question}}}$

Three solid metal cubes of edges 6 cm, 8 cm and 10 cm are melted and recasted into a single solid cube. Find the length of the edge of the cube so obtained

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$\underline{\boxed{\bold{Answer}}}$

$\huge\red{12 \: cm}$

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$\underline{\boxed{\bold{Given}}}$

Edge of a first solid cube= 6cm

Volume=(6)³=216cm³

Edge of a second cube =8cm

volume=(8)³=512cm³

Edge of a third cube =10cm

volume=(10)³=1000

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$\underline{\boxed{\bold{To \: find}}}$

The edge of the resulting cube

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$\underline{\boxed{\bold{Solution}}}$

Let x be the edge of the formed cube volume.

Sum of three volumes= 216+512+1000

${\huge{\underline{\sf{\blue{1728cm²}}}}}$

Edge of the volume be =a³

a³=1728

a=(12)³

a=12

Hence ,the of the cube is 12cm.

[Hope this helps you.../]

Answer 2

GIVEN :

Edge of a first cube = 6 cm.

Edge of a second cube = 8 cm.

Edge of a third cube = 10 cm.

TO FIND :

The edge of the cube formed ?

The length of the cube obtained ?

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SOLUTION :

Edge of a first solid cube,

Volume = a³

=> (6)³

=> 216 cm³

Edge of a second solid cube,

=> (8)³

=> 512 cm³

Edge of a third solid cube,

=> (10)³

=> 1000 cm³

Hence,

• Volume of a first cube = 216 cm³

• Volume of a second cube = 512 cm³

• Volume of a third cube = 1000 cm³

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Now,

To find the edge of the cube formed = ?

Sum of the volumes of the three solid cubes = 216 + 512 + 1000

=> 1728.

Volume = a³

=> a³ = 1728

=> a = 12³

=> a = 12 cm

•°• Therefore, the edge of a cube formed is 12 cm.

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