Question

Three cubes of iron whose edges are 3cm, 4cm, 5cm respectively are melted and formed into a single cube.What was edge of the new cube formed

Answer 1

Vol(R1+R2+R3)=vol(r)

4/3*3^3+4^3+5^3=4/3r^3

27+64+125=r^3

216=r^3

r=√216

r=6

Answer 2

Answer:

6 cm

Step-by-step explanation:

Volume of cube = $side^3$

Volume of cube of side 3 cm= $3^3=27cm^3$

Volume of cube of side 4 cm=$4^3=64 cm^3$

Volume of cube of side 5 cm=$5^3=125 cm^3$

Total Volume of Cube = 27+64+125=216 cubic cm

Now we are given that Three cubes of iron whose edges are 3cm, 4cm, 5cm respectively are melted and formed into a single cube.

Let the side of the single cube be x

Total volumes of cube = Volume of new cube

So,$216=x^3$

$\sqrt[3]{216}=x$

$6=x$

Hence the edge of new cube is 6 cm