Three solid iron cubes of edges 8 cm, 10 cm and 12 cm are melted together to make a new cube. It is observed that 496 cm³ of the melted material is lost due to improper handling. The area (in cm²) of the whole surface of the newly formed cube is:
- A1331
- B1176
- C2197
- D3375
Solution & Step-by-step Explanation
First, find the total initial volume of the three iron cubes.
Volume of a cube with side a is V=a
3
.
Volume of the first cube (V
1
) = 8
3
=512cm
3
Volume of the second cube (V
2
) = 10
3
=1000cm
3
Volume of the third cube (V
3
) = 12
3
=1728cm
3
Total initial volume:
Total Volume=512+1000+1728=3240cm
3
Material lost due to improper handling = 496cm
3
Volume of the new cube (V
new
):
V
new
=3240−496=2744cm
3
Let the side of the newly formed cube be A.
A
3
=2744
A=
3
2744
=14cm
The total surface area of a cube is given by the formula 6A
2
:
Total Surface Area=6×(14)
2
=6×196=1176cm
2
Volume of a cube with side a is V=a
3
.
Volume of the first cube (V
1
) = 8
3
=512cm
3
Volume of the second cube (V
2
) = 10
3
=1000cm
3
Volume of the third cube (V
3
) = 12
3
=1728cm
3
Total initial volume:
Total Volume=512+1000+1728=3240cm
3
Material lost due to improper handling = 496cm
3
Volume of the new cube (V
new
):
V
new
=3240−496=2744cm
3
Let the side of the newly formed cube be A.
A
3
=2744
A=
3
2744
=14cm
The total surface area of a cube is given by the formula 6A
2
:
Total Surface Area=6×(14)
2
=6×196=1176cm
2