Question

The dimensions of a piece of iron in the shape of a cuboid are 270 cm, 100 cm and 64 cm. If it is melted and recast into a cube, then the surface area of the cube will be:

Answer 1

given dimensions of the cuboid = 270cm × 100cm × 64cm

ATQ, the cuboid is melted and recasted into a cube.

therefore volume of the cuboid = volume of the cube

volume of the cuboid = l × b × h

= 270 × 100 × 64

= 1728000cm³

➡ volume of the cube = 1728000cm³

we know that side³ = volume of a cube

➡ side³ = 1728000cm³

➡ side = ³√1728000

➡ side = 120cm

the side of the cube is 120cm.

now it's surface area = 6 × side²

= 6 (120)²

= 6 × 14400

= 86400cm²

hence, the surface area of the cube is 86400cm².

Answer 2

We know that volume of cuboid = volume of cube.

so,volume of cuboid =l×b×h

=270*100*64=1728000 cm^3.

volume of cube =1728000cm^3.

(side)^3=1728000cm^3

side= 3√1728000

side= 120 cm

Then,total surface area= 6l^2

6x 120^2 = 86400cm^2.

hope it helps.