Question

Three iron small cubes with sides 3 cm., 4 cm. and 5 cm respectively are melted and cast into a big cube. What is the ratio of total surface areas of sum of three cubes and a big cube?​

Answer 1

Step-by-step explanation:

Answer = 1 : 1

see the attachment

Answer 2

The ratio of total surface areas of sum of three cubes and a big cube is given by,

Volume of bigger cube = Sum of volumes of smaller cubes

Volume of a cube = a³  (a = side of the cube)

Given,

Three iron small cubes with sides 3 cm, 4 cm, and 5 cm

Volume of bigger cube = 3³ + 4³ + 5³

= 27 + 64 + 125

= 216

V = a³ = 216

Therefore, the side of a bigger cube is,

a³ = 216

a³ = 6³

a = 6

Now, consider, surface area of the cube,

Surface area of cube = S = 6a²

The ratio of total surface areas of sum of three cubes and a big cube

= [ S3 + S4 + S5 ] / S6

= [ 6 × 3² + 6 × 4² + 6 × 5² ] / (6 × 6²)

= [ 3² + 4² + 5² ] / 6²

= 50 / 36

= 25 / 28

Hence the required ratio.