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If a cuboid of dimensions 32 cm × 12 cm × 9 cm is melted and recast into two equal cubes of the same size, what will be the ratio of the total surface area of the cuboid to the total surface area of the two cubes?

If a cuboid of dimensions 32 cm × 12 cm × 9 cm is melted and recast into two equal cubes of the same size, what will be the ratio of the total surface area of the cuboid to the total surface area of the two cubes? Correct Answer 65 ∶ 72

As we know

Volume of cuboid = lbh

Total surface of area of cuboid = 2(lb + bh + hl)

Volume of cube = a3

Total surface cube = 6a2

Dimensions of cuboid = 32 cm × 12 cm × 9 cm

Let the side of the cube = a

According to the question

⇒ 2a3 = 32 × 12 × 9

⇒ a3 = (32 × 12 × 9)/2

⇒ a3 = 16 × 12 × 9

⇒ a = 12

Total surface area of cuboid = 2(lb + bh + hl) = 2(32 × 12 + 12 × 9 + 9 × 32) = 2(384 + 108 + 288) = 2 × 780 = 1560

Total surface area of two cubes = 2 × 6a2 = 2 × 6 × 12 × 12 = 1728

Required ratio = 1560 ∶ 1728 = 65 ∶ 72 

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