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A solid metallic cube is melted to form a cuboid of sides in the ratio 1 : 2 : 4. The percentage by which the sum of the surface areas of the cuboid exceeds the surface area of the cube is nearest to?

A solid metallic cube is melted to form a cuboid of sides in the ratio 1 : 2 : 4. The percentage by which the sum of the surface areas of the cuboid exceeds the surface area of the cube is nearest to? Correct Answer 17%

Calculation:

Volume of the cube of side length a = a3

Let r, 2r and 4r are the sides of cuboid,

Volume of cuboid = r × 2r × 4r

⇒ 8r3

∴ a3 = 8r3

⇒ a = 2r

Total surface area of cuboid =2( r × 2r + r × 4r + 2r × 4r)

⇒ 28r2

\Total surface area of cube = 6a2

⇒ 6 × 4r2 = 24r2

Required percentage = ((28r2 - 24r2)/(24r2)) × 100

⇒ 16.66% ≈ 17%

Additional Information

The surface area of a cuboid of length a breadth b and height c is given by 2(ab + bc + ac).

The surface area of a cube of side length 'a' will be 6a2.

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