Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes.

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Solution

Let a be the side of three equal cubes
∴ Surface area of 3 cubes
= 3 x 6a² = 18a²
and length of so formed cuboid = 3a
Breadth = a
and height = a
∴ Surface area = 2(lb + bh + hl)
= 2[3a x a + a x a+a x 3a] = 2[3a² + a² + 3a²] = 2 x 7a² = 14a²
∴ Ratio in the surface areas of cuboid and three cubes
= 14a² : 18a²= 7 : 9

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