Question

Three equal cubes are placed adjacently in a row. The ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes, is
A. 7 : 9
B. 49 : 81
C. 9 : 7
D. 27 : 23

Answer 1

Given: Three equal cubes are placed adjacently in a row.

Let the edge of cube be 'a'.

Surface area of a cube = 6a²

Sum of the surface area of 3 cubes = 3 × 6a² = 18a²

Sum of the surface area of 3 cubes = 18a²

So, length of resulting cuboid thus formed , l = a + a + a = 3a

Breadth of resulting cuboid , b = a

Height of resulting cuboid , h = a

Surface area of cuboid = 2(lb + bh + hl)  

= 2(3a × a + a × a + a × 3a)

= 2(3a² + a² + 3a²)

= 14a²

Total surface area of resulting cuboid = 14a²

Ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes :  

= 14a²/18a²

= 14/18

= 7/9  

= 7 : 9

Hence, the ratio of the total surface area of the resulting cuboid to that of the sum of the surface areas of three cubes is 7 : 9.

Option (A) 7 : 9 is correct.

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