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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

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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

Options

  • 7 : 9

  • 49 : 81

  • 9 : 7

  • 27 : 23

MCQ
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Solution

Let, a → Side of each cube

So, the dimensions of the resulting cuboid are,

Length(l) = 3a

Breadth (b) = a

Height (h) = a

Total surface area of the cuboid,

=2(lb + bh + hl)

=2[(3a) a + a × a + a (3a)]

= 14 a2

Sum of the surface areas of the three cubes,

= 3 (6a2) 

= 18 a2                                                                            Required ratio,

=`(14a^2)/(18a^2)`

=7:9

Thus, the required ratio is 7: 9 .

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Chapter 18: Surface Areas and Volume of a Cuboid and Cube - Exercise 18.4 [Page 35]

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R.D. Sharma Mathematics [English] Class 9
Chapter 18 Surface Areas and Volume of a Cuboid and Cube
Exercise 18.4 | Q 4 | Page 35

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