# 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 - Mathematics

MCQ

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

• 7 : 9

• 49 : 81

• 9 : 7

• 27 : 23

#### Solution

Let, a → Side of each cube

So, the dimensions of the resulting cuboid are,

Length(l) = 3a

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 .

Concept: Surface Area of a Cuboid
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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 18 Surface Areas and Volume of a Cuboid and Cube
Exercise 18.4 | Q 4 | Page 35

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