Question

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

• A. $5:9$
• B. $3:5$
• C. $2:5$
• D. $7:9$

Answer 1

Answer: (D)

$7:9$

Total surface area of a cube of edge a $=6a^2$
Length of the resulting cuboid $a+a+a=3a$
Height and breadth remain same as $a$

Total surface area of a cuboid of length l, breadth b and height h $=2(l\times b+b\times h+l\times h)$

So, the total surface area of this cuboid $=2(3a\times a+a\times a+3a\times a)$

$=14a^2$

Required ratio =$14a^2:18a^2$

$=14:18$

$=7:9$