Question

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 total surface areas of the three cubes.

• A. $7:13$
• B. $7:3$
• C. $7:9$
• D. $7:5$

Answer 1

Answer: (C)

Given, three cube are placed adjacent in a row to form a cuboid.
Let the side of cube be $a$, three cube placed in row then breadth of cuboid be $a$, length of cuboid is 3$a$, height of cuboid is $a$.
Sum of total surface area of three cubes $=6a^2+6a^2+6a^2=18a^2$
Total surface area of resulting cuboid $=2(lb+bh+lh)=2(3a\times a+a\times a+3a\times a)=2\left(7a^2\right)=14a^2$
Ratio of total surface area of cuboid to that of cube $=\frac{{14a}^2}{18a^2}=\frac{7}{9}=7:9$