Question

Two equal cubes are placed adjoint in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of two cubes.

• A. $\frac{5}{2}$
• B. $\frac{5}{6}$
• C. $\frac{5}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

$\frac{5}{6}$

$\frac{\text{total surface area of cuboid}}{\text{sum of surface areas of 2 cubes}}=\frac{2(2a\times a+a\times a+2a\times a)}{6a^2+6a^2}$

$=\frac{2(2a^2+2a^2+a^2)}{12a^2}$

$=\frac{10a^2}{12a^2}$

$=\frac{5}{6}$.