If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?
Options:
A . 1 : 3
B . 2 : 3
C . 5 : 9
D . 7 : 9
Answer: Option D Let the length of each edge of each cube be a Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a Surface area of the cuboid : $$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$ Sum of surface area of 3 cubes : $$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$
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