Question

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

Answer 1

\[\text { Suppose that the side of the cube  }= x cm\]

\[\text { Surface area of the cube = 6  }\times \text { (side  })^2 = 6 \times x^2 = 6 x^2 {cm}^2 \]

\[\text { i . e . , the sum of the surface areas of three such cubes  }= 6 x^2 + 6 x^2 + 6 x^2 = 18 x^2 {cm}^2 \]

\[\text { Now, these three cubes area placed together to form a cuboid . } \]

\[\text { Then the length of the new cuboid will be 3 times the edge of the cube } = 3 \times x = 3x cm\]

\[\text { Breadth of the cuboid = x cm }\]

\[\text { Height of the cuboid = x cm }\]

\[ \therefore\text {  Total surface area of the cuboid = 2 } \times\text {  (length  }\times\text {  breadth + breadth  }\times\text {  height + length } \times\text {  height) }\]

\[ = 2 \times (3x \times x + x \times x + 3x \times x)\]

\[ = 2 \times (3 x^2 + x^2 + 3 x^2 )\]

\[ = 2 \times (7 x^2 )\]

\[ = 14 x^2 cm\]

2

i.e., the ratio of the total surface area cuboid to the sum of the surface areas of the three cubes =

\[14 x^2 c m^2 : 18 x^2 c m^2 \]

\[ = 7: 9\]