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

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Solution

Let the length of side be $a$
Surface area of $3$ cubes is $3 \times 6 a^{2} = 18 a^{2}$
The length of resultant cuboid $3 a$
Surface area of cuboid is $2 (a^{2} + 3 a^{2} + 3 a^{2}) 14 a^{2}$
The ratio is given as $14 a^{2} : 18 a^{2} 7 : 9$

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

Let the length of side be aSurface area of 3 cubes is 3×6a2=18a2The length of resultant cuboid 3aSurface area of cuboid is 2(a2+3a2+3a2)14a2The ratio is given as 14a2:18a27:9

Let the length of side be $a$
Surface area of $3$ cubes is $3 \times 6 a^{2} = 18 a^{2}$
The length of resultant cuboid $3 a$
Surface area of cuboid is $2 (a^{2} + 3 a^{2} + 3 a^{2}) 14 a^{2}$
The ratio is given as $14 a^{2} : 18 a^{2} 7 : 9$