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.

Explanation

Let breadth of the cuboid `= a `
Then, length of the new cuboid `= 3a`
Height of the new cuboid = a
Now, Total surface area of the new cuboid `(TSA) = 2(lb+bh+hl) `
` = 2(3a xx a + a xx a + a xx 3a) `
` = 14a^2 `
Total Surface area of three cubes `= 3 xx (6 side^2)`
` = 3 xx 6a^2 = 18a^2 `
Therefore, ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes` = (14a^2)/(18a^2) = 7/9 or 7:9 `
Therefore, required ratio is` 7:9`.