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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.
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Solution
Length of new cuboid= 3a
Breadth of cuboid=a
Height of new cuboid= a
The total surface area of new cuboid
`⇒(TSA)_1= 2[lb+bh+hl]`
`⇒(TSA)_1=2[3axxa+axxa+3axxa]`
`⇒(TSA)= 14 a^2`
Total surface area of three cubes
`⇒(TSA)_2=3xx6a^2=18a^2`
`∴(TSA)_2/(TSA_2)=(14a^2)/(18a^2)= 7/9`
`Ratio is 7:9`
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