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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.
Given:
Three equal cubes are placed adjacently in a row.
To do:
We have to find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let each side of the cube be $s\ cm$
This implies,
Surface area of the cube $= 6s^2\ cm^2$
Surface area of three such cubes $= 3 \times 6s^2$
$= 18s^2\ cm^2$
By placing three cubes side by side, we get a cuboid of
Length $(l) = s \times 3 = 3s$
Breadth $(b) =s$
Height $(h) = s$
Therefore,
Total surface area of the cuboid $= 2(lb + bh + lh)$
$= 2(3s \times s+s \times s+s \times 3s)$
$= 2(3s^2 + s^2 + 3s^2)$
$= 14\ s^2$
The ratio between their surface areas $= 14s^2 : 18s^2$
$= 7 : 9$
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