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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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