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.

Answer 1

Let each side of a cube = a cm

Then surface area = $6a^{2}c{m}^2$

and surface area of 3 such cubes

$=3\times 6a^2=18a^{2}c{m}^2$

By placing three cubes side by side we get a cuboid whose,

Length (l) $a\times 3=3a$

Breadth (b) = a

Height (h) = a

$\therefore \text{Total surface area=2(lb+bh+hl)}$

$=2[3a\times a+a\times a+a\times 3a]c{m}^2$

$=2[3a^2+a^2+3a^2]=14a^2$

$\therefore \text{Ratio between their surface areas}$

$=14a^2:18a^2=7:9$