Question

Four cubes of volume $125c{m}^3$ each are joined end to end in a row. Find the surface area of the resulting cuboid.

Answer 1

Let side of each cube be $a$

Given volume $=125c{m}^3$

So, $a^3=125c{m}^3$

$⟹a=5cm$

Length of the resulting cuboid$=4acm$

Breadth of the resulting cuboid$=acm$

Height of the resulting cuboid$=acm$

Surface area of cuboid$=2(length\times breadth+breadth\times height+height\times length)$

$=2(4a\times a+a\times a+a\times 4a)$

$=2(4a^2+a^2+4a^2)$

$=18a^2$

$=18\times 5^{2}c{m}^2$

$=450c{m}^2$

Hence, the surface area of the resulting cuboid is $450c{m}^2$