Question

Three cubes of metal with edges 3 cm, 4 cm, and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is

$\left(a\right)72c{m}^2\left(b\right)144c{m}^2\left(c\right)128c{m}^2\left(d\right)256c{m}^2$

Answer 1

Solution:

Given the edges of three metal cubes are 3 cm, 4 cm and 5 cm.

The volume of new cube = sum of the volume of three cubes

Let the edge of the new cube be x.

Recall that volume of cube = $l^3$ cubic units

The volume of the 1^st cube = $3^3$ = 27 $c{m}^3$

The volume of the 2^nd cube = $4^3$ = 64 $c{m}^3$

The volume of the 3^rd cube = $5^3$ = 125 $c{m}^3$

Volume of new cube = $x^3$ = 27 + 64 + 125 = 216 $c{m}^3$

$x^3$ = 216 $c{m}^3$

x = 6 cm

Hence the length of the edge of the new cube is 6 cm.

Lateral surface area of the cube of edge "a" = 4 $a^2$

Lateral surface area of the cube of edge "6" = 4 $\times 6^2$ = 144 $c{m}^2$

Therefore, Option b is correct answer.