Question

Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.

Answer 1

Edge of first cube = 3 cm
$\therefore$ Volume = $a^3=(3{)}^3=27c{m}^3$
Edge of second cube = 4 cm
$\therefore$ Volume = $a^3=(4{)}^3=64c{m}^3$
Edge of third cube = 5 cm
$\therefore$ Volume = $a^3=(5{)}^3=125c{m}^3$
Volume of three cubes together
= 27+64+125 = 216 $c{m}^3$
$\therefore$ Volume of the new cube = 216 $c{m}^3$
$\therefore$ Edge = $\sqrt[3]{3216}cm=\sqrt[3]{36\times 6\times 6}=\sqrt[3]{36^3}$
$=6cm$
Surface area of the new cube = $6a^2$
$=6\times (6{)}^2=6\times 6\times 6c{m}^2=216c{m}^2$