Question

Three metal cubes whose edges measure $3$ cm, $4$ cm and $5$ cm respectively are melted to form a single cube. Find (i) side-length (ii) total surface area of the new cube. What is the difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes?

Answer 1

Three metal cubes whose edges measure $3cm$, $4cm$ and $5cm$ respectively are melted to form a single cube.

Let $a,b$ and $c$ be the side length of cubes with edges $3cm,4cm$ and $5cm$ respectively.

Let $l$ be the side length of new cube.

(i)

Volume of new cube $=$ Sum of volume of all three cubes

$=3^3+4^3+5^3$

$=27+64+125=216c{m}^3$

$\therefore$ Volume of new cube $=216=l^3$

$⟹l=6cm$

(ii) Total surface area of the new cube

$=6a^2$
$=6\left(6^2\right)$

$=216c{m}^2$

(iii)
Total surface area of the first cube with edge $3cm$

$=6a^2=6\times 9=54c{m}^2$

Total surface area of the $2^{nd}$ cube with edge $4cm$

$=6b^2=6\times 16=96c{m}^2$
Total surface area of the $3^{rd}$ cube with edge $5cm$

$=6c^2=6\times 25=150c{m}^2$
Sum of total surface area of these three cubes

$=54+96+150$
$=300c{m}^2$
The difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes $=300-216=84c{m}^2$

Hence, the difference is $84c{m}^2$.