Cube Surface Area Increase.

let $S_1$ be the surface area of the original cube and $a$ be its side length

let $S_2$ be the surface area of the larger cube and $1.2a$ be its side length

$\dfrac{S_2}{S_1}=\dfrac{6(1.2a)^2}{6a^2}=\dfrac{8.64}{6}=1.44$

$S_2=1.44S_1$

Therefore, the increase in surface area is $(1.44-1)(100\%)=$ $\color{#D61F06}\large \boxed{44\%}$

Couldn't write with proper formatting but here is a picture

Let the new variables be denote by ' on old variables . $\large\dfrac{A'}{A}=(\dfrac{r'}{r})^2=\dfrac{36}{25}$ Subtracting 1 from both sides: $\large \dfrac{\triangle A}{A}=\dfrac{11}{25}$ In percentage $\dfrac{11}{25}=\boxed{44\%}$

we let the side of the first cube to be 1 unit

$S_i$ $=$ $6a^2$ $=$ $6*1^2$ $=$ $6$

then the side of the second cube is 1.2 units

$S_f$ $=$ $6a^2$ $=$ $6*1.2^2$ $=$ $8.64$

$increased$ $in$ $surface$ $area=$ $\frac{S_f-S_i}{S_i}$ $=$ $44$ $percent$

Let $a$ be the edge length of the cube, then the edge length of the new cube is $1.2a$ .

Formula: $s = 6a^2$ where $s$ is the surface area of the cube and $a$ is the edge length of the cube

The surface area of the new cube is $6(1.2a)^2=8.64a^2$

The percentage increased is $\dfrac{(8.64-6)}{6}(100\%) = \boxed{44\%}$