Question

Each edge of a cube is increased by $50$%. Find the percentage increase in the surface area.

Answer 1

Let edge length$=a$

Total surface area$=6a^2$

with $50\%$ increase in edge, side$=a+50\%$ of a

$=a+\frac{a}{2}=\frac{3a}{2}$

Total surface area$=6{\left(\frac{3a}{2}\right)}^2$$=\frac{54a^2}{4}=\frac{27}{2}{a}^2$

Percentage increase$=\frac{\frac{27}{2}{a}^2-6a^2}{6a^2}\times 100$

$=\frac{15a^2}{12a^2}\times 100$

$=125$

$\Rightarrow 125\%$ increase in the surface area.