Question

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

Answer 1

Let the length of each edge of the cube be $l$ units.Then,

$S_1=TSA=6l^2$

Length of each edge when each edge is increased by $50$%

$=l+\frac{50}{100}\times l=\frac{3l}{2}$

$S_2=$ TSA of new cube $=6\times {\left(\frac{3l}{2}\right)}^2=\frac{27}{2}{l}^2$

Percentage increase in surface area $=\frac{S_2-S_1}{S_1}\times 100$

$=\frac{\frac{27}{2}{l}^2-6l^2}{6l^2}=\frac{15}{12}\times 100=125$