If each edge of a cube is increased by $50\%$, then find the percentage increase in its surface area.

The correct option is A $125\%$
Let the edge $=a\text{cm}$
New edge after increase $=a+50\%a=a+\frac{a}{2}=\frac{3a}{2}$
Total surface area of original cube $=6a^2$

TSA of new cube $=6\times {\left(\frac{3a}{2}\right)}^2=6\times \frac{9a^2}{4}=2.25\times 6a^2$
$TSA=2.25\times Totalsurfaceareaoforiginalcube$

Increase in area $=[2.25\times 6a^2]-\left[6a^2\right]=(2.25-1)\times 6a^2=1.25\times 6a^2$
$\therefore$ Increase in area
$=1.25\times$ TSA of original cube
$\Rightarrow$ Increase $\%=125\%$