Question

Each edge of a cube is increased by $50\%$, then percentage increase in its surface area =.

• A. $125\%$
• B. $110\%$
• C. $150\%$

Answer 1

The correct option is A $125\%$

Let the edge $=acm$.
New measure of the edges after increase $=a+50\%a=a+\frac{a}{2}=\frac{3a}{2}$
Total surface area of the original cube $=6a^2$

Total surafec area of the new cube
$=6\times {\left(\frac{3a}{2}\right)}^2=6\times \frac{9a^2}{4}=\frac{9}{4}\times 6a^2=2.25\times 6a^2$

$=2.25\times$Total surface area of the original cube

Increase in the area
= New total surface area - orignal suface area
$=(2.25-1)\times 6a^2=1.25\times 6a^2$

Total surface area of the original cube
$=1.25\times$ Total surface area of the original cube.

$\therefore$ Percentage increase in surface area $=125\%$.