Question:
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let 'a' be the edge of the cube
Therefore the surface area of the cube $=6 a^{2}$
i.e., $\mathrm{s}_{1}=6 \mathrm{a}^{2}$
We get a new edge after increasing the edge by 50%
The new edge = a + 50/100∗a
= 3/2∗a
Considering the new edge, the new surface area is $=6 *(3 / 2 a)^{2}$
i.e., $\mathrm{S}_{2}=6 * \frac{9}{4} \mathrm{a}^{2}$
$\mathrm{S}_{2}=\frac{27}{2} \mathrm{a}^{2}$
Therefore, increase in the Surface Area
$=\frac{27}{2} a^{2}-6 a^{2}$
$=\frac{15}{2} a^{2}$
So, increase in the surface area
$=\frac{\frac{15}{2} a^{2}}{6 a^{2}} * 100$
= 15/12∗100
= 125%