Question

Each edge of a cube is increased by $\left(\frac{27}{2}\right)x^2-6x^2$.Find the percentage increase in the surface area of the cube.

Answer 1

Explanation:

Let x be the edge of a cube.

Surface area of the cube having edge $x=6x^2$ ………..(1)

As given, a new edge after increasing the existing edge by $50\%$,we get

The new edge =$x+(\frac{50x}{100})$

The new edge = $\frac{3x}{2}$

Surface area of the cube having edge$\frac{3x}{2}=6x{\left(\frac{3x}{2}\right)}^2=\left(\frac{27}{2}\right)x^2$……..(2)

Subtract equation (1) from (2) to find the increase in the Surface Area:

Increase in the Surface Area =$\left(\frac{27}{2}\right)x^2-6x^2$

Increase in the Surface Area = $\frac{15}{2}{x}^2$

Now,Percentage increase in the surface area

$=\left(\right(\frac{15}{12})x^2/6x^2)\times 100$

$$\begin{gathered} =\frac{15}{12}\times 100 \\ =125\% \end{gathered}$$

Hence, the percentage increase in the surface area of a cube is $125\%$