Question

Two cubes have their volumes in the ratio $1:27$. Find the ratio of their surface areas.

Answer 1

Step 1: Find the relation between sides of both cubes:

Let the side of first cube C₁ be $s_1$ and that of second cube C₂ be $s_2$.

As, $\frac{Volumeof{C}_1}{Volumeof{C}_2}=\frac{1}{27}$

$$\begin{gathered} \Rightarrow \frac{{\left(s_1\right)}^3}{{\left(s_2\right)}^3}=\frac{1}{27}[\because Volumeofcube={\left(side\right)}^3] \\ \Rightarrow \frac{s_1}{s_2}=\frac{1}{3} \\ \Rightarrow s_2=3s_1...\left\{i\right\} \end{gathered}$$

Step 2: Find the relation between surface area of both cubes:

We know that, surface area of cube = $6\times {\left(side\right)}^2$

Therefore,

$$\begin{gathered} \frac{Surfaceareaof{C}_1}{Surfaceareaof{C}_2}=\frac{6{\left(s_1\right)}^2}{6{\left(s_2\right)}^2} \\ =\frac{6{\left(s_1\right)}^2}{6{\left(3s_1\right)}^2}[U\sin g(i\left)\right] \\ =\frac{1}{9} \end{gathered}$$

Final answer: Therefore, the ratio of the surface area of two cubes having their volume in the ratio $1:27$ is $\mathbf{1}\mathbf{:}\mathbf{9}$.