Question

The diameter of the moon is one fourth of the diameter of the earth.

Find the ratio of their surface areas. Also find the percentage increase intheir surface areas.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{earth}=\mathrm{d} \\ \mathrm{So},\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{moon}=\frac{\mathrm{d}}{4} \\ \mathrm{Thus},\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{earth}\left(\mathrm{r}\right)=\frac{\mathrm{d}}{2} \\ \mathrm{And},\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{moon}\left(\mathrm{R}\right)=\frac{\mathrm{d}}{8} \\ \mathrm{Now}, \\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{earth}=4{\mathrm{\pi r}}^2=4\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2={\mathrm{\pi d}}^2 \\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{moon}=4{\mathrm{\pi R}}^2=4\mathrm{\pi }{\left(\frac{\mathrm{d}}{8}\right)}^2=\frac{{\mathrm{\pi d}}^2}{16} \\ \mathrm{Thus}, \\ \mathrm{Required}\mathrm{ratio}=\frac{\mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{moon}}{\mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{earth}}=\frac{\frac{{\mathrm{\pi d}}^2}{16}}{{\mathrm{\pi d}}^2}=\frac{1}{16} \\ \mathbf{Hence}\mathbf{,}\mathbf{}\mathbf{ratio}\mathbf{}\mathbf{of}\mathbf{}\mathbf{their}\mathbf{}\mathbf{surface}\mathbf{}\mathbf{area}\mathbf{}\mathbf{is}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{16} \end{gathered}$$