Let the radius of both solids be r and their heights be h units,
According to condition,
\(\frac{r}{h}=\frac{4}{3}\\\therefore h=\frac{3r}{4}\)
TSA of cylinder = \(2\pi r(r+h)\)
TSA of cone = \(\pi r (r+l)\)
On comparing the ratios,
\(=\frac {2\pi r (r+h)}{\pi r(r+l)}\\=\frac{2(r+h)}{r+\sqrt{h²+r²}} (Since \ l² =\ \sqrt{r²+h²}) \\\)
\(= \frac{2(r + \frac{3}{4}r)}{r + \sqrt{\frac{9}{16}r^2 + r^2}}\)
\(= \frac{2\left(\frac{7r}{4}\right)}{r + \sqrt{\frac{25}{16}r^2}}\)
\(= \cfrac{\frac{7r}{2}}{r + \frac{5}{4}r}\)
\(= \cfrac{\frac{7r}{2}}{\frac{9r}{4}}\)
\(=\frac{7}{2} \times \frac{4}{9}\)
\(=\frac{14}{9}.\)
Therefore the ratio of total surface area of cylinder to that of cone is 14:9
