Question

The height of a right circular cylinder is 10.5 cm. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathit{}r\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circular}\mathrm{cylinder}. \\ \mathrm{Height},\mathit{}h=10.5\mathrm{cm} \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{curved}\mathrm{surface},S_1=2\mathrm{\pi }rh \\ S\mathrm{um}\mathrm{of}\mathrm{the}\mathrm{areas}\mathrm{of}\mathrm{its}\mathrm{two}\mathrm{circular}\mathrm{faces},S_2=2\mathrm{\pi }{r}^2 \\ \mathrm{According}\mathrm{to}\mathrm{question}: \\ 3S_2=2S_1 \\ 3\times 2{\mathrm{\pi r}}^2=2\times 2\mathrm{\pi }rh \\ 6r=4h \\ 3r=2h \\ \mathit{}\mathit{}r=\frac{2}{3}\times 10.5\mathrm{cm} \\ =7\mathrm{cm} \end{gathered}$$