Câu hỏi 30

Ta có: \(\left\{ \begin{array}{l}V = \dfrac{1}{3}\pi {R^2}h = 12\pi \\{S_{xq}} = \pi Rl = 15\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}h = \dfrac{{36}}{{{R^2}}}\\l = \dfrac{{15}}{R}\end{array} \right.\)

Do \({R^2} = {l^2} - {h^2}\) nên \({R^2} = {\left( {\dfrac{{15}}{R}} \right)^2} - {\left( {\dfrac{{36}}{{{R^2}}}} \right)^2} \Leftrightarrow {R^6} - 225{R^2} + 1296 = 0 \Leftrightarrow R = 3\).

Chọn: B