$$\eqalign{
& Goi\,M\,la\,\,trung\,\,diem\,\,BC \cr
& H\,\,la\,\,trong\,\,tam\,\,\Delta ABC \cr
& \Rightarrow SH \bot \left( {ABC} \right) \Rightarrow SH = h \cr
& AM = {{a\sqrt 3 } \over 2} \Rightarrow HM = {1 \over 3}AM = {{a\sqrt 3 } \over 6} \cr
& \Rightarrow SM = \sqrt {S{H^2} + H{M^2}} = \sqrt {{h^2} + {{{a^2}} \over {12}}} = {{\sqrt {12{h^2} + {a^2}} } \over {2\sqrt 3 }} \cr
& \Rightarrow {S_{\Delta SBC}} = {1 \over 2}SM.BC = {1 \over 2}{{\sqrt {12{h^2} + {a^2}} } \over {2\sqrt 3 }}.a = {{a\sqrt {12{h^2} + {a^2}} } \over {4\sqrt 3 }} \cr
& \Rightarrow {S_{mat\,\,ben}} = 3{S_{\Delta SBC}} = {{\sqrt 3 a\sqrt {12{h^2} + {a^2}} } \over 4} \cr
& {S_{\Delta ABC}} = {{{a^2}\sqrt 3 } \over 4} \cr
& \Rightarrow {S_{tp}} = {S_{mat\,\,ben}} + {S_{\Delta ABC}} \cr
& = {{\sqrt 3 a} \over 4}\left( {\sqrt {12{h^2} + {a^2}} + a} \right) \cr
& V = {1 \over 3}h.{{{a^2}\sqrt 3 } \over 4} = {{\sqrt 3 } \over 4}.{1 \over 3}{a^2}h \cr
& \Rightarrow r = {{3{V_{chop}}} \over {{S_{tp}}}} = {{{{\sqrt 3 } \over 4}{a^2}h} \over {{{\sqrt 3 } \over 4}a\left( {\sqrt {12{h^2} + {a^2}} + a} \right)}} = {{ah} \over {\sqrt {12{h^2} + {a^2}} + a}} \cr
& Chon\,\,A. \cr} $$
