Cho khối lăng trụ tam giác đều $ABC.{A}'{B}'{C}'$ có cạnh đáy bằng...

Gọi \)">M{B}'{C}'\widehat{\left( \left( A{B}'{C}' \right),\left( {A}'{B}'{C}' \right) \right)}=\widehat{\left( {A}'M,AM \right)}=\widehat{{A}'MA}=60{}^\circ {A}'{B}'{C}'{A}'M=\dfrac{a\sqrt{3}}{2};{{S}_{\Delta {A}'{B}'{C}'}}=\dfrac{{{a}^{2}}\sqrt{3}}{4}AM{A}'{A}'A{A}'={A}'M.\tan 60{}^\circ =\dfrac{3a}{2}V={{S}_{\Delta {A}'{B}'{C}'}}.A{A}'=\dfrac{{{a}^{2}}\sqrt{3}}{4}.\dfrac{3a}{2}=\dfrac{3\sqrt{3}{{a}^{3}}}{8}$.