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Câu hỏi: Cho hình lăng trụ tam giác đều $ABC.{A}'{B}'{C}'$ có góc giữa hai mặt phẳng $\left( {A}'BC \right)$ và $\left( ABC \right)$ bằng $60{}^\circ ,$ cạnh $AB=a.$ Thể tích của khối lăng trụ $ABC.{A}'{B}'{C}'$ bằng
A. $\dfrac{{{a}^{3}}\sqrt{3}}{8}.$
B. $\dfrac{{{a}^{3}}\sqrt{3}}{4}.$
C. $\dfrac{3{{a}^{3}}\sqrt{3}}{4}.$
D. $\dfrac{3{{a}^{3}}\sqrt{3}}{8}.$
Kẻ $AH\bot BC\Rightarrow \widehat{\left( \left( {A}'BC \right);\left( ABC \right) \right)}=\widehat{{A}'HA}={{60}^{0}}$
$\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}={A}'A.{{S}_{ABC}}=\dfrac{3a}{2}.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{3{{a}^{3}}\sqrt{3}}{8}.$
$\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}={A}'A.{{S}_{ABC}}=\dfrac{3a}{2}.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{3{{a}^{3}}\sqrt{3}}{8}.$
A. $\dfrac{{{a}^{3}}\sqrt{3}}{8}.$
B. $\dfrac{{{a}^{3}}\sqrt{3}}{4}.$
C. $\dfrac{3{{a}^{3}}\sqrt{3}}{4}.$
D. $\dfrac{3{{a}^{3}}\sqrt{3}}{8}.$
Kẻ $AH\bot BC\Rightarrow \widehat{\left( \left( {A}'BC \right);\left( ABC \right) \right)}=\widehat{{A}'HA}={{60}^{0}}$
$\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}={A}'A.{{S}_{ABC}}=\dfrac{3a}{2}.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{3{{a}^{3}}\sqrt{3}}{8}.$
$\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}={A}'A.{{S}_{ABC}}=\dfrac{3a}{2}.\dfrac{{{a}^{2}}\sqrt{3}}{4}=\dfrac{3{{a}^{3}}\sqrt{3}}{8}.$
Đáp án D.