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Câu hỏi: Cho lăng trụ tam giác đều $ABC.{A}'{B}'{C}'$ có các cạnh đều bằng a. Gọi $M$ là điểm đối xứng của $A$ qua ${B}'C$. Thể tích khối tứ diện $AB{C}'M$ bằng
A. $\dfrac{3 \sqrt{3} a^{3}}{16}$.
B. $\dfrac{\sqrt{3} a^{3}}{8}$.
C. $\dfrac{\sqrt{3} a^{3}}{12}$.
D. $\dfrac{\sqrt{3}{{a}^{3}}}{16}$.
Gọi $H$, $I$, $N$ lần lượt là trung điểm của $AC$, ${B}'C$, $AM$.
Ta có $\vartriangle A{B}'C$ cân tại ${B}'$ nên ${B}'{{H}^{2}}={B}'{{A}^{2}}-A{{H}^{2}}={{\left( a\sqrt{2} \right)}^{2}}-{{\left( \dfrac{a}{2} \right)}^{2}}=\dfrac{7{{a}^{2}}}{4}\Rightarrow {B}'H=\dfrac{a\sqrt{7}}{2}$.
Mà $AN.{B}'C={B}'H.AC\Rightarrow AN=\dfrac{{B}'H.AC}{{B}'C}=\dfrac{\dfrac{a\sqrt{7}}{2}.a}{a\sqrt{2}}=\dfrac{a\sqrt{7}}{2\sqrt{2}}$.
$C{{N}^{2}}=A{{C}^{2}}-A{{N}^{2}}={{a}^{2}}-{{\left( \dfrac{a\sqrt{7}}{2\sqrt{2}} \right)}^{2}}=\dfrac{{{a}^{2}}}{8}\Rightarrow CN=\dfrac{a\sqrt{2}}{4}$.
$\dfrac{\text{d}\left( N,(A{C}'B) \right)}{\text{d}\left( C,(A{C}'B) \right)}=\dfrac{NI}{CI}=\dfrac{CI-CN}{CI}=\dfrac{\dfrac{a\sqrt{2}}{2}-\dfrac{a\sqrt{2}}{4}}{\dfrac{a\sqrt{2}}{2}}=\dfrac{1}{2}$.
$\Rightarrow \text{d}\left( N,(A{C}'B) \right)=\dfrac{1}{2}\text{d}\left( C,(A{C}'B) \right)$.
$\text{d}\left( M,(A{C}'B) \right)=2\text{d}\left( N,(A{C}'B) \right)\text{=d}\left( C,(A{C}'B) \right)$.
Thể tích khối chóp $AB{C}'M$ là
$V=\dfrac{1}{3}{{S}_{\vartriangle {C}'AB}}.\text{d}\left( M,({C}'AB) \right)=\dfrac{1}{3}.{C}'C.\dfrac{A{{B}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{3}}{12}$.
A. $\dfrac{3 \sqrt{3} a^{3}}{16}$.
B. $\dfrac{\sqrt{3} a^{3}}{8}$.
C. $\dfrac{\sqrt{3} a^{3}}{12}$.
D. $\dfrac{\sqrt{3}{{a}^{3}}}{16}$.
Ta có $\vartriangle A{B}'C$ cân tại ${B}'$ nên ${B}'{{H}^{2}}={B}'{{A}^{2}}-A{{H}^{2}}={{\left( a\sqrt{2} \right)}^{2}}-{{\left( \dfrac{a}{2} \right)}^{2}}=\dfrac{7{{a}^{2}}}{4}\Rightarrow {B}'H=\dfrac{a\sqrt{7}}{2}$.
Mà $AN.{B}'C={B}'H.AC\Rightarrow AN=\dfrac{{B}'H.AC}{{B}'C}=\dfrac{\dfrac{a\sqrt{7}}{2}.a}{a\sqrt{2}}=\dfrac{a\sqrt{7}}{2\sqrt{2}}$.
$C{{N}^{2}}=A{{C}^{2}}-A{{N}^{2}}={{a}^{2}}-{{\left( \dfrac{a\sqrt{7}}{2\sqrt{2}} \right)}^{2}}=\dfrac{{{a}^{2}}}{8}\Rightarrow CN=\dfrac{a\sqrt{2}}{4}$.
$\dfrac{\text{d}\left( N,(A{C}'B) \right)}{\text{d}\left( C,(A{C}'B) \right)}=\dfrac{NI}{CI}=\dfrac{CI-CN}{CI}=\dfrac{\dfrac{a\sqrt{2}}{2}-\dfrac{a\sqrt{2}}{4}}{\dfrac{a\sqrt{2}}{2}}=\dfrac{1}{2}$.
$\Rightarrow \text{d}\left( N,(A{C}'B) \right)=\dfrac{1}{2}\text{d}\left( C,(A{C}'B) \right)$.
$\text{d}\left( M,(A{C}'B) \right)=2\text{d}\left( N,(A{C}'B) \right)\text{=d}\left( C,(A{C}'B) \right)$.
Thể tích khối chóp $AB{C}'M$ là
$V=\dfrac{1}{3}{{S}_{\vartriangle {C}'AB}}.\text{d}\left( M,({C}'AB) \right)=\dfrac{1}{3}.{C}'C.\dfrac{A{{B}^{2}}\sqrt{3}}{4}=\dfrac{{{a}^{3}}\sqrt{3}}{12}$.
Đáp án C.