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Câu hỏi: Cho lăng trụ đứng $ABC.{A}'{B}'{C}'$ có $AB=a$, $AC=2a$, $\widehat{BAC}=120{}^\circ $. Gọi O, I lần lượt là tâm của các mặt bên $\left( BC{C}'{B}' \right)$, $\left( AB{B}'{A}' \right)$ và M là trung điểm $C{C}'$. Biết rằng hai mặt phẳng $\left( AC{B}' \right)$, $\left( AB{C}' \right)$ tạo với nhau góc thỏa mãn $\cos \alpha =\dfrac{\sqrt{10}}{5}$. Thể tích khối đa diện $ABC.OIM$ bằng
A. $\dfrac{{{a}^{3}}}{2}$.
B. $\dfrac{7{{a}^{3}}}{16}$.
C. $\dfrac{5{{a}^{3}}}{8}$.
D. $\dfrac{9{{a}^{3}}}{16}$.
Ta có ${{V}_{ABC.OIM}}={{V}_{O.ABC}}+{{V}_{A.IOM}}+{{V}_{A.BIO}}+{{V}_{A.COM}}$
${{V}_{O.ABC}}=\dfrac{1}{6}{{V}_{ABC.{A}'{B}'{C}'}}$
${{V}_{A.IOM}}=\dfrac{1}{4}.\dfrac{1}{2}.\dfrac{1}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{24}{{V}_{ABC.{A}'{B}'{C}'}}\left( \text{do }{{S}_{\Delta OIM}}=\dfrac{1}{4}{{S}_{\Delta ABC}} \right)$
${{V}_{A.BIO}}=\dfrac{1}{2}{{V}_{A.B{B}'O}}=\dfrac{1}{2}.\dfrac{1}{4}{{V}_{A.BC{C}'{B}'}}=\dfrac{1}{2}.\dfrac{1}{4}.\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{12}{{V}_{ABC.{A}'{B}'{C}'}}$
${{V}_{A.COM}}=\dfrac{1}{8}{{V}_{A.BC{C}'{B}'}}=\dfrac{1}{8}.\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{12}{{V}_{ABC.{A}'{B}'{C}'}}$
Vậy ${{V}_{ABC.OIM}}=\left( \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{12} \right){{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{3}{8}{{V}_{ABC.{A}'{B}'{C}'}}$.
Ta có $\widehat{\left( \left( AC{B}' \right),\left( AB{C}' \right) \right)}=\alpha ,\dfrac{\sqrt{15}}{5}=\sin \alpha =\dfrac{d\left( B,\left( AC{B}' \right) \right)}{d\left( B,AO \right)}$.
Dựng $BH\bot AC,BK\bot {B}'H$ ta có $BH=\dfrac{a\sqrt{3}}{2}$. Đặt $\Rightarrow d\left( B,\left( AC{B}' \right) \right)=BK=\dfrac{x.\dfrac{a\sqrt{3}}{2}}{\sqrt{{{x}^{2}}+\dfrac{3{{a}^{2}}}{4}}}$.
Xét $\Delta ABC$ có $AB=a,AC=2a,\measuredangle BAC=120{}^\circ \Rightarrow BC=a\sqrt{7}\Rightarrow \left\{ \begin{aligned}
& BN=\dfrac{a\sqrt{7}}{2} \\
& AN=\dfrac{a\sqrt{3}}{2} \\
\end{aligned} \right.$
$\Rightarrow A{{B}^{2}}+A{{N}^{2}}=B{{N}^{2}}\Rightarrow \Delta ABN$ vuông tại A, suy ra $AB\bot \left( OAN \right)\Rightarrow d\left( B,AO \right)=BA=a$.
Suy ra $\dfrac{x.\dfrac{\sqrt{3}}{2}}{\sqrt{{{x}^{2}}+\dfrac{3{{a}^{2}}}{4}}}=\dfrac{\sqrt{15}}{5}\Leftrightarrow x=a\sqrt{3}$. Vậy ${{V}_{ABC.OIM}}=\dfrac{3}{8}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{3}{8}.B{B}'.{{S}_{\Delta ABC}}=\dfrac{3}{8}.a\sqrt{3}.\dfrac{{{a}^{2}}\sqrt{3}}{2}=\dfrac{9{{a}^{3}}}{16}$.
A. $\dfrac{{{a}^{3}}}{2}$.
B. $\dfrac{7{{a}^{3}}}{16}$.
C. $\dfrac{5{{a}^{3}}}{8}$.
D. $\dfrac{9{{a}^{3}}}{16}$.
Ta có ${{V}_{ABC.OIM}}={{V}_{O.ABC}}+{{V}_{A.IOM}}+{{V}_{A.BIO}}+{{V}_{A.COM}}$
${{V}_{O.ABC}}=\dfrac{1}{6}{{V}_{ABC.{A}'{B}'{C}'}}$
${{V}_{A.IOM}}=\dfrac{1}{4}.\dfrac{1}{2}.\dfrac{1}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{24}{{V}_{ABC.{A}'{B}'{C}'}}\left( \text{do }{{S}_{\Delta OIM}}=\dfrac{1}{4}{{S}_{\Delta ABC}} \right)$
${{V}_{A.BIO}}=\dfrac{1}{2}{{V}_{A.B{B}'O}}=\dfrac{1}{2}.\dfrac{1}{4}{{V}_{A.BC{C}'{B}'}}=\dfrac{1}{2}.\dfrac{1}{4}.\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{12}{{V}_{ABC.{A}'{B}'{C}'}}$
${{V}_{A.COM}}=\dfrac{1}{8}{{V}_{A.BC{C}'{B}'}}=\dfrac{1}{8}.\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{1}{12}{{V}_{ABC.{A}'{B}'{C}'}}$
Vậy ${{V}_{ABC.OIM}}=\left( \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{12} \right){{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{3}{8}{{V}_{ABC.{A}'{B}'{C}'}}$.
Ta có $\widehat{\left( \left( AC{B}' \right),\left( AB{C}' \right) \right)}=\alpha ,\dfrac{\sqrt{15}}{5}=\sin \alpha =\dfrac{d\left( B,\left( AC{B}' \right) \right)}{d\left( B,AO \right)}$.
Dựng $BH\bot AC,BK\bot {B}'H$ ta có $BH=\dfrac{a\sqrt{3}}{2}$. Đặt $\Rightarrow d\left( B,\left( AC{B}' \right) \right)=BK=\dfrac{x.\dfrac{a\sqrt{3}}{2}}{\sqrt{{{x}^{2}}+\dfrac{3{{a}^{2}}}{4}}}$.
Xét $\Delta ABC$ có $AB=a,AC=2a,\measuredangle BAC=120{}^\circ \Rightarrow BC=a\sqrt{7}\Rightarrow \left\{ \begin{aligned}
& BN=\dfrac{a\sqrt{7}}{2} \\
& AN=\dfrac{a\sqrt{3}}{2} \\
\end{aligned} \right.$
$\Rightarrow A{{B}^{2}}+A{{N}^{2}}=B{{N}^{2}}\Rightarrow \Delta ABN$ vuông tại A, suy ra $AB\bot \left( OAN \right)\Rightarrow d\left( B,AO \right)=BA=a$.
Suy ra $\dfrac{x.\dfrac{\sqrt{3}}{2}}{\sqrt{{{x}^{2}}+\dfrac{3{{a}^{2}}}{4}}}=\dfrac{\sqrt{15}}{5}\Leftrightarrow x=a\sqrt{3}$. Vậy ${{V}_{ABC.OIM}}=\dfrac{3}{8}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{3}{8}.B{B}'.{{S}_{\Delta ABC}}=\dfrac{3}{8}.a\sqrt{3}.\dfrac{{{a}^{2}}\sqrt{3}}{2}=\dfrac{9{{a}^{3}}}{16}$.
Đáp án D.