Google
Câu hỏi: Cho lăng trụ tam giác $ABC.{A}'{B}'{C}'$ có đáy là tam giác vuông tại $A$, $AB=2;AC=\sqrt{3}$. Góc $\widehat{CA{A}'}={{90}^{0}},\widehat{BA{A}'}={{120}^{0}}$. Gọi $M$ là trung điểm cạnh $B{B}'$. Biết $CM$ vuông góc với ${A}'B$, tính thể khối lăng trụ đã cho.
A. $V=\dfrac{1+\sqrt{33}}{8}$.
B. $V=\dfrac{1+\sqrt{33}}{4}$.
C. $V=\dfrac{3\left( 1+\sqrt{33} \right)}{8}$.
D. $V=\dfrac{3\left( 1+\sqrt{33} \right)}{4}$.
Ta có: $\left\{ \begin{aligned}
& CA\bot AB \\
& CA\bot AA' \\
\end{aligned} \right.\Rightarrow CA\bot \left( ABB'A' \right)$
Lại có: $\left\{ \begin{aligned}
& A'B\bot CM \\
& A'M\bot CA\left( V\grave{i}CA\bot \left( ABB'A' \right) \right) \\
\end{aligned} \right.\Rightarrow A'B\bot \left( ACM \right)$
$\Rightarrow A'B\bot AM$
* Đặt $AA'=2x\Rightarrow BM=x$
$\Rightarrow $ Xét $\Delta ABM:AM=\sqrt{{{x}^{2}}+{{2}^{2}}-2.x.2.\cos 60}=\sqrt{{{x}^{2}}-2x+4}$
$\Rightarrow AO=\dfrac{2}{3}.AM=\sqrt{{{x}^{2}}-2x+4}.\dfrac{2}{3}\Rightarrow BO=\sqrt{{{2}^{2}}-\dfrac{4}{9}\left( {{x}^{2}}-2x+4 \right)}$
$\left. \begin{aligned}
& {{S}_{\Delta ABA'}}=\dfrac{1}{2}.2.2x.\sin 120=x\sqrt{3} \\
& {{S}_{\Delta ABA'}}=\dfrac{1}{2}.AO.A'B=\dfrac{1}{2}.\left( \dfrac{2}{3}\sqrt{{{x}^{2}}-2x+4} \right).3\sqrt{{{2}^{2}}-\dfrac{4}{9}\left( {{x}^{2}}-2x+4 \right)} \\
\end{aligned} \right\}\Rightarrow x=\dfrac{\sqrt{33}+1}{4}$
$\Rightarrow {{V}_{LT}}=3{{V}_{CABA'}}=\dfrac{3\left( \sqrt{33}+1 \right)}{4}$
A. $V=\dfrac{1+\sqrt{33}}{8}$.
B. $V=\dfrac{1+\sqrt{33}}{4}$.
C. $V=\dfrac{3\left( 1+\sqrt{33} \right)}{8}$.
D. $V=\dfrac{3\left( 1+\sqrt{33} \right)}{4}$.
& CA\bot AB \\
& CA\bot AA' \\
\end{aligned} \right.\Rightarrow CA\bot \left( ABB'A' \right)$
Lại có: $\left\{ \begin{aligned}
& A'B\bot CM \\
& A'M\bot CA\left( V\grave{i}CA\bot \left( ABB'A' \right) \right) \\
\end{aligned} \right.\Rightarrow A'B\bot \left( ACM \right)$
$\Rightarrow A'B\bot AM$
* Đặt $AA'=2x\Rightarrow BM=x$
$\Rightarrow $ Xét $\Delta ABM:AM=\sqrt{{{x}^{2}}+{{2}^{2}}-2.x.2.\cos 60}=\sqrt{{{x}^{2}}-2x+4}$
$\Rightarrow AO=\dfrac{2}{3}.AM=\sqrt{{{x}^{2}}-2x+4}.\dfrac{2}{3}\Rightarrow BO=\sqrt{{{2}^{2}}-\dfrac{4}{9}\left( {{x}^{2}}-2x+4 \right)}$
$\left. \begin{aligned}
& {{S}_{\Delta ABA'}}=\dfrac{1}{2}.2.2x.\sin 120=x\sqrt{3} \\
& {{S}_{\Delta ABA'}}=\dfrac{1}{2}.AO.A'B=\dfrac{1}{2}.\left( \dfrac{2}{3}\sqrt{{{x}^{2}}-2x+4} \right).3\sqrt{{{2}^{2}}-\dfrac{4}{9}\left( {{x}^{2}}-2x+4 \right)} \\
\end{aligned} \right\}\Rightarrow x=\dfrac{\sqrt{33}+1}{4}$
$\Rightarrow {{V}_{LT}}=3{{V}_{CABA'}}=\dfrac{3\left( \sqrt{33}+1 \right)}{4}$
Đáp án D.