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Câu hỏi: Cho hình chóp $S.ABCD$ có đáy là hình thang vuông tại $A$ và $B$, $SA\bot \left( ABCD \right)$, $AD=3a$, $SA=AB=BC=a$. Gọi $S'$ là điểm thỏa mãn $\overrightarrow{SS'}=\dfrac{1}{2}\overrightarrow{AB}$. Tính thể tích khối đa diện $SS'ABCD$.
A. $\dfrac{13{{a}^{3}}}{10}$.
B. $\dfrac{11{{a}^{3}}}{12}$.
C. $\dfrac{11{{a}^{3}}}{10}$.
D. $\dfrac{13{{a}^{3}}}{12}$.
Gọi $E$ là điểm trên cạnh $AD$ sao cho $DE=2AE$.
Do $\overrightarrow{SS'}=\dfrac{1}{2}\overrightarrow{AB}\Rightarrow SS'=\dfrac{a}{2}$.
Ta có: $\left\{ \begin{aligned}
& BC\bot AB \\
& BC\bot SA \\
\end{aligned} \right.\Rightarrow BC\bot \left( SABS' \right)$.
${{V}_{SS'ABCD}}={{V}_{S.ABCD}}+{{V}_{C.BSS'}}+{{V}_{D.CSS'}}$
Trong đó:
+) ${{V}_{S.ABCD}}=\dfrac{1}{3}{{S}_{ABCD}}.SA=\dfrac{1}{3}.\dfrac{1}{2}.\left( BC+AD \right).AB.SA=\dfrac{1}{6}.\left( a+3a \right).a.a=\dfrac{2{{a}^{3}}}{3}$ (đvtt).
+) ${{V}_{C.BSS'}}=\dfrac{1}{3}.{{S}_{BSS'}}.CB=\dfrac{1}{3}.\dfrac{1}{2}.SS'.d\left( B,SS' \right).CB=\dfrac{1}{6}.SS'.SA.CB=\dfrac{1}{6}.\dfrac{a}{2}.a.a=\dfrac{{{a}^{3}}}{12}$ (đvtt).
+) Do $d\left( D,(CSS') \right)=2d\left( A,(CSS') \right)$ nên suy ra
${{V}_{D.CSS'}}=2{{V}_{A.CSS'}}=2{{V}_{C.ASS'}}=2.\dfrac{1}{3}.{{S}_{ASS'}}.CB=
\dfrac{2}{3}.\dfrac{1}{2}.SA.SS'.CB=\dfrac{1}{3}a.\dfrac{a}{2}.a=\dfrac{{{a}^{3}}}{6}$ (đvtt).
Vậy ${{V}_{SS'ABCD}}=\dfrac{2{{a}^{3}}}{3}+\dfrac{{{a}^{3}}}{12}+\dfrac{{{a}^{3}}}{6}=\dfrac{11{{a}^{3}}}{12}$ (đvtt).
A. $\dfrac{13{{a}^{3}}}{10}$.
B. $\dfrac{11{{a}^{3}}}{12}$.
C. $\dfrac{11{{a}^{3}}}{10}$.
D. $\dfrac{13{{a}^{3}}}{12}$.
Gọi $E$ là điểm trên cạnh $AD$ sao cho $DE=2AE$.
Do $\overrightarrow{SS'}=\dfrac{1}{2}\overrightarrow{AB}\Rightarrow SS'=\dfrac{a}{2}$.
Ta có: $\left\{ \begin{aligned}
& BC\bot AB \\
& BC\bot SA \\
\end{aligned} \right.\Rightarrow BC\bot \left( SABS' \right)$.
${{V}_{SS'ABCD}}={{V}_{S.ABCD}}+{{V}_{C.BSS'}}+{{V}_{D.CSS'}}$
Trong đó:
+) ${{V}_{S.ABCD}}=\dfrac{1}{3}{{S}_{ABCD}}.SA=\dfrac{1}{3}.\dfrac{1}{2}.\left( BC+AD \right).AB.SA=\dfrac{1}{6}.\left( a+3a \right).a.a=\dfrac{2{{a}^{3}}}{3}$ (đvtt).
+) ${{V}_{C.BSS'}}=\dfrac{1}{3}.{{S}_{BSS'}}.CB=\dfrac{1}{3}.\dfrac{1}{2}.SS'.d\left( B,SS' \right).CB=\dfrac{1}{6}.SS'.SA.CB=\dfrac{1}{6}.\dfrac{a}{2}.a.a=\dfrac{{{a}^{3}}}{12}$ (đvtt).
+) Do $d\left( D,(CSS') \right)=2d\left( A,(CSS') \right)$ nên suy ra
${{V}_{D.CSS'}}=2{{V}_{A.CSS'}}=2{{V}_{C.ASS'}}=2.\dfrac{1}{3}.{{S}_{ASS'}}.CB=
\dfrac{2}{3}.\dfrac{1}{2}.SA.SS'.CB=\dfrac{1}{3}a.\dfrac{a}{2}.a=\dfrac{{{a}^{3}}}{6}$ (đvtt).
Vậy ${{V}_{SS'ABCD}}=\dfrac{2{{a}^{3}}}{3}+\dfrac{{{a}^{3}}}{12}+\dfrac{{{a}^{3}}}{6}=\dfrac{11{{a}^{3}}}{12}$ (đvtt).
Đáp án B.