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Câu hỏi: Cho hình hộp chữ nhật $ABCD.{A}'{B}'{C}'{D}'$ có $AB=2a$, $AD=3a$, $A{A}'=3a$. $E$ thuộc cạnh ${B}'{C}'$ sao cho ${B}'E=3{C}'E$. Thể tích khối chóp $E.BCD$ bằng:
A. $2{{a}^{3}}$.
B. ${{a}^{3}}$.
C. $3{{a}^{3}}$.
D. $\dfrac{{{a}^{3}}}{2}$.
${{V}_{ABCD.A'B'C'D'}}=2a.3a.3a=18{{a}^{3}}.$
${{V}_{E.BCD}}=\dfrac{1}{3}d\left( E;\left( BCD \right) \right).{{S}_{BCD}}.$
Vì $B'C'//\left( ABCD \right)$ nên $d\left( E;\left( BCD \right) \right)=d\left( B';\left( BCD \right) \right)=d\left( B';\left( ABCD \right) \right).$
${{S}_{BCD}}=\dfrac{1}{2}{{S}_{ABCD}}.$
Do đó: ${{V}_{E.BCD}}=\dfrac{1}{3}d\left( B';\left( ABCD \right) \right).\dfrac{1}{2}.{{S}_{ABCD}}=\dfrac{1}{2}{{V}_{B'.ABCD}}=\dfrac{1}{2}.\dfrac{1}{3}{{V}_{ABCD.A'B'C'D'}}$
$\Rightarrow {{V}_{E.BCD}}=\dfrac{1}{6}.18{{a}^{3}}=3{{a}^{3}}.$
A. $2{{a}^{3}}$.
B. ${{a}^{3}}$.
C. $3{{a}^{3}}$.
D. $\dfrac{{{a}^{3}}}{2}$.
${{V}_{ABCD.A'B'C'D'}}=2a.3a.3a=18{{a}^{3}}.$
${{V}_{E.BCD}}=\dfrac{1}{3}d\left( E;\left( BCD \right) \right).{{S}_{BCD}}.$
Vì $B'C'//\left( ABCD \right)$ nên $d\left( E;\left( BCD \right) \right)=d\left( B';\left( BCD \right) \right)=d\left( B';\left( ABCD \right) \right).$
${{S}_{BCD}}=\dfrac{1}{2}{{S}_{ABCD}}.$
Do đó: ${{V}_{E.BCD}}=\dfrac{1}{3}d\left( B';\left( ABCD \right) \right).\dfrac{1}{2}.{{S}_{ABCD}}=\dfrac{1}{2}{{V}_{B'.ABCD}}=\dfrac{1}{2}.\dfrac{1}{3}{{V}_{ABCD.A'B'C'D'}}$
$\Rightarrow {{V}_{E.BCD}}=\dfrac{1}{6}.18{{a}^{3}}=3{{a}^{3}}.$
Đáp án C.