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Câu hỏi: Cho hình chóp ${S.ABCD}$ có đáy là hình vuông cạnh ${a}$, ${SA\bot (ABCD),}$ ${SB=a\sqrt{3}}$. Tính thể tích ${V}$ của khối chóp ${S.ABCD}$ theo ${a.}$
A. ${V=\dfrac{{{a}^{3}}\sqrt{3}}{3}}$.
B. ${V={{a}^{3}}\sqrt{2}}$.
C. ${V=\dfrac{{{a}^{3}}\sqrt{2}}{3}}$.
D. ${V=\dfrac{{{a}^{3}}\sqrt{2}}{6}.}$
Ta có: $SA=\sqrt{S{{B}^{2}}-A{{B}^{2}}}=\sqrt{{{\left( a\sqrt{3} \right)}^{2}}-{{a}^{2}}}=a\sqrt{2}$
${{S}_{ABCD}}={{a}^{2}}$
${{V}_{S.ABCD}}=\dfrac{1}{3}.SA.{{S}_{ABCD}}=\dfrac{1}{3}a\sqrt{2}.{{a}^{2}}=\dfrac{{{a}^{3}}\sqrt{2}}{3}$
A. ${V=\dfrac{{{a}^{3}}\sqrt{3}}{3}}$.
B. ${V={{a}^{3}}\sqrt{2}}$.
C. ${V=\dfrac{{{a}^{3}}\sqrt{2}}{3}}$.
D. ${V=\dfrac{{{a}^{3}}\sqrt{2}}{6}.}$
Ta có: $SA=\sqrt{S{{B}^{2}}-A{{B}^{2}}}=\sqrt{{{\left( a\sqrt{3} \right)}^{2}}-{{a}^{2}}}=a\sqrt{2}$
${{S}_{ABCD}}={{a}^{2}}$
${{V}_{S.ABCD}}=\dfrac{1}{3}.SA.{{S}_{ABCD}}=\dfrac{1}{3}a\sqrt{2}.{{a}^{2}}=\dfrac{{{a}^{3}}\sqrt{2}}{3}$
Đáp án C.