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Câu hỏi: Tính thể tích khối chóp tứ giác S.ABCD biết $AB=a,SA=a.$
A. $\dfrac{{{a}^{3}}\sqrt{2}}{2}$
B. $\dfrac{{{a}^{3}}\sqrt{2}}{6}$
C. $\dfrac{{{a}^{3}}}{3}$
D. ${{a}^{3}}$
A. $\dfrac{{{a}^{3}}\sqrt{2}}{2}$
B. $\dfrac{{{a}^{3}}\sqrt{2}}{6}$
C. $\dfrac{{{a}^{3}}}{3}$
D. ${{a}^{3}}$
Ta có $SO=\sqrt{S{{A}^{2}}-O{{A}^{2}}}=\sqrt{{{a}^{2}}-\dfrac{{{a}^{2}}}{2}}=\dfrac{a\sqrt{2}}{2}$
Ta có ${{V}_{S.ABC\text{D}}}=\dfrac{1}{3}SO.{{S}_{ABC\text{D}}}$
$=\dfrac{1}{3}.\dfrac{a\sqrt{2}}{2}{{a}^{2}}=\dfrac{{{a}^{3}}\sqrt{2}}{6}$ (đvtt)
Ta có ${{V}_{S.ABC\text{D}}}=\dfrac{1}{3}SO.{{S}_{ABC\text{D}}}$
$=\dfrac{1}{3}.\dfrac{a\sqrt{2}}{2}{{a}^{2}}=\dfrac{{{a}^{3}}\sqrt{2}}{6}$ (đvtt)
Đáp án B.