Cho hình chóp S.ABCD có đường thẳng SA vuông góc v...

D

- Hướng dẫn giải

Ta có:

\(\begin{array}{l}
{S_{ABCD}} = \frac{{\left( {AD + BC} \right).AB}}{2} = \frac{{\left( {2a + a} \right).a}}{2} = \frac{{3{a^2}}}{2};\\
{S_{ABD}} = \frac{1}{2}AB.AD = \frac{1}{2}.a.2a = a{}^2\\
 \Rightarrow {S_{BCD}} = {S_{ABCD}} - {S_{ABD}} = \frac{3}{2}{a^2} - {a^2} = \frac{{{a^2}}}{2}\\
 \Rightarrow {V_{S.ABCD}} = \frac{1}{3}SA.{S_{ABCD}} = \frac{1}{3}.a\sqrt 2 .\frac{{{a^2}}}{2} = \frac{{{a^3}\sqrt 2 }}{6}.
\end{array}\)