Thể tích khối tam diện vuông O.ABC vuông tại O có $OA = a,{\rm{ }}OB = OC = 2a$ là
A.\(\frac{{2{a^3}}}{3} \cdot \)
B.\(\frac{{{a^3}}}{2} \cdot \)
C. \(\frac{{{a^3}}}{6} \cdot \)
D. \(2{a^3}\).
\(\begin{array}{l}\left\{ \begin{array}{l}{S_{OBC}} = \frac{1}{2}OB.OC = 2{a^2}\\h = OA = a\end{array} \right.{\rm{ }}\\ \Rightarrow {V_{O.ABC}} = \frac{1}{3}OA \cdot {S_{OBC}} = \frac{{2{a^3}}}{3}\end{array}\)
A.\(\frac{{2{a^3}}}{3} \cdot \)
B.\(\frac{{{a^3}}}{2} \cdot \)
C. \(\frac{{{a^3}}}{6} \cdot \)
D. \(2{a^3}\).
\(\begin{array}{l}\left\{ \begin{array}{l}{S_{OBC}} = \frac{1}{2}OB.OC = 2{a^2}\\h = OA = a\end{array} \right.{\rm{ }}\\ \Rightarrow {V_{O.ABC}} = \frac{1}{3}OA \cdot {S_{OBC}} = \frac{{2{a^3}}}{3}\end{array}\)