# Find the volume of the tetrahedron whose vertices are the given points: (0,0,0), (2,0,0), (0,2,0), (0,0,2).

Finding the volume of the tetrahedron with vertices (0,0,0), (2,0,0), (0,2,0), (0,0,2). I get 8; answer is 4/3.
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors A, B and C then the area of the object is $A\cdot \left(B×C\right)$. Let V be the volume we are trying to find.
$\begin{array}{rl}{x}^{2}& =6-{y}^{2}-{z}^{2}\\ A& =\left(2,0,0\right)-\left(0,0,0\right)=\left(2,0,0\right)\\ B& =\left(0,2,0\right)-\left(0,0,0\right)=\left(0,2,0\right)\\ C& =\left(0,0,2\right)-\left(0,0,0\right)=\left(0,0,2\right)\\ V& =|\begin{array}{ccc}{a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\\ {c}_{1}& {c}_{2}& {c}_{3}\end{array}|=|\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}|\\ & =2|\begin{array}{cc}2& 0\\ 0& 2\end{array}|=2\left(4-0\right)\\ & =8\end{array}$
However, the book gets $\frac{4}{3}$
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Kendrick Jacobs
Explanation:
Note that the given volume is a cone with the height 2 and a right isosceles triangle of side 2 as the base. Thus, its volume can be calculated as $\frac{1}{3}Are{a}_{base}\cdot Height=\frac{1}{3}\left(\frac{1}{2}\cdot 2\cdot 2\right)2=\frac{4}{3}$.
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Lorena Lester
Explanation:
Your tetrahedron is also a pyramid. with the volume of $\frac{1}{3}\cdot \frac{2\cdot 2}{2}\cdot 2=\frac{4}{3}$
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