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This excellent solution came from Ruth from Manchester High School for Girls.
In the tetrahedron $ABCD$, let $ABC$ and $ACD$ be right angled. If you position the tetrahedron so that $ABC$ is the base, then the vertex $D$ is directly above the edge $AC$. This means that the height of the tetrahedron is the height of the triangle $ACD$ which is $\frac{1}{\sqrt{2}}$ and the area of the base of the tetrahedron is the area of the triangle $ABC$ which is $\frac{1}{2}$. The volume of a pyramid is one third base times height. Therefore
$$\begin{eqnarray} V &=& \frac{1}{3}\frac{1}{2} \frac{1}{\sqrt{2}} \\ &=&\frac{1}{6 \sqrt{2}} \\ &=& \frac{\sqrt{2}}{12} \end{eqnarray}$$
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This excellent solution came from Ruth from Manchester High School for Girls.
In the tetrahedron $ABCD$, let $ABC$ and $ACD$ be right angled. If you position the tetrahedron so that $ABC$ is the base, then the vertex $D$ is directly above the edge $AC$. This means that the height of the tetrahedron is the height of the triangle $ACD$ which is $\frac{1}{\sqrt{2}}$ and the area of the base of the tetrahedron is the area of the triangle $ABC$ which is $\frac{1}{2}$. The volume of a pyramid is one third base times height. Therefore
$$\begin{eqnarray} V &=& \frac{1}{3}\frac{1}{2} \frac{1}{\sqrt{2}} \\ &=&\frac{1}{6 \sqrt{2}} \\ &=& \frac{\sqrt{2}}{12} \end{eqnarray}$$
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