Reach for Polydron
A tetrahedron has two identical equilateral triangles faces, of side length 1 unit. The other two faces are right angled isosceles triangles. Find the exact volume of the tetrahedron.
Age
16 to 18
Challenge level
A tetrahedron has two identical equilateral triangles faces, of side length 1 unit. The other two faces are right angled isosceles triangles. Find the exact volume of the tetrahedron.
To discover the easy way to find this volume you can very quickly make for yourself a model of the tetrahedron. Take a square $ABCD$ of stiff paper and fold it along the diagonal $AC$ . Open it until the distance between $B$ and $D$ is equal to the length of the sides of the square and then the four vertices are the vertices of the tetrahedron with two equilateral faces and two isosceles faces. You can either stand this 'tetrahedron' on an equilateral face ( $ABD$ or $CBD$ ) as its base or on an isosceles face ( $ABC$ or $ADC$ ) as its base. One choice makes it easy to find the area of the base and the height and so to find the volume.
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
$$\begin{eqnarray} V &=& \frac{1}{3}\frac{1}{2} \frac{1}{\sqrt{2}} \\ &=&\frac{1}{6 \sqrt{2}} \\ &=& \frac{\sqrt{2}}{12} \end{eqnarray}$$