Four Points on a Cube

What is the surface area of the tetrahedron with one vertex at O the vertex of a unit cube and the other vertices at the centres of the faces of the cube not containing O?

Tetra Inequalities

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.

In a right-angled tetrahedron prove that the sum of the squares of the areas of the 3 faces in mutually perpendicular planes equals the square of the area of the sloping face. A generalisation of Pythagoras' Theorem.

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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

$$\begin{eqnarray} V &=& \frac{1}{3}\frac{1}{2} \frac{1}{\sqrt{2}} \\ &=&\frac{1}{6 \sqrt{2}} \\ &=& \frac{\sqrt{2}}{12} \end{eqnarray}$$

Cartesian equations of circles. Pythagoras' theorem. Pythagorean triples. Tetrahedra. Volume and capacity. Circles. Mathematical reasoning & proof. Cosine rule. chemistry. Similarity.

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Four Points on a Cube

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Reach for Polydron

Stage: 5 Challenge Level: