Mensuration

Two sectors of the same disc with equal angles have equal areas: a rotation about the centre carries one onto the other. Cut the disc into \(360\) sectors of \(1^\circ\): they all have the same area, the area of the disc shared out equally, \(\dfrac{\pi r^2}{360}\). A sector whose angle is a whole number of degrees, \(\theta = n^\circ\), is made of \(n\) of them: $$\begin{aligned}[t]A &= n \times \frac{\pi r^2}{360} && (n \text{ sectors of } 1^\circ)\\ &= \frac{n^\circ}{360^\circ} \times \pi r^2 = \frac{\theta}{360^\circ} \times \pi r^2.\end{aligned}$$ Cutting the disc into smaller equal sectors — tenths, hundredths of a degree — gives the same formula for the angles measured in tenths, in hundredths of a degree. Any other angle lies between two such angles as close to each other as we wish, and its sector lies between their two sectors: its area is squeezed between two values of the formula that are as close as we wish, so it obeys the formula too. The area of a sector is proportional to its angle. Check: \(\theta = 90^\circ\) gives \(\dfrac{1}{4}\pi r^2\), the quarter disc, and \(\theta = 360^\circ\) the whole disc.

The octahedron is two square-based pyramids \(SABCD\) and \(TABCD\) whose edges all measure \(a\). In the square \(ABCD\), \(AC^2 = a^2 + a^2 = 2a^2\), so \(AC = a\sqrt{2}\) and the half-diagonal is \(OA = \frac{a\sqrt{2}}{2}\). The triangle \(SOA\) has a right angle at \(O\): $$\begin{aligned}[t] SO^2 &= SA^2 - OA^2 && \text{(Pythagoras' theorem)}\\ &= a^2 - \frac{2a^2}{4} = \frac{a^2}{2}, \end{aligned}$$ so \(SO = \frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}\). Then $$\begin{aligned}[t] V_{\text{octahedron}} &= 2 \times \frac{1}{3} \times a^2 \times \frac{a\sqrt{2}}{2} && \text{(volume of a pyramid, twice)}\\ &= \frac{\sqrt{2}}{3}\,a^3. \end{aligned}$$ The tetrahedron. Take a cube of edge \(s = \frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}\) and join four of its vertices, no two on a common edge, as \(ACFH\) in the cube \(ABCDEFGH\): the six segments are diagonals of faces, of length \(\sqrt{s^2 + s^2} = s\sqrt{2} = a\), so they form a regular tetrahedron of edge \(a\). Two regular tetrahedra with the same edge are congruent, so they have the same volume, and it is enough to compute this one. It is the cube minus four corner pyramids, each with a right-angled triangle as its base, whose two sides of the right angle measure \(s\), and with height \(s\): $$\begin{aligned}[t] V_{\text{tetrahedron}} &= s^3 - 4 \times \frac{1}{3} \times \frac{s^2}{2} \times s && \text{(the cube minus the four corners)}\\ &= s^3 - \frac{2}{3}s^3 = \frac{1}{3}s^3\\ &= \frac{1}{3} \times \frac{2\sqrt{2}\,a^3}{8} && \left(s^3 = \left(\tfrac{a\sqrt{2}}{2}\right)^3\right)\\ &= \frac{\sqrt{2}}{12}\,a^3. \end{aligned}$$ The quotient \(\dfrac{V_{\text{octahedron}}}{V_{\text{tetrahedron}}} = \dfrac{\sqrt{2}}{3} \times \dfrac{12}{\sqrt{2}} = 4\).
Check with the octahedron of edge \(6\) cm and the tetrahedron of edge \(6\sqrt{2}\) cm computed with pyramids: \(\frac{\sqrt{2}}{3} \times 6^3 = 72\sqrt{2}\) cm\(^3\) and \(\frac{\sqrt{2}}{12} \times \left(6\sqrt{2}\right)^3 = \frac{\sqrt{2}}{12} \times 432\sqrt{2} = 72\) cm\(^3\).