A regular tetrahedron has four faces, each of which is an equilateral triangle - Leaving Cert Mathematics - Question 9 - 2014

Applying the sine rule in triangle AOB:

$\frac{|OA|}{\sin 30^{\circ}} = \frac{|AB|}{\sin 120^{\circ}}$

This implies:

$|OA| = 2a \cdot \frac{\sin 30^{\circ}}{\sin 120^{\circ}} = 2a \cdot \frac{1/2}{\frac{\sqrt{3}}{2}} = \frac{2a}{\sqrt{3}}.$

Thus, the radius r of the cylinder is ( r = \frac{2a}{\sqrt{3}} ).

By drawing a vertical line from point D (the top vertex of the tetrahedron) to point O, we create a right triangle with height h. Using Pythagoras' theorem:

$h^2 + \left(\frac{2a}{\sqrt{3}}\right)^2 = (2a)^2,$

this simplifies to:

$h^2 = 4a^2 - \frac{4a^2}{3} = \frac{8a^2}{3}.$

Thus, we find ( h = \sqrt{\frac{8}{3}} imes a ).

The volume V of a cylinder is given by:

$V = \pi r^2 h.$

Substituting in for r and h, we have:

$V = \pi \left(\frac{2a}{\sqrt{3}}\right)^2 \left(\sqrt{\frac{8}{3}} a\right) = \pi \cdot \frac{4a^2}{3} \cdot \left(\frac{2a \sqrt{8}}{3}\right) = \frac{8a^3 \sqrt{6}}{9} \pi,$

which matches the required volume.