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Let the length of a side of a triangle be \(2L\). Using \(2L\) instead of \(L\) gets rid of fractions in the later calculations.
By Pythagoras' Theorem, the height of a triangle is \(L\sqrt{3}\).
Using Pythagoras' Theorem again, the height of the square-based pyramid is \(L\sqrt{2}\).
Therefore, the volume of the square-based pyramid is:
$$\frac{1}{3}\times (2L)^2\times L\sqrt{2}$$
$$=\frac{4}{3}\sqrt{2}L^3$$
Next, we find the area of the tetrahedron.
Call the point on the base of the tetrahedron, directly below the vertex at the top, \(A\).
Using cosine in the triangle made by \(A\), the corner of the base and the midpoint of a side of the base, the distance from the corner to \(A\) is \frac{2}{\sqrt{3}}L\).
Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is \(\frac{2\sqrt{2}}{\sqrt{3}}L\).
Therefore, the volume of the tetrahedron is:
$$\frac{1}{3}\times\frac{1}{2}\times 2L\times L\sqrt{3}\times \frac{2\sqrt{2}}{\sqrt{3}}L$$
$$= \frac{2}{3}\sqrt{2}L^3$$
Finally, the ratio of volume of the square based pyramid to the tetrahedron is:
$$\frac{4}{3}\sqrt{2}L^3:\frac{2}{3}\sqrt{2}L^3$$
$$2:1$$
Extension
What would the ratio be if they were isosceles triangles?
