Pyramid vs Pyramid

Let x be the side length of the square.

The volume of pyramid A is simply one-third of the cube's volume: $\frac{x^3}{3}$

The volume of pyramid B = $\frac{1}{3}$ $\times B$ $\times H$

The base B area is an equilateral triangle of x $\sqrt{2}$ : $\frac{\sqrt{3}}{4}$ $[x(\sqrt{2})]^{2}$ = $\frac{\sqrt{3}}{2}$ $x^{2}$

For height H, we can set right triangle OPQ, where O is a point directly below the pyramid's apex, P is the apex, and Q is one of the base points, as shown.

Then by taking a tetrahedron side PQ as the hypotenuse side of a right triangle, we can find H (OP) by Pythagorean theorem; all we need is to find the length of OQ. When dividing the triangular base area with 3 lines from point O to the other base points, the area will be divided into 3 equal areas. Therefore, the height of the divided triangle is also one third of the base height, making OQ two-thirds of the base height.

We know that the tetrahedral side equals to x $\sqrt{2}$ , so the base height = $\frac{\sqrt{3}}{2}$ x $\sqrt{2}$

Then OQ = $\frac{2}{3}$ $\frac{\sqrt{3}}{2}$ x $\sqrt{2}$ = $\frac{x(\sqrt{6})}{3}$

Hence, $PQ^2$ = $OP^2$ + $OQ^2$ ; $H^2$ = $[x(\sqrt{2})]^{2}$ - $[(\frac{x(\sqrt{6})}{3})]^2$ = 2 $x^{2}$ - $\frac{2}{3}$ $x^{2}$ = $\frac{4}{3}$ $x^{2}$

Thus, H = $\frac{2x}{\sqrt{3}}$ .

As a result, the volume of pyramid B = $\frac{1}{3}$ $\frac{\sqrt{3}}{2}$ $x^{2}$ $\frac{2x}{\sqrt{3}}$ = $\frac{x^3}{3}$ .

Therefore, the volume of pyramid B = volume of pyramid A = $\frac{x^3}{3}$ .

$\dfrac { 1 }{ 3 } (1\cdot (1\cdot 1))=1-4\left( \dfrac { 1 }{ 3 } (1\cdot \frac { 1 }{ 2 } \left( 1\cdot 1 \right) ) \right)$