Volume of tetrahedron Solid

Question

To obtain the geometric solid shown below, a $ L $ edge cube was started and a triangular base pyramid with the lateral edges measuring $ \frac {L} {4} $ was removed from each of the vertices of that cube. , as shown in the figure. Calling $ V $ the volume of the cube from which the solid was obtained, it can be concluded that the volume of the solid is: Answer: $\frac{47}{48}$

Attemp: I had made the volume of the cube $ L ^ 3 $ minus the volume of the tetrahedron, knowing that the volume of the tetrahedron is $ \frac {a ^ 3 \sqrt {2}} {12} $. How do I continue? I had seen that there is a way to solve it using the tetrahedron volume formula, but I don't know if it is true, and if so, why?

Answer 1

Please note that you have $8$ tri-rectangular tetrahedrons (wiki) taken out and each of their volume is given by

$\displaystyle V = \frac{1}{6} \ abc \ , $ where $a, b, c$ are edges to the base, making right angles with each other.

In this case, $a = b = c = \frac{L}{4}$.

So volume of resulting solid is $ = \displaystyle \small L^3 - 8 \cdot \frac{1}{6} \cdot \big(\frac{L}{4}\big)^3 = \frac{47 L^3}{48}$

Answer 2

The volume of a single tetrahedron is equal to the area of one right triangular base $B = \frac{1}{2}\cdot \frac{L}{4} \cdot \frac{L}{4}$, times the perpendicular height from that base, $h = \frac{L}{4}$, times $\frac{1}{3}$: $$V_{\text{tetrahedron}} = \frac{1}{3}Bh = \frac{1}{6} \cdot \frac{L^3}{4^3} = \frac{L^3}{384}.$$ Since there are eight such tetrahedra, the total volume that is cut off is $L^3/48$, hence the remaining volume is $$V = L^3 - \frac{L^3}{48} = \frac{47}{48} L^3.$$