Fraction of a Cubical volume

Question

There are a lot of cube cutting problems in different texts. This one is an original thought, and actually I have been able to solve it too. But the answer I am getting, is a bit hard to accept from actual shape of solid. Want to validate my answer here.

In the figure below, a plane cuts along one corner of a cube such it reaches at half of each edge. What is the volume of this part, as a fraction of the original cube?

Considering it as a pyramid and applying the formula, $V=1/3~×$ Base Area × Vertical Height is giving me 1/48. That is a bit difficult to digest, since if I consider the smaller cube of side length half of the original cube formed by completing this tetrahedron/ pyramid from points where it touches the cube, then its volume is only one-sixth of that cube. From the physical appearance, it seems to be larger than that fraction.

Here is my working of getting 1/48:

Assume the side length of cube = $a$. Each side of the triangular base is the hypotenuse of a right isosceles triangle that lies along the face of original cube, each of whose legs are $a/2$ each. Hence, the base is an equilateral triangle of side $a/√2$

Now we come to the isosceles triangle along lateral faces. The longer, equal sides are $a/2$ each. I am referring here to the length from apex to any vertex of the base triangle. Consider the triangle (not explicitly drawn in this figure) joining the vertical height to this length, through the circumradius of the base triangle. Since we found the side above as $a/√2$, hence circumradius is $a/√2~×1/√3 = a/√6.$ Thus, we finally get vertical height as $a/2√3$ by applying Pythagoras theorem in this triangle.

Finally that is giving me Volume = $1/3× √3/4 *a^2/2 * a/2√3 = a^3/48$. So, is it really this value? If yes, can somebody make me fathom how it is such a small tiny fraction? If no, please find my fault in the calculation.

Answer 1

This is correct.

First part : A slightly simpler way of finding the answer

Note : In this answer, I am using this drawing as a representation of our "cut corner", which is named a trirectangular tetrahedron :

(Trirectangular tetrahedron image courtesy of this answer)

Another way of seeing this cut corner is as the volume of a trirectangular tetrahedron where all the edges along the right angles are the same length $AB = AC = AD = \ell$.

Coming from this, we can see that, if we take $ABD$ as the base of our tetrahedron, the altitude is $AC$, which is quite good.

Thus, we have :

$$ \mathcal{V}_{ABCD} = \frac{1}{3} \mathcal A_{ABD} \times AC $$

$ABD$ is a right triangle, so we can say that $\mathcal A_{ABD} = \frac{AB \times AD}{2}= \frac{\ell^2}{2}$

Thus we have :

$$ \mathcal{V}_{ABCD} = \frac{1}{3} \mathcal A_{ABD} \times AC = \frac{1}{3} \frac{\ell^2}{2} \times \ell = \frac{\ell^3}{6} $$

This also shows that the ratio between a trirectangular tetrahedron and its holding parallelepiped is $1/6$.

Finally, in your case, we can plug in $\ell = \frac{a}{2}$, which gives us :

$$ \mathcal{V} = \frac{\ell^3}{6} = \frac{\left(a/2\right)^3}{6} = \frac{a^3}{48} $$

Second part : Why $\frac{1}{48}$?

Another way to find this result is to see your cut corner as the cut along the corner of a cube of side length $\frac{a}{2}$ , such that it reaches the other vertex of the edge, like this :

(Image courtesy of Wikimedia Commons, released under CC BY-SA 4.0)

In our case, the volume of the cut corner is $1/6^{th}$ of the volume of the cube (see part 1). But our smaller cube itself has a volume equal to $1/8^{th}$ of our larger cube, so the ratio between the volume of the cut corner and the larger cube is equal to : $1/6 \times 1/8 = 1/48$