Finding fraction of a volume by visualization of cube being cut into separate pieces

Question

Cube $ABCDEFGH$ is cut into four pieces by cutting along planes $BCHE$ and $BDHF.$ Find the fraction of the volume occupied by the piece containing the vertex $A.$

I'm not really able to visualize how the cut along BDHF would affect the piece containing the vertex A.

Answer 1

Observe that the fraction containing the vertex A is a pyramid with the square base AEHD of area $1$ and the height AB = $1$. Thus, its volume is $\frac13$.

Answer 2

First of all, the first cut would divide the cube into an upper triangular brism and a lower triangular prism (hopefully you can imagine that) The shape under consideration is the upper triangular brism $ADCBEH$ Now, to see why it affects it, take a vertical plane to cut the cube on $BD$. (Like take $BD$ and extend it into a rectangle vertically like it cuts the shape underneath it)

Answer 3

Let the side of the cube =$1$.

$BCHE$ cuts the cube into two equal prisms, whose faces are two faces of the original cube, i.e. square $ABCD$ and $AEHD$, two isosceles right triangles from two bisected faces of the cube, i.e. $EAB$ and $HDC$, and one rectangular face $BCHE$ having the base of one of the isosceles triangles for length and the side of the cube for width.

Focussing on the prism that has point $A$ on it, what $BDHF$ does is

1) remove isosceles right triangular face $HDC$,

2) leave face $EAB$ unchanged,

3) bisect square $ABCD$ leaving isosceles right triangle $BAD$,

4) bisect rectangle $BCHE$ leaving right triangle $BEH$ with sides $EH=1$, $EB=\sqrt 2$, and (by Pythagoras) $BH=\sqrt 3$, and

5) create a new face: right triangle $BDH$ (right angle at $D$) also with sides $1, \sqrt 2, \sqrt 3$.

Thus we are left with a pyramid with square base $ADHE$, whose faces $EAB$, $BAD$, $BEH$, $BDH$, are two isosceles right triangle equal to half the square, and two congruent scalene right triangles with sides $1$, $\sqrt 2$, and $\sqrt 3$.

Since the volume of the cube is $1$, and the volume of a pyramid on a square base is one-third the area of the base times the height, and the area of base $ADHE=1$, and height $AB=1$, the volume of the pyramid here is$$1\cdot \frac{1}{3}=\frac{1}{3}$$the volume of the cube.