3-D geometry

The volume of a regular tetrahedron is $\frac{a^3}{6 \sqrt{2}}$ where a is our side length in this case $\sqrt{2}$ so we now have $\frac{2 \sqrt{2}}{6 \sqrt{2}}$ or $\frac{1}{3}$ .

This can be done without the formula of a regular tetrahedron. We can easily observe that in order to obtain the tetrahedron we want, we need to remove 4 smaller tetrahedrons (one is at the back), and each of this 'small' tetrahedrons are rather easy to count, as their base is half the area of the cube's base, while their height is equal to the cube's height. Therefore each of these small tetrahedrons' volume is $\frac{base \times height}{3} = \frac{\frac{1}{2} \times 1}{3} = \frac{1}{6}$ There are four of them, so the volume of the tetrahedron we want is $1-4 \times \frac{1}{6} = \frac{1}{3}$

Collecting the parts together form a pyramid!!

If we see the figure, we need to cut four equal solids to form the new geometric shape. The volume of that one solid is $\dfrac{1}{3}A_bh$ where $A_b$ = area of the base and $h$ = height. That is equal to $\dfrac{1}{3}\left(\dfrac{1}{2} \right)(1^2)(1)=\dfrac{1}{6}$ . So the volume of the new geometric shape is

$1^3-4\left(\dfrac{1}{6}\right)=1-\dfrac{2}{3}=$ $\large \boxed{\color{#D61F06}\dfrac{1}{3}}$