Shell Game 3

We can get the result by "slicing off" the floor of the original regular tetrahedron which will reduce the height by 6 inches producing smaller regular tetrahedron. Its height can be obtained using the formula for the height of regular tetrahedron of side $a$ : $h=\sqrt{\frac{2}{3}} a$ which gives the height of
$h = 10 \sqrt{ \frac{2}{3} }-0.5$ . Repeating it 3 more times to slice off the other 3 faces should give the height of $h_2 = 10 \sqrt{ \frac{2}{3 } } - 2$ and edge of $a_2 = \sqrt{ \frac{3}{2} }(10 \sqrt{ \frac{2}{3} } - 2) = 10 - \sqrt{6}$ .

Outside tetrahedron has the volume of:

$V=\sqrt{2} a^3/ 12 = 250 \sqrt{2}/ 3$

Inside tetrahedron has the volume of :

$V_2 = \sqrt{2} a_{2} ^3/ 12 = \sqrt{2} (10 - \sqrt{6})^3 / 12$

Subtracting volume of "inside" tetrahedron from the volume of the "outside " tetrahedron gives the result of : $V - V_2 = 51 \sqrt{3} - 15 \sqrt{2} \approx \boxed{67.1}$

Solution 2

Formula for the insphere radius of a tetrahedron is $r = a / \sqrt{24}$ . The insphere of the tetrahedron obtained by removing the walls will have radius $r_2= 10 / \sqrt{24} - 1/2$ and the new tetrahedron's edge will be $r_2 * \sqrt{24}=10-\sqrt{6}$ .

Subtracting volume of "inside" tetrahedron from the volume of the "outside " tetrahedron gives the result of : $\sqrt{2} *10^3/ 12 -( \sqrt{2} (10 - \sqrt{6})^3 / 12) = 51 \sqrt{3} - 15 \sqrt{2} \approx \boxed{67.1}$