Tetrahedral Water Tank - Part 2

If the tetrahedron has side $a$ , the new altitude (the distance between the midpoints of opposite sides) is $h = \tfrac{1}{\sqrt{2}}a$ , and the volume of the tetrahedron is $V = \tfrac{1}{6\sqrt{2}}a^3 = \tfrac13h^3$ .

In the new configuration, the cross-section of the tetrahedron at height $z$ is a rectangle of sides $\tfrac{a}{h}z$ and $\tfrac{a}{h}(h-z)$ . Thus the cross-sectional area at height $z$ is $\big(\tfrac{a}{h}\big)^2z(h-z)$ , and hence the volume up to height $z$ is $V(z) \; = \; \big(\tfrac{a}{h}\big)^2 \int_0^z u(h-u)\,du \; = \; \big(\tfrac{a}{h}\big)^2 \times \tfrac16z^2(3h - 2z) \; = \; \tfrac13z^2(3h - 2z) \; = \; \big(\tfrac{z}{h}\big)^2\big[3 - 2\tfrac{z}{h}\big]V$ The volume of water in the tetrahedron is $V(1 - 0.6^3)$ , and so we want to determine $z$ such that $\big(\tfrac{z}{h}\big)^2\big[3 - 2\tfrac{z}{h}\big] \; = \; 1 - 0.6^3$ which gives the answer $\tfrac{z}{h} = \boxed{0.7}$ .

Another way of looking at this problem is to observe that the tetrahedron is now a ruled prism, and so its cross-sectional area is a quadratic function of the height which is zero at $z=0$ and $z=h$ . Thus $A(z) = \alpha z(h-z)$ , so that $V(z) = \tfrac16\alpha z^2(3h - 2z)$ . Since $V(h) = \tfrac13h^3$ we can deduce that $\alpha = 2$ .