Martini Glass Conundrum

Suppose we completely fill a conical martini glass. If we then pour the entire contents equally into two other identical martini glasses, what will the depth of liquid be in the two glasses?

(A) 49% of the original glass

(B) 59% of the original glass

(C) 69% of the original glass

(D) 79% of the original glass

Martini Glass
The solution is (D).  Easiest way to see this is to consider the volume of the full glass as $V_f = \frac{1}{3}\pi r^2h$. As the ratio between the height and the radius is constant, then let r = $\alpha h$, which gives $$ V_f = \frac{1}{3}\pi \alpha^2h^3 $$ The volume of the half filled glasses are therefore $$ \frac{1}{2}V_f = \frac{1}{2}\frac{1}{3}\pi \alpha^2 h^3 $$ $$ \frac{1}{2}V_f = \frac{1}{3}\pi \alpha^2 \left(\frac{h}{^3\!\sqrt{2}}\right)^3 $$ The height of liquid in a half full glass is therefore $\frac{h}{^3\!\sqrt{2}} \approx 0.79h$