The (Martini) Glass Is Half Full

Relevant wiki: Volume of a Cone

Let the volume, depth, side length and surface radius be $V$ , $h$ , $x$ and $r$ respectively. Then we have:

$\begin{aligned} V & \propto \color{#3D99F6}{r}^2h \quad \quad \small \color{#3D99F6}{\text{Since } r \propto h} \\ V & \propto h^3 \quad \quad \small \color{#3D99F6}{\text{And as } x \propto h} \\ V & \propto x^3 \\ \Rightarrow \frac{V_\frac{1}{2}}{V_{full}} & = \left(\frac{x_\frac{1}{2}}{x_{full}}\right)^3 \\ \left(\frac{x_\frac{1}{2}}{x_{full}}\right)^3 & = \frac{1}{2} \\ \frac{x_\frac{1}{2}}{x_{full}} & = \frac{1}{\sqrt [3]{2}} \approx \boxed{79} \, \% \end{aligned}$

The cone formed by the liquid is geometrically similar to the cone of the glass. The ratio of the volumes of similar solids is the cube of the ratio of any corresponding linear dimensions. So the liquid is the cube-root of $\frac{1}{2}$ up the side. That's about $\boxed{79}\%$ .

For a much tougher (but very interesting) challenge, consider the same question when we tip the glass .