7. As shown in the figure, a cross-section of the cube $A B C D-E F G H$ passes through vertices $A, C$ and a point $K$ on edge $E F$, dividing the cube into two parts with a volume ratio of $3: 1$. Then the value of $\frac{E K}{K F}$ is $\qquad$ .
Answer: $\sqrt{3}$.
Solution: Let $\alpha$ be the plane of the section. Extend $A K, B F$ to intersect at point $P$, then $P$ is on $\alpha$, so line $C P$ is the intersection line of $\alpha$ and plane $B C G F$. Let $C P$ intersect $F G$ at point $L$, then quadrilateral $A K L C$ is the section.
Since plane $A B C$ is parallel to plane $K F L$, and $A K, B F, C L$ intersect at point $P$, $A B C-K F L$ is a frustum. Assume the edge length of the cube is 1, then the volume of the cube is 1, and according to the condition, the volume $V$ of the frustum $A B C-K F L$ is $\frac{1}{4}$.

Let $P F=h$, then $\frac{K F}{A B}=\frac{F L}{B C}=\frac{P F}{P B}=\frac{h}{h+1}$. Note that $P B, P F$ are the heights of the pyramids $P-A B C$ and $P-K F L$, respectively, thus
$$
\begin{aligned}
\frac{1}{4} & =V=V_{P-A B C}-V_{P-K F L}=\frac{1}{6} A B \cdot B C \cdot P B-\frac{1}{6} K F \cdot F L \cdot P F \\
& =\frac{1}{6}(h+1)\left(1-\left(\frac{h}{h+1}\right)^{3}\right)=\frac{3 h^{2}+3 h+1}{6(h+1)^{2}} .
\end{aligned}
$$

Simplifying gives $3 h^{2}=1$, hence $h=\frac{1}{\sqrt{3}}$. Therefore, $\frac{E K}{K F}=\frac{A E}{P F}=\frac{1}{h}=\sqrt{3}$.