In the cube shown, $L$ and $K$ are midpoints of adjacent edges $A D$ and $A B$. The perpendicular distance from $F$ to the line segment $L K$ is 10. What is the volume of the cube, to the nearest integer?
(A) 323
(D) 326
(B) 324
(C) 325
(E) 327

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-125.jpg?height=328&width=285&top_left_y=584&top_left_x=1359)

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-125.jpg?height=32&width=239&top_left_y=968&top_left_x=913)
We label the cube as shown in the diagram.

Notice that since $A L=A K$ that $\triangle A L K$ is isosceles and right-angled. This means that $L K=\sqrt{2} x$.

Next, we draw the line from $F$ which is perpendicular to $L K$ and meets $L K$ at $Q$. By symmetry, $Q$ is the midpoint of $L K$. Thus we can label $F Q=10$ and $L Q=\frac{\sqrt{2} x}{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-207.jpg?height=428&width=374&top_left_y=241&top_left_x=1366)

Using Pythagoras in $\triangle D G F, D F^{2}=(2 x)^{2}+(2 x)^{2}=8 x^{2}$.

Also, $\triangle F D L$ is right-angled at $D$, so using Pythagoras again, $F L^{2}=8 x^{2}+x^{2}=9 x^{2}$. Lastly, using Pythagoras in $\triangle L Q F$,

$$
\begin{aligned}
L F^{2} & =L Q^{2}+Q F^{2} \\
9 x^{2} & =\left(\frac{\sqrt{2} x}{2}\right)^{2}+10^{2} \\
9 x^{2} & =\frac{1}{2} x^{2}+100 \\
\frac{17}{2} x^{2} & =100 \\
x & \approx 3.429
\end{aligned}
$$

The volume is thus $8 x^{3} \approx 322.82$, which rounding off is 323 .