Question 6

Explanation

To find the distance from point $Q$ to line $BD$, we will form triangle $BDQ$ and calculate its height from the vertex $Q$. The first step is to calculate the lengths of all three sides of this triangle.

Calculating Side $BD$ (Base Diagonal)

Line $BD$ is the diagonal of the cube's base with edge length $s = 4$.

Calculating Side $DQ$

Point $Q$ is at the midpoint of $FG$. We can find the length of $DQ$ by considering the right-angled triangle $DHQ$. First, we calculate the length of $HQ$ on the top face of the cube:

Next, use the Pythagorean theorem on triangle $DHQ$:

Calculating Side $BQ$

The length of $BQ$ can be calculated from the right-angled triangle $BFQ$, with $BF = 4$ and $FQ = 2$:

Calculating Distance from $Q$ to Line $BD$

Now we have triangle $BDQ$ with sides $BD = 4\sqrt{2}$, $DQ = 6$, and $BQ = 2\sqrt{5}$. Let $O$ be the perpendicular projection of $Q$ onto line $BD$.

Let the length $DO = x$, then the length $OB = 4\sqrt{2} - x$. The height of the triangle $QO$ (let's call it $y$) is the distance we are looking for. Using the two formed right-angled triangles, we obtain the equation:

Substituting the side values:

After finding $x = 3\sqrt{2}$, we substitute it back to find $y$:

Thus, the distance from point $Q$ to line $BD$ is $3\sqrt{2}$ units.