When an altitude is drawn from the right angle to the hypotenuse of a right triangle, the altitude length satisfies $QS^2 = PS \cdot RS$. Use $QS = 6$ and $RS = 4$ to compute $PS$.

After you know $PS$ and $RS$, find $PR = PS + RS$. Then, use the fact that the leg next to $PS$ (which is $PQ$) satisfies $PQ^2 = PS \cdot PR$. Plug in your values for $PS$ and $PR$ and simplify.

Use the relationship $QS^2 = PS \cdot RS$. Since $QS = 6$ and $RS = 4$, type 36/4 into Desmos. The value shown is $PS$.

Now compute the full hypotenuse: type 9+4 (using your value for $PS$ plus 4 for $RS$). The result is the length $PR$.

To get $PQ$, use $PQ^2 = PS \cdot PR$. In Desmos, type sqrt(9*13) (or more generally sqrt(PS*PR) using your computed values). The output is the length of $PQ$; compare this value with the answer choices to select the matching expression.

Draw right triangle $\triangle PQR$ with $\angle Q$ a right angle, and altitude $QS$ dropping to hypotenuse $PR$.

• $QS$ is perpendicular to $PR$, so $S$ lies on $PR$.
• The hypotenuse $PR$ is split into two segments: $PS$ and $RS$, and we are given $QS = 6$ and $RS = 4$.
• This construction creates three right triangles: $\triangle PQR$, $\triangle PQS$, and $\triangle QRS$. All three are similar to each other (same angles, different sizes).

In a right triangle, when you drop an altitude from the right angle to the hypotenuse, the altitude is the geometric mean of the two segments of the hypotenuse:

Substitute the known values $QS = 6$ and $RS = 4$:

Solve for $PS$:

So $PS = 9$ units.

The hypotenuse $PR$ is made of the two segments $PS$ and $RS$:

Now we know both the full hypotenuse $PR = 13$ and the adjacent segment $PS = 9$. Next, we will relate these to leg $PQ$. In this configuration, each leg of the original right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse next to that leg.

Because $\triangle PQR$ and $\triangle PQS$ are similar, leg $PQ$ is the geometric mean of the whole hypotenuse $PR$ and the adjacent segment $PS$:

Substitute $PS = 9$ and $PR = 13$:

Take the square root:

So the length of $\overline{PQ}$ is $3\sqrt{13}$, which corresponds to choice B.