Find $\text{QR}$ where $\text{PQ} = 39$ and $\text{PR} = 17$ and altitude of $\triangle \text{PQR}$ is $15$?

Stumped on the second part of this question for a tutoring student. The problem states:

In $$\triangle \text{PQR}$$, $$\text{PQ} = 39 \text{ in.}$$, $$\text{PR} = 17 \text{ in.}$$, and the altitude $$\text{PN} = 15 \text{ in.}$$ Find $$\text{QR}$$. Consider all cases.

The first case, which is more obvious, is that $$\text{PN}$$ and $$\text{NQ}$$ satisfy the Pythagorean Theorem because $$\triangle \text{QPN}$$ and $$\triangle \text{PNR}$$ are right triangles. Therefore:

$$15^2+\text{QN}^2=39^2 \implies \text{QN}=36$$

$$15^2+\text{NR}^2=17^2 \implies \text{NR}=8$$

$$\text{QR}=\text{QN}+\text{NR}=36+8=44$$

The system allows my student to fill in two possible answers. Of course, the first answer is $$44$$ in., but what is the case that would give the second answer? At first I thought that we could consider the case where $$\angle \text{QPR}=90^{\circ}$$ and use the Pythagorean Theorem, but I'm pretty sure that would mean that $$\text{PN}\neq 15$$. What would the other case be?

(keep in mind this student has not covered trig yet)

• I'm also not sure how there could be multiple cases. I guess perhaps one possible such consideration would be whether $PN$ is in the triangle or not (visual), but it's pretty easy to see the left-hand case leads to impossibilities. Jan 12 at 2:27
• The problem appears to be fully constrained with PN given. There is no ambiguity, not even the possibility of a degenerate triangle, in which case the altitude’s length would equal that of both adjacent sides. Jan 12 at 3:01
• We've found the other case! See the answers below. :) Jan 12 at 3:55

Without many words .... $${}{}{}{}{}{}{}{}$$

As @ACB rightfully pointed out, there are two cases. The first case, which I have described in the question, is where $$\text{N} \in \overline{\text{QR}}$$. The second case is where $$\text{N} \notin \overline{\text{QR}}$$, $$\angle \text{QRP}$$ is obtuse, and $$\overline{\text{PN}}$$ is outside of $$\triangle \text{PQR}$$ (picture not to scale):

As is pictured, let $$x=\text{QR}$$ and $$y=\text{RN}$$.

Using the Pythagorean Theorem, we can create a system of equations to solve $$x$$ and $$y$$:

$$\Bigg\{\begin{array}[c] 115^2 + (x+y)^2 = 39^2\; (1) \\ 15^2+y^2=17^2 \; (2) \end{array}$$

Solving equation $$(2)$$ for $$y$$, we get $$y=8$$. Then by substitution we can write equation $$(1)$$ as:

\begin{align} 15^2+(x+8)^2&=39^2 \\ 225+(x+8)^2&=1521 \\ (x+8)^2&=1296 \\ x+8&=36 \\ x&=28 \end{align}

Therefore, $$x=\text{QR}=28$$ and the second answer is $$28$$ in.

Let the base, part $$1$$, be $$\space QX=\sqrt{QP^2-PX^2}=\sqrt{39^2-15^2}=36.\space$$

Let the base, part $$2$$, be $$\space XR=\sqrt{PR^2-PX^2} =\sqrt{17^2-15^2}=8.\space$$

Then, it follows

$$\quad QR=QX+XR=36+8=44$$

or

$$\quad QR=QX-XR=36-8=28$$