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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)

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  • $\begingroup$ 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. $\endgroup$ Commented Jan 12, 2022 at 2:27
  • $\begingroup$ 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. $\endgroup$ Commented Jan 12, 2022 at 3:01
  • $\begingroup$ We've found the other case! See the answers below. :) $\endgroup$ Commented Jan 12, 2022 at 3:55

3 Answers 3

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

can you see two triangles?

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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):

enter image description here

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.

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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

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

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