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:

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

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


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)

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

3 Answers 3


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

can you see two triangles?


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.


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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.