# The height and hypotenuse of a right triangle

So I’ve recently started a three semester teachers program at the Stockholm university in order to be formally certified to teach within the Swedish educational system. It’s mostly reading and informal cognition theory, becoming more aware of how students approach materials and how to critically analyse the sequencing and structure of instruction, what values to impart and all that good stuff.

In reading the material this week however I came across a pretty neat problem which I’d like to present some solutions for. So in Skott et al’s Matematik för lärare $\delta$ Didaktik (Mathematics for teachers), page 270 we find a suggested seminar acticity which mentions the following (meta problem).

“A math book contains the following exercise: “Consider the right triangle with hypotenuse 8, and altitude relative to the hypotenuse 5. What is the area?” This exercise contains an error. Find it.”

So superficially this is about the concept of computing the area of a triangle using base(foot) and height(altitude)and the student will be reminded of the important idea that a triangle has more than one height depending on what is considered the base (or “foot”), but in actuality it about first reaching a conclusion about the hypothenuse and hypothenuse-height not being independent quantities but in fact connected by mathematical relations.

Since the reader might want to solve or engage with the problem themselves I’ve inserted the read-more breaker here to act as a loose spoiler-marker but the point here will be to investigate some different ways of presenting the solution.

Posing the problem in formal terms:

As the exercise contains only two numerical quantities and only one piece of structural information (the triangle being right) one should quickly realize that the two quantities must somehow be incompatible. How you arrive at this conclusion is personal but the basic conclusion is that the altitude must be less than half the hypotenuse.

$h < c/2$

Since 5 is more than half of 8 the triangle being described in the original problem cannot the real. Why that is the case or why one would suspect it becomes linked to ones representation of the problem.

Approach 1 (Algebraic)

One approach is to try and find the values of the legs of the right triangle $a,b$ and identify an algebraic hurdle to finding them which exposes the inconsistency of the problem. Using the pythagorean theorem and the definition of area we get two equations

$\begin{cases}a^2 + b^2 = c^2 & \\ ab/2 = ch/2 &\end{cases}$

Which I then rewrite as

$\begin{cases}a^2 + b^2 = c^2 & \\ a^2b^2 = c^2h^2 &\end{cases}$

Now this is a system of equations of a sum-product-type which can be solved in various ways but most directly by observing that $a^2,b^2$ must be solutions to the quadratic equation.

$x^2 - c^2x + c^2 h^2 = 0$

For this equation to have real solutions it’s discriminant must be positive

$c^4 - 4c^2 h^2 > 0$

but

$c^4 - 4c^2 h^2 > 0 \Leftrightarrow c^2 > 4h^2 \Leftrightarrow c > 2h$

and thus

$c/2 > 0$

This is fairly direct and transparent and could have been approached using the explicit quantities c = 8, h = 5 had one not been comfortable with generalitites.

Approach 2 (Using circumcise)

If one is aware that every triangle has an excircle and that the excircle of a right triangle has the hypotenuse as it’s diameter one could draw the following diagram which exposes the inequality fairly directly

Approach 3 (Using the midpoint of the hypotenuse)

A property of right triangles is that the line segment connecting the midpoint of the hypotenuse to the point at the right angle has the same length as half the hypotenuse.

If one draws such a line it will expose a small right triangle for which the original altitude is a leg and this new segment is a line.

The stament will now follow from the hypotenuse being the longest side in a right triangle.

Okay that’s it for now,.