Problem-solving with ratios and similar triangles.

Question

The following question is designed to test problem-solving and reasoning skills with ratios and similar triangles. Without using Pythagoras Theorem or Trigonometry this question is supposed to be solvable. But how?

I'm trying to figure out if there is an error in this question. Can this be answered using ratio and scale factor only?

A string 50m long is pegged to the ground and tied to the top of a flag pole. It just touches the head of Maureen, who is 5 meters away from the point where the string is held to the ground. If Maureen is 1.5 meters tall, show that the height, h, of the flagpole is 14.37m

Answer 1

Let $CD$ be an altitude of $\Delta ABC$ and $\measuredangle ACB=90^{\circ}.$

Thus, $$\Delta ACD\sim\Delta ABC,$$ which gives $$\frac{AC}{AB}=\frac{AD}{AC}$$ or $$AC^2=AD\cdot AB.$$ Similarly from similarity of triangles $BCD$ and $BAD$ we obtain $$BC^2=BD\cdot AB.$$ Id est, $$AC^2+BC^2=AD\cdot AB+BD\cdot AB=(AD+BD)AB=AB^2.$$ Now, let $h=BC$.

Thus, $$\frac{h}{1.5}=\frac{AC}{5},$$ which gives $$AC=\frac{10h}{3},$$ $$h^2+\left(\frac{10h}{3}\right)^2=50^2$$ or $$h=\frac{150}{\sqrt{109}}.$$

Answer 2

Here is my take on solving the problem without an appeal to Pythagoras.

Referring to the first figure below, it's obvious from the similarity of triangles that ${H}/{h} = (5+x)/{5}, \text{that is, }$

$$ H=h+ \frac{xh}{5} \tag{1} $$

From the figure of the square formed with four triangles congruent to the triangle in the problem, the following is obvious:

$$ 50^2 = 2H(5+x) + (5+x-H)^2 $$

Substituting (1) in the above equation we get a quadratic in $x$:

$$ 1.09x^2 + 10.9x - 2471.75 = 0. $$

The positive solution is $x=42.882$, from which the result follows.

This proof refers to the well known arrangement of right triangles used to prove Pythagoras theorem, however, the theorem is not referenced in any way.