Here is a math question i got from school:
On a horizontal plane, there are two flagpoles. One is 20m, and the other is 10m. There is a wire connected from the top of each flagpole, to the bottom of the other one so that they cross each other . How high is the point they cross from the ground?
How do I solve it without using pythagoras
$\endgroup$2 Answers
$\begingroup$Let $h$ be the desired height, and let $a$ and $b$ be the horizontal distances from the point where the wires cross to the left and right flagpoles, respectively. Then using similar triangles, we have: $$ \frac{h}{a} = \frac{10}{a + b} \qquad\text{and}\qquad \frac{h}{b} = \frac{20}{a + b} $$ Solving for $h$, we can combine the above to get that: $$ \frac{10a}{a+b} = h = \frac{20b}{a + b} \iff a = 2b $$ Substituting into the second equation, we get: $$ \frac{h}{b} = \frac{20}{(2b) + b} \iff \boxed{h = \dfrac{20}{3}} $$
$\endgroup$ $\begingroup$This is the definition of the harmonic mean.
Let's call the left flag pole $h_1$ and the the right one $h_2$. The distance from $h_1$ to $x$ is $a$, from $x$ to $h_2$ is $b$.
Then using similar right triangles: $$\begin{align} \frac{a+b}{h_1} &= \frac{b}{x}\\ \frac{a+b}{h_2} &= \frac{a}{x}\\ \text{Adding them together}\\ \frac{a+b}{h_1} + \frac{a+b}{h_2} &= \frac{a+b}{x}\\ \text{Divide through by $a+b$}\\ \frac{1}{h_1} + \frac{1}{h_2} &= \frac{1}{x}\\ \end{align} $$
Plug in 10 and 20 for $h_1$ and $h_2$.
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