How do I find the lengths of the two legs of a right triangle if one leg is half the size of the other and I'm given the hypotenuse?
I've tried squaring the hypotenuse and then dividing up that number into 3 parts. I then give 2 of those parts to the bigger leg and one to the smaller leg. Afterward, I find the square root of each part of the hypotenuse squared. Whenever I do this, I end up with two numbers, but neither of them are half of the other.
(P.S. I'm sorry if this question is confusing to read.) (I'm not sure if I put down the right tag either. This is my first time on this website.)
$\endgroup$ 12 Answers
$\begingroup$Let the hypotenuse be given as $h$. Let the short leg be $x$ and the long leg be $2x$. By the Pythagorean theorem,\begin{align}x^2+(2x)^2&=h^2\\x^2+4x^2&=h^2\\5x^2&=h^2\\x^2&=\frac{h^2}{5}\\x&=\frac h{\sqrt 5}\end{align}
$\endgroup$ 1 $\begingroup$The proportion of the sides should be $(1,2, \sqrt5)$. Even if not required after seeing $\sqrt 5$ was curious to find how this right triangle whose sides can be related to the Golden Ratio $ \phi = \dfrac{\sqrt5 +1}{2}$ using Pythagorean triples... even if not asked for strictly in the question.
In the Pythagorean triplet let
$$(m^2+n^2)^2=(2mn)^2+(m^2-n^2)^2,\; \text{when} \;2mn > m^2-n^2\tag1 $$
$$2mn= 2(m^2-n^2),\; m^2-mn-n^2 =0 \tag 2 $$
Solving the quadratic,
$$ m = n \dfrac{1\pm \sqrt5}{2} = n \phi \tag3$$
We should discard negative sign of a line segment.
Plugging into first part of(1)
$$ n^2 \left((\phi^2+1),(\phi^2-1),2\phi \right ) \tag4$$
By the Golden Ratio property Equation (2) has the form $\; \phi^2= \phi+1$ we have
$$ n^2 \left((\phi+2), \phi, 2\phi \right ) \tag5$$
and wlog the proportion of three sides turn out related to $\phi$ as expected.
$$ \left(\phi+2, \phi, 2\phi \right ) \tag6$$
which are essentially in the proportion
$$ (\sqrt 5, 1,2). \tag7 $$
