Finding perimeter when the hypotenuse is 1 and is the sum of the altitude and side of a triangle

Question

Let us say the hypotenuse of a triangle is 1, which is also equal to the sum of side $a$ and the altitude $h$ when taking the hypotenuse as the base. How would I find the length of side $a$ and $b$ (the other side), and thus the triangle's perimeter?

I've tried making a triangle inscribed in a circle, where the hypotenuse is the diameter of the circle so it's right-angled, and making $a$ and $b$ chords. But it leads nowhere for me. Also tried using the geometric mean theorem. I split the hypotenuse into two lengths, $p$ and $q$, where $$h^2+q^2=a^2$$ Then I defined $h=pq$ based on the theorem, which can be rewritten as $h=(1-q)q=q-q^2$. Therefore$$(q-q^2)^2+q^2=a^2\\q^4-2q^3+2q^2=a^2\\\sqrt{q^4-2q^3+2q^2}=a$$ And because $a+h=1$, we can replace them with what we derived $$\sqrt{q^4-2q^3+2q^2}+q-q^2=1\\(\sqrt{q^4-2q^3+2q^2})^2=(q^2-q+1)^2\\q^4-2q^3+2q^2=q^4-2q^3+3q^2-2q+1\\q^2-2q+1=0\\(q-1)^2,\;q=1$$ I've gone wrong somewhere but I don't know where, and I also have no idea how to solve this problem. I don't think any angle related formulas like Law of Sines or Law of Cosines would help as I also don't know how to find any angles.

Answer 1

You have three equations relating $a$, $b$, and $h$: $$\begin{align} a+h&=1\\a^2+b^2&=1^2\\ab&=1\cdot h \end{align}$$ The first one is given, the second is Pythagoras' theorem, the third one is the area of the triangle. $$\begin{align}h&=1-a\\b&=\frac ha=\frac{1-a}{a}\\b&=\sqrt{1-a^2}\end{align}$$ This yields $$a^3+a^2+a-1=0$$ You can solve this numerically or use the cubic equation formalism. You will get $$a\approx 0.54369$$The perimeter is $$p=a+b+1=a+\frac 1a-1+1=a+\frac 1a\approx 2.382973$$

Answer 2

The area of the triangle is $\frac{1}{2}\times 1\times h$ (considering hypotenus as base). And is also $\frac{1}{2}\times a\times b$. Thus $ab=h$ and $a^2+b^2=1$. Therefore $a+b=\sqrt{a^2+b^2+2ab}$. Thus $a+b+1=1+\sqrt{1+2h}$ is the perimeter. Hope its correct.

Answer 3

You made a mistake with $h=pq$, which should be $h^2=qp=q(1-q)$. Then, you have

$$q^2+q(1-q) = a^2$$ which leads to

$$q =a^2=\left(1-\sqrt{q(1-q)}\right)^2$$

Then, you could solve for $q$.

Answer 4

From given, $a+h = 1$ and $a^2 + b^2 = 1$.

From similar triangles inside the right-angled triangle,

$$\begin{align*}\frac b1 &= \frac ha\\ ab &= h\\ a+ab &= 1\\ b &= \frac{1-a}{a}\\ a^2 + \left(\frac{1-a}{a}\right)^2 &= 1\\ a^4 + 1 - 2a + a^2 &= a^2\\ a^4 - 2a + 1 &= 0\\ (a-1)(a^3 +a^2+a-1)&=0 \end{align*}$$

There is one real root where $0<a<1$: $$a\approx 0.543689012692076$$