Length $x$ of part of a triangle

Question

I can't calculate the length $x$ in this diagram. What is $x$?

I think some people misunderstood the question. Please see below picture.

To be frank, I tried to use two Pitagoras' formula as below (like others). $$x^2 + y^2 = 3$$ $$x^2 + (y+\sqrt 3)^2 = 2^2$$ substituting $y^2 = 3 - x^2$, (or $y = \sqrt(3-x^2)$, $$x^2+y^2+2\sqrt3y+3 = 4$$ $$x^2+3-x^2+2\sqrt3\sqrt{3-x^2}=1$$ $$2+2\sqrt3\sqrt{3-x^2}=0$$ $$1+\sqrt{9-3x^2}=0$$ This is unsolvable? What's wrong?

EDIT : I realized too late that the triangle is acute. when three sides are equal, they are 60 degrees. Even if two sides are $\sqrt3$, they are still acute(by comparison of $c^2$ and $a^2+b^2$ as A. Pongrácz told me. How stupid I was..

Answer 1

It is just two pythagorean theorem:

Answer 2

It is the altitude. There are a couple of ways to compute it. An elegant one: compute the are of the triangle using Heron's formula, and then use the well-known formula that involves the area, a side and an altitude.

By the way, the picture is not really correct. The triangle should be acute, so the altitude intersects the opposite side between the vertices.

Answer 3

You will get two equations

$$\sqrt{3}^2=y^2+x^2$$ $$(\sqrt{3}-y)^2+x^2=4$$ Can you solve this? The second equation is

$$3+x^2+y^2-2\sqrt{3}y=4$$