Theodorus spiral, a sequence of right triangles with unit legs, demonstrating the growth of hypotenuse lengths.

We are given a geometric figure composed of right triangles, all sharing a common vertex at the center. The figure is made up of several right triangles, each with legs of length 1 or previously computed hypotenuses. The goal is to find the value of $ x $, which is the length of the outermost segment (the hypotenuse of the largest triangle).

This construction is known as the "Theodorus spiral" (or square root spiral), named after the ancient Greek mathematician Theodorus of Cyrene.

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Step-by-step solution:

Each triangle is a right triangle, and we will use the Pythagorean theorem:
$$
a^2 + b^2 = c^2
$$

We’ll compute the hypotenuses step by step: $ c_1, c_2, c_3, c_4 $, and finally $ x $.

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#### Step 1: Find $ c_1 $

The first triangle has two legs of length 1.

So,
$$
c_1^2 = 1^2 + 1^2 = 1 + 1 = 2 \\
\Rightarrow c_1 = \sqrt{2}
$$

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#### Step 2: Find $ c_2 $

The next triangle has one leg of length 1 and the other leg equal to $ c_1 = \sqrt{2} $. So:
$$
c_2^2 = 1^2 + (\sqrt{2})^2 = 1 + 2 = 3 \\
\Rightarrow c_2 = \sqrt{3}
$$

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#### Step 3: Find $ c_3 $

Now, one leg is 1, and the other is $ c_2 = \sqrt{3} $:
$$
c_3^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \\
\Rightarrow c_3 = \sqrt{4} = 2
$$

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#### Step 4: Find $ c_4 $

Next triangle: one leg is 1, the other is $ c_3 = 2 $:
$$
c_4^2 = 1^2 + 2^2 = 1 + 4 = 5 \\
\Rightarrow c_4 = \sqrt{5}
$$

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#### Step 5: Find $ x $

Now, the final triangle has one leg of length 1 and the other leg $ c_4 = \sqrt{5} $:
$$
x^2 = 1^2 + (\sqrt{5})^2 = 1 + 5 = 6 \\
\Rightarrow x = \sqrt{6}
$$

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✔ Final Answer:

$$
\boxed{\sqrt{6}}
$$

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Summary:

- $ c_1 = \sqrt{2} $
- $ c_2 = \sqrt{3} $
- $ c_3 = 2 $
- $ c_4 = \sqrt{5} $
- $ x = \sqrt{6} $

This pattern continues in the Theodorus spiral, where each new hypotenuse is $ \sqrt{n} $ for $ n = 2, 3, 4, \dots $, and here we have gone up to $ \sqrt{6} $.