To solve this problem using geometric corollaries related to right triangles, we can use the relationships between the segments formed by the altitude from the right angle to the hypotenuse.

Given:

• A right triangle with legs $x$ and $x + 5$
• The hypotenuse of this right triangle is 20
• An altitude from the right angle to the hypotenuse forms two smaller right triangles

Steps to solve:

1. Identify the segments on the hypotenuse formed by the altitude: Let $h$ be the altitude from the right angle to the hypotenuse, dividing the hypotenuse into two segments $d$ and $20 - d$.

2. Use the corollary of the geometric mean (altitude) theorem: The altitude to the hypotenuse of a right triangle is the geometric mean of the segments of the hypotenuse: $h = \sqrt{d(20 - d)}$

3. Apply the corollary of the geometric mean (leg) theorem: Each leg of the original triangle is the geometric mean of the hypotenuse and the adjacent segment: $x = \sqrt{d \cdot 20}$ $x + 5 = \sqrt{(20 - d) \cdot 20}$

We will use these relationships to form equations and solve for $x$.

Solving the equations:

Since: $x = \sqrt{20d}$ $x + 5 = \sqrt{20(20 - d)}$

Square both sides of the equations to remove the square roots: $x^2 = 20d$ $(x + 5)^2 = 20(20 - d)$

Expanding the second equation: $x^2 + 10x + 25 = 400 - 20d$

From $x^2 = 20d$, we substitute $20d$ in the expanded second equation: $x^2 + 10x + 25 = 400 - x^2$

Combine like terms: $2x^2 + 10x + 25 = 400$ $2x^2 + 10x + 25 - 400 = 0$ $2x^2 + 10x - 375 = 0$

Divide the entire equation by 2 to simplify: $x^2 + 5x - 187.5 = 0$

This is a quadratic equation of the form $ax^2 + bx + c = 0$. We can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation, $a = 1$, $b = 5$, and $c = -187.5$: $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-187.5)}}{2 \cdot 1}$ $x = \frac{-5 \pm \sqrt{25 + 750}}{2}$ $x = \frac{-5 \pm \sqrt{775}}{2}$ $x = \frac{-5 \pm \sqrt{25 \cdot 31}}{2}$ $x = \frac{-5 \pm 5\sqrt{31}}{2}$ $x = \frac{-5 + 5\sqrt{31}}{2} \quad \text{or} \quad x = \frac{-5 - 5\sqrt{31}}{2}$

Since $x$ represents a length, we take the positive value: $x = \frac{-5 + 5\sqrt{31}}{2}$

Simplifying further: $x = \frac{5(\sqrt{31} - 1)}{2}$

Thus, the value of $x$ is: $x \approx 6.1$

Would you like further details or have any questions?

Here are five more questions you can ask:

1. How do I solve another quadratic equation?
2. Can you explain the steps of the quadratic formula?
3. What are the properties of right triangles?
4. How is the Pythagorean theorem used in different types of problems?
5. How can I verify the solution to a geometric problem?

Tip: Always double-check your algebraic manipulations when solving equations to avoid small errors that can lead to incorrect solutions.