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Let's define the variables:

- Let \( x \) be the length of the shorter leg.
- The length of the longer leg is given as \( 2x + 20 \) feet.
- The length of the hypotenuse is \( 2x + 22 \) feet.

We can apply the Pythagorean theorem to solve for \( x \):

\[
x^2 + (2x + 20)^2 = (2x + 22)^2
\]

### Expanding both sides:

1. \( (2x + 20)^2 = 4x^2 + 80x + 400 \)
2. \( (2x + 22)^2 = 4x^2 + 88x + 484 \)

### Substituting these into the equation:

\[
x^2 + 4x^2 + 80x + 400 = 4x^2 + 88x + 484
\]

### Simplifying:

\[
5x^2 + 80x + 400 = 4x^2 + 88x + 484
\]

Move all terms to one side:

\[
5x^2 + 80x + 400 - 4x^2 - 88x - 484 = 0
\]

This simplifies to:

\[
x^2 - 8x - 84 = 0
\]

### Solving this quadratic equation:

Using the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-84)}}{2(1)}
\]

\[
x = \frac{8 \pm \sqrt{64 + 336}}{2}
\]

\[
x = \frac{8 \pm \sqrt{400}}{2}
\]

\[
x = \frac{8 \pm 20}{2}
\]

Thus, the possible solutions are:

\[
x = \frac{8 + 20}{2} = 14 \quad \text{or} \quad x = \frac{8 - 20}{2} = -6
\]

Since the length cannot be negative, \( x = 14 \).

### Final side lengths:

- Shorter leg \( = 14 \) feet.
- Longer leg \( = 2(14) + 20 = 48 \) feet.
- Hypotenuse \( = 2(14) + 22 = 50 \) feet.

Would you like further clarification or a detailed explanation of any step?

Here are 5 related questions to expand on this topic:
1. How does the Pythagorean theorem apply to different types of triangles?
2. What are the possible methods to solve quadratic equations?
3. Can you derive a formula for the hypotenuse in any right triangle?
4. What are some real-life applications of the Pythagorean theorem?
5. How would the equation change if the triangle were isosceles?

**Tip:** Always check the discriminant (\( b^2 - 4ac \)) when solving quadratic equations to see if real solutions exist.

The length of the longer leg of a right triangle is 20 ft more than twice the length of the shorter leg. The length of the hypotenuse is 22 ft more than twice the length of the shorter leg. Find the side lengths of the triangle.

Learn how to calculate the side lengths of a right triangle where the longer leg is 20 ft more than twice the shorter leg and the hypotenuse is 22 ft more. This problem involves using the Pythagorean theorem and solving a quadratic equation, suitable for Grades 8-10.

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