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A large right triangle is going to be a part of a geometric sculpture, as shown below. The hypotenuse will be 
13
13​ feet long. The length of one leg of the triangle is 
2
2​​ feet​ more​ than twice the other leg. Find the length of each leg, in feet, and separate them with a comma.

Given the right triangle, we need to find the lengths of the two legs \(x\) and \(2x + 2\), where the hypotenuse is 13 feet.

We can use the Pythagorean theorem, which states:
\[
\text{(Leg 1)}^2 + \text{(Leg 2)}^2 = \text{Hypotenuse}^2
\]

In this case:
\[
x^2 + (2x + 2)^2 = 13^2
\]

Let's solve this equation step by step.

1. Expand the squared term:
\[
(2x + 2)^2 = (2x + 2)(2x + 2) = 4x^2 + 8x + 4
\]

2. Substitute into the equation:
\[
x^2 + 4x^2 + 8x + 4 = 169
\]

3. Combine like terms:
\[
5x^2 + 8x + 4 = 169
\]

4. Move 169 to the left side of the equation to set it to zero:
\[
5x^2 + 8x + 4 - 169 = 0
\]

5. Simplify further:
\[
5x^2 + 8x - 165 = 0
\]

Now we solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \(a = 5\), \(b = 8\), and \(c = -165\):
\[
x = \frac{-8 \pm \sqrt{8^2 - 4(5)(-165)}}{2(5)}
\]
\[
x = \frac{-8 \pm \sqrt{64 + 3300}}{10}
\]
\[
x = \frac{-8 \pm \sqrt{3364}}{10}
\]
\[
x = \frac{-8 \pm 58}{10}
\]

This gives us two possible values for \(x\):
1. \(x = \frac{50}{10} = 5\)
2. \(x = \frac{-66}{10} = -6.6\) (not physically possible since length cannot be negative)

Thus, \(x = 5\) feet.

Substituting \(x = 5\) back into \(2x + 2\):
\[
2(5) + 2 = 10 + 2 = 12 \text{ feet}
\]

So, the lengths of the legs are 5 feet and 12 feet.

Do you want details or have any questions?

Here are some related questions:
1. How would the problem change if the hypotenuse were a different length?
2. What if the relationship between the legs was \(3x + 3\) instead of \(2x + 2\)?
3. How can we verify the solution using a different method?
4. What is the area of the triangle with the given dimensions?
5. What are the angles in this triangle if the sides are as calculated?

**Tip:** When solving quadratic equations, always check whether both solutions are valid in the context of the problem.

A large right triangle is going to be a part of a geometric sculpture. The hypotenuse will be 13 feet long. The length of one leg of the triangle is 2 feet more than twice the other leg. Find the length of each leg, in feet, and separate them with a comma.

Learn how to find the lengths of the legs of a right triangle when the hypotenuse is 13 feet and one leg is 2 feet more than twice the other. This problem involves using the Pythagorean theorem and solving a quadratic equation. Suitable for students in Grades 8-10.

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