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The problem provided involves a right triangle where one leg is \( x \) and the other leg is \( x+2 \), and the hypotenuse is given as 10. The question asks which of the given equations can be used to find the lengths of the legs of the triangle.

To solve this, we can apply the **Pythagorean theorem**:

\[
a^2 + b^2 = c^2
\]

Where:
- \( a = x \)
- \( b = x + 2 \)
- \( c = 10 \)

Substituting these into the equation:

\[
x^2 + (x+2)^2 = 10^2
\]

Now, expanding \( (x+2)^2 \):

\[
x^2 + (x^2 + 4x + 4) = 100
\]

Simplifying:

\[
2x^2 + 4x + 4 = 100
\]

Subtracting 100 from both sides:

\[
2x^2 + 4x - 96 = 0
\]

Dividing the entire equation by 2:

\[
x^2 + 2x - 48 = 0
\]

Now, let's match this result with the given choices:

- **Option A**: \( 0.5 \cdot (x)(x+2) = 24 \) — This does not match.
- **Option B**: \( x(x+2) = 24 \) — This is incorrect.
- **Option C**: \( x^2 + 2x - 24 = 0 \) — Close, but does not match exactly.
- **Option D**: \( x^2 + 2x - 48 = 0 \) — This matches the equation derived from the Pythagorean theorem.
- **Option E**: \( x^2 + (x+2)^2 = 100 \) — This is the equation before simplifying, which is correct.

So, the correct answers are **D** and **E**.

Do you have any further questions or want more details?

Here are five related questions:
1. What is the solution to the quadratic equation \( x^2 + 2x - 48 = 0 \)?
2. How would you apply the quadratic formula to solve this equation?
3. What are the steps to expand \( (x + 2)^2 \)?
4. How does the Pythagorean theorem apply to non-right triangles?
5. Can you solve this problem by factoring instead of using the quadratic formula?

**Tip:** Always simplify the expressions as much as possible before matching them with given options.

Which equation can be used to find the lengths of the legs in a right triangle with legs x, x + 2, and hypotenuse 10?

Learn how to solve a right triangle problem using the Pythagorean theorem. This problem involves finding the lengths of the legs x and x + 2 when the hypotenuse is 10. Suitable for students in Grades 8-10.

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