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This problem involves finding the hypotenuse \( x \) of a right triangle using the Pythagorean theorem. The sides given are:

- One leg: \( 5 \)
- Another leg: \( 6 + 1 = 7 \)

### Solution:

For a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \), the Pythagorean theorem states:
\[
a^2 + b^2 = c^2
\]

In this case:
\[
5^2 + 7^2 = x^2
\]

Calculating each term:
\[
5^2 = 25
\]
\[
7^2 = 49
\]

Adding these:
\[
25 + 49 = 74
\]

So,
\[
x^2 = 74
\]

To solve for \( x \), take the square root of both sides:
\[
x = \sqrt{74}
\]

Since \( \sqrt{74} \) is already in its simplest radical form, the answer is:
\[
x = \sqrt{74}
\]

Would you like more details or have any questions?

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Here are five related questions that build on this concept:

1. How would you solve for a missing leg if the hypotenuse and one leg are known?
2. What are some methods to approximate \( \sqrt{74} \) to a decimal value?
3. How can you simplify a square root if the number under the root has factors that are perfect squares?
4. What are some real-life applications of the Pythagorean theorem?
5. How does the Pythagorean theorem extend to three dimensions?

**Tip:** Always double-check the lengths of all sides in a triangle to ensure they meet the conditions needed for applying the Pythagorean theorem.

Solve for x. Leave your answer in simplest radical form. (Triangle with sides 5 and 6, hypotenuse x)

Learn how to solve for the hypotenuse of a right triangle with sides 5 and 6 using the Pythagorean theorem. This problem involves calculating the hypotenuse and expressing the answer in simplest radical form, ideal for students in Grades 8-10.

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