https://file.aiquickdraw.com/mathbot/file/17267608571828qqquv8a.jpg

This problem shows a right triangle, where one leg is labeled as 5, the hypotenuse is labeled as 8, and the other leg is labeled as \( x \), which is the value we need to solve for.

We can apply the **Pythagorean theorem**, which is given by:

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

Where:
- \( a \) and \( b \) are the lengths of the legs,
- \( c \) is the length of the hypotenuse.

In this case:
- \( a = 5 \),
- \( b = x \),
- \( c = 8 \).

Using the Pythagorean theorem:

\[
5^2 + x^2 = 8^2
\]

Let's solve for \( x \).

\[
25 + x^2 = 64
\]

Subtract 25 from both sides:

\[
x^2 = 64 - 25
\]

\[
x^2 = 39
\]

Now, take the square root of both sides:

\[
x = \sqrt{39}
\]

Since \( 39 \) is not a perfect square, this is already in its simplest radical form. Therefore, the solution is:

\[
x = \sqrt{39}
\]

Would you like more details or have any other questions? Here are some related questions to consider:

1. What are the properties of right triangles and how do they help solve problems like this?
2. Can the Pythagorean theorem be used for non-right triangles?
3. How can the square root of a number be simplified when it's not a perfect square?
4. What are some real-world applications of the Pythagorean theorem?
5. How would the equation change if the hypotenuse were unknown?

**Tip**: Always check if the radical can be simplified by factoring out perfect squares from the number under the square root.

Solve for x. Leave your answer in simplest radical form. A right triangle has sides labeled 5 and x, and a hypotenuse labeled 8.

Solve for x in a right triangle where one leg is 5, the hypotenuse is 8, and the answer is left in simplest radical form. This solution uses the Pythagorean theorem and involves simplifying square roots, ideal for students in Grades 8-10.

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