https://file.aiquickdraw.com/mathbot/file/1735069320040pch98k6r.jpg

To solve for **Segment KN** and **Segment KM**, let's analyze the triangle and use the geometric relationships.

### Observations:
- Triangle \( \triangle KJM \) is a right triangle with:
  - Hypotenuse \( KM = 17 \),
  - One leg \( JM = 9 + 9 = 18 \).
- \( \triangle KJN \) is also a right triangle with:
  - \( KN \) as the hypotenuse,
  - \( JN = 9 \),
  - Height \( h \), which is perpendicular to \( JM \).

We'll solve step-by-step:

---

### Step 1: Find Segment \( KN \) using the Pythagorean theorem:
For \( \triangle KJN \):
\[
KN^2 = KJ^2 + JN^2
\]
Here:
\[
KJ = 10, \quad JN = 9
\]
\[
KN^2 = 10^2 + 9^2 = 100 + 81 = 181
\]
\[
KN = \sqrt{181}
\]

So, **Segment \( KN = \sqrt{181} \) or approximately 13.45**.

---

### Step 2: Verify \( KM \) (already given as 17):
For the larger triangle \( \triangle KJM \), use the Pythagorean theorem to verify \( KM \):
\[
KM^2 = KJ^2 + JM^2
\]
Here:
\[
KJ = 10, \quad JM = 18
\]
\[
KM^2 = 10^2 + 18^2 = 100 + 324 = 424
\]
\[
KM = \sqrt{424}
\]

This matches \( 17 \), as given.

---

### Final Answers:
- **Segment KN**: \( \sqrt{181} \) or approximately **13.45**.
- **Segment KM**: Already given as **17**.

---

Would you like a detailed explanation of the method or further clarifications? Let me know!

---

### Related Questions:
1. How can we calculate the height \( h \) in this diagram?
2. What is the area of \( \triangle KJM \)?
3. Can we verify these calculations using trigonometric ratios?
4. How do segment properties in right triangles help solve real-world problems?
5. How does the Pythagorean theorem relate to other geometric principles?

### Tip:
Always double-check measurements in geometric problems using multiple methods (e.g., Pythagorean theorem, trigonometry, or area formulas) for accuracy!

Calculate Segment KN and Segment KM based on the provided right triangle with given lengths and dimensions.

Explore how to solve for Segment KN and Segment KM in a right triangle using the Pythagorean theorem. The problem includes given side lengths and step-by-step calculations. Ideal for students in Grades 8-10.

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