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ABCD is a kite, so start overline, A, C, end overline
AC
is perpendicular to⊥ start overline, D, B, end overline
DB
and D, E, equals, E, BDE=EB. Calculate the length of start overline, A, C, end overline
AC
, to the nearest tenth of a centimeter.

To calculate the length of \( AC \), we use the Pythagorean theorem. Here’s the step-by-step explanation:

### Given:
1. \( AC \) is perpendicular to \( DB \).
2. \( D \) is 8 cm away from \( B \), so the total \( DB = 8 + 8 = 16 \, \text{cm} \).
3. \( DE = EB = 5 \, \text{cm} \), and \( AC \) bisects \( DB \) at \( E \).

### Solution:
1. **Determine \( AE \) and \( EC \):**
   Since \( AC \) bisects \( DB \), the lengths are split equally. Hence, \( AE = EC \).

2. **Apply the Pythagorean theorem to triangle \( AED \):**
   \[
   AD^2 = AE^2 + DE^2
   \]
   Substitute \( AD = 8 \, \text{cm} \) and \( DE = 5 \, \text{cm} \):
   \[
   8^2 = AE^2 + 5^2
   \]
   \[
   64 = AE^2 + 25
   \]
   \[
   AE^2 = 39
   \]
   \[
   AE = \sqrt{39} \approx 6.2 \, \text{cm}.
   \]

3. Since \( AE = EC \), the total length of \( AC \) is:
   \[
   AC = 2 \times AE = 2 \times 6.2 = 12.4 \, \text{cm}.
   \]

### Final Answer:
\[
AC \approx 12.4 \, \text{cm}.
\]

Let me know if you need further clarification! 

Here are 5 related questions:
1. How is the Pythagorean theorem applied to triangles in kites?
2. Why is \( AC \) split equally in the context of this problem?
3. What are the properties of a kite that help solve this problem?
4. How would the calculation change if \( DE \neq EB \)?
5. Can this problem be generalized to other quadrilaterals?

**Tip:** Always double-check which line segments are perpendicular or bisected when working with kites!

ABCD is a kite, so AC ⊥ DB and DE = EB. Calculate the length of AC, to the nearest tenth of a centimeter.

Solve a kite geometry problem by calculating the length of AC using the Pythagorean theorem. Learn key concepts in geometry and properties of kites, suitable for students in Grades 8-10.

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