8) In the diagram below of circle O, chord DF Bisects chord BC at E. If BC = 12 and FE is 5 more than DE, then FE is 1) 13 2) 9 3) 6 4) 4

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### Geometry Problem

**Question:**

8) In the diagram below of circle O, chord DF bisects chord BC at E. If \( BC = 12 \) and \( FE \) is \( 5 \) more than \( DE \), then \( FE \) is:
1) \( 13 \)
2) \( 9 \)
3) \( 6 \)
4) \( 4 \)

**Diagram Explanation:**

The diagram depicts a circle with center \( O \). Within the circle:
- Chord \( DF \) bisects chord \( BC \) at point \( E \).
- Chord \( BC \) is labeled with points \( B \) and \( C \) on the circumference of the circle.
- Point \( E \) is the midpoint of chord \( BC \), and thus, \( BE = EC = \frac{BC}{2} = \frac{12}{2} = 6 \).
- The lengths \( DE \) and \( FE \) are part of chord \( DF \), with point \( E \) lying somewhere along the length of \( DF \).

To find \( FE \):
- Note that \( FE \) is \( 5 \) more than \( DE \).
- Let \( DE = x \).
- Therefore, \( FE = x + 5 \).
- Using the properties of the circle and the Pythagorean Theorem in triangles:

Given (from diagram geometry):
- \( BE = 6 \)
- \( EC = 6 \)

Note: \( E \) is the midpoint, so \( BE = EC = 6 \).

**Calculation:**

Since \( DF \) bisects \( BC \) perpendicularly at \( E \) (or using geometric properties), we have:
- Triangle \( DEF \) gives \( DE^2 + BE^2 = FE^2 \).
  
By setting up the equations:
- \( x^2 + 6^2 = (x + 5)^2 \)
- Simplifying this will help find the value of \( FE \).

This setup allows us to determine the correct answer among the provided multiple-choice options.

**Answer:**
2) \( 9 \)
Transcribed Image Text:### Geometry Problem **Question:** 8) In the diagram below of circle O, chord DF bisects chord BC at E. If \( BC = 12 \) and \( FE \) is \( 5 \) more than \( DE \), then \( FE \) is: 1) \( 13 \) 2) \( 9 \) 3) \( 6 \) 4) \( 4 \) **Diagram Explanation:** The diagram depicts a circle with center \( O \). Within the circle: - Chord \( DF \) bisects chord \( BC \) at point \( E \). - Chord \( BC \) is labeled with points \( B \) and \( C \) on the circumference of the circle. - Point \( E \) is the midpoint of chord \( BC \), and thus, \( BE = EC = \frac{BC}{2} = \frac{12}{2} = 6 \). - The lengths \( DE \) and \( FE \) are part of chord \( DF \), with point \( E \) lying somewhere along the length of \( DF \). To find \( FE \): - Note that \( FE \) is \( 5 \) more than \( DE \). - Let \( DE = x \). - Therefore, \( FE = x + 5 \). - Using the properties of the circle and the Pythagorean Theorem in triangles: Given (from diagram geometry): - \( BE = 6 \) - \( EC = 6 \) Note: \( E \) is the midpoint, so \( BE = EC = 6 \). **Calculation:** Since \( DF \) bisects \( BC \) perpendicularly at \( E \) (or using geometric properties), we have: - Triangle \( DEF \) gives \( DE^2 + BE^2 = FE^2 \). By setting up the equations: - \( x^2 + 6^2 = (x + 5)^2 \) - Simplifying this will help find the value of \( FE \). This setup allows us to determine the correct answer among the provided multiple-choice options. **Answer:** 2) \( 9 \)
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