Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question
![The problem is to find the value of \( x \) in Figure 5.12, given that \( DE \) is parallel to \( AC \).
**Description of Figure 5.12:**
- The figure shows a triangle \( \triangle ABC \) with a line \( DE \) inside it, parallel to side \( AC \).
- Points \( D \) and \( E \) are on sides \( AB \) and \( BC \), respectively.
- Segment \( AB \) is divided into \( AD = x \) and \( DB = x + 3 \).
- Segment \( BE = 8 \).
- Segment \( EC = 2x \).
**Explanation:**
Since \( DE \) is parallel to \( AC \), by the basic proportionality theorem (or Thales's theorem), the ratios of the segments are equal:
\[
\frac{AD}{DB} = \frac{BE}{EC}
\]
Substitute the given segment lengths:
\[
\frac{x}{x+3} = \frac{8}{2x}
\]
Cross-multiply to solve for \( x \):
\[
x \cdot 2x = 8(x + 3)
\]
\[
2x^2 = 8x + 24
\]
Simplify the equation:
\[
2x^2 - 8x - 24 = 0
\]
Divide the whole equation by 2:
\[
x^2 - 4x - 12 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -4 \), and \( c = -12 \).
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}
\]
\[
x = \frac{4 \pm \sqrt{16 + 48}}{2}
\]
\[
x = \frac{4 \pm \sqrt{64}}{2}
\]
\[
x = \frac{4 \pm 8}{2}
\]
The possible solutions are:
\[
x = \frac{12}{2} \quad \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98b6e310-08ba-4e1d-a9bc-704b45d2ce6c%2Fe3ee8967-7388-4b50-b28b-2d948004d27a%2Fu14xz6r_processed.png&w=3840&q=75)
Transcribed Image Text:The problem is to find the value of \( x \) in Figure 5.12, given that \( DE \) is parallel to \( AC \).
**Description of Figure 5.12:**
- The figure shows a triangle \( \triangle ABC \) with a line \( DE \) inside it, parallel to side \( AC \).
- Points \( D \) and \( E \) are on sides \( AB \) and \( BC \), respectively.
- Segment \( AB \) is divided into \( AD = x \) and \( DB = x + 3 \).
- Segment \( BE = 8 \).
- Segment \( EC = 2x \).
**Explanation:**
Since \( DE \) is parallel to \( AC \), by the basic proportionality theorem (or Thales's theorem), the ratios of the segments are equal:
\[
\frac{AD}{DB} = \frac{BE}{EC}
\]
Substitute the given segment lengths:
\[
\frac{x}{x+3} = \frac{8}{2x}
\]
Cross-multiply to solve for \( x \):
\[
x \cdot 2x = 8(x + 3)
\]
\[
2x^2 = 8x + 24
\]
Simplify the equation:
\[
2x^2 - 8x - 24 = 0
\]
Divide the whole equation by 2:
\[
x^2 - 4x - 12 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -4 \), and \( c = -12 \).
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}
\]
\[
x = \frac{4 \pm \sqrt{16 + 48}}{2}
\]
\[
x = \frac{4 \pm \sqrt{64}}{2}
\]
\[
x = \frac{4 \pm 8}{2}
\]
The possible solutions are:
\[
x = \frac{12}{2} \quad \
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