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Find the value of x in Figure 5.12 if DE is parallel to AC. x + 3 A D X B 8 E Figure 5.12 2x

Find the value of x in Figure 5.12 if DE is parallel to AC. x + 3 A D X B 8 E Figure 5.12 2x

Elementary Geometry For College Students, 7e
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
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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 \
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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