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QUESTION IMAGE

** this is a 2 - page document! ** use the diagram below to answer ques…

Question

this is a 2 - page document!
use the diagram below to answer questions 1 and 2.

  1. if (lm = 22) and (mn = 15), find (ln).
  2. if (ln = 54) and (lm = 31), find (mn).
  3. if (rt = 36), find the value of (x).
  4. if (df = 9x-39), find (ef).
  5. if (uw = 6x - 35), find (uw).
  6. if (hj = 7x - 27), find the value of (x).
  7. if (bd = 7x - 10), (bc = 4x - 29), and (cd = 5x - 9), find each value.
  8. if (overline{bd}congoverline{bc}), (bd = 5x - 26), (bc = 2x + 1), and (ac = 43), find (ab).

© gina wilson (all things algebra®, llc), 2014 - 2017

Explanation:

Step1: Use segment - addition postulate

For a line segment with points \(L\), \(M\), and \(N\) in that order, \(LN=LM + MN\).

  1. Given \(LM = 22\) and \(MN = 15\), then \(LN=22 + 15=37\).
  2. Given \(LN = 54\) and \(LM = 31\), using \(LN=LM + MN\), we can solve for \(MN\) as \(MN=LN - LM\). So \(MN=54 - 31 = 23\).
  3. Given \(RT=36\), and \(RT=RS+ST\) where \(RS = 6x + 1\) and \(ST=x + 7\). Then \(6x+1+x + 7=36\). Combine like - terms: \(7x+8 = 36\). Subtract 8 from both sides: \(7x=36 - 8=28\). Divide both sides by 7: \(x = 4\).
  4. Given \(DF=9x - 39\), and \(DF=DE + EF\) where \(DE = 47\) and \(EF=3x + 10\). Then \(9x-39=47+3x + 10\). Combine like - terms: \(9x-3x=47 + 10+39\). \(6x=96\), so \(x = 16\). Then \(EF=3x + 10=3\times16+10=48 + 10=58\).
  5. Given \(UW=6x - 35\), and \(UW=UV+VW\) where \(UV = 19\) and \(VW=4x - 20\). Then \(6x-35=19+4x - 20\). Combine like - terms: \(6x-4x=19 - 20 + 35\). \(2x=34\), so \(x = 17\). Then \(UW=6x - 35=6\times17-35=102 - 35=67\).
  6. Given \(HJ=7x - 27\), and \(HJ=HI+IJ\) where \(HI = 3x - 5\) and \(IJ=x - 1\). Then \(7x-27=3x - 5+x - 1\). Combine like - terms: \(7x-(3x + x)=-5 - 1+27\). \(7x - 4x=21\), \(3x=21\), so \(x = 7\).
  7. Given \(BD=7x - 10\), \(BC=4x - 29\), and \(CD=5x - 9\), and \(BD=BC + CD\). Then \(7x-10=(4x - 29)+(5x - 9)\). Combine like - terms: \(7x-10=4x+5x-29 - 9\). \(7x-10=9x-38\). Subtract \(7x\) from both sides: \(-10=9x-7x-38\). \(2x=38 - 10=28\), so \(x = 14\). Then \(BC=4x - 29=4\times14-29=56 - 29 = 27\), \(CD=5x - 9=5\times14-9=70 - 9 = 61\), \(BD=7x - 10=7\times14-10=98 - 10=88\).
  8. Given \(\overline{BD}\cong\overline{BC}\), so \(BD = BC\). Then \(5x-26=2x + 1\). Subtract \(2x\) from both sides: \(5x-2x-26=1\). \(3x=27\), so \(x = 9\). Then \(BC=2x + 1=2\times9+1=19\). Given \(AC = 43\), and \(AC=AB + BC\), so \(AB=AC - BC=43 - 19=24\).

Answer:

  1. \(LN = 37\)
  2. \(MN = 23\)
  3. \(x = 4\)
  4. \(EF = 58\)
  5. \(UW = 67\)
  6. \(x = 7\)
  7. \(x = 14\), \(BC = 27\), \(CD = 61\), \(BD = 88\)
  8. \(AB = 24\)