Question

quiz practice mj maddox
use the diagram below to answer questions 1 and 2.

1. if lm = 22 and mn = 15, find ln. ln = 37
2. if ln = 54 and lm = 31, find mn. mn = 23
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. m is the mid - point of $overline{pq}$. pm = 5x + 8 and pq = 76, find x.
8. m is the mid - point of $overline{pq}$. pm = 9x + 3 and mq = 11x - 17, find x and pq.
9. m is the mid - point of $overline{pq}$. mq = 5x + 8 and pq = 6x + 56, find pm.
10. $overline{de}$ bisects $overline{ab}$ at c. if ac = 3x - 5 and cb = 5x + 13, find ab.

Explanation:

Step1: Use segment - addition postulate for question 1

If \(LN = LM+MN\), given \(LM = 22\) and \(MN = 15\), then \(LN=22 + 15=37\).

Step2: Use segment - addition postulate for question 2

If \(LN=LM + MN\), then \(MN=LN - LM\). Given \(LN = 54\) and \(LM = 31\), so \(MN=54 - 31 = 23\).

Step3: Use segment - addition postulate for question 3

If \(RT=RS+ST\), then \(36=(6x + 1)+(x + 7)\). Combine like - terms: \(36=7x+8\). Subtract 8 from both sides: \(36−8 = 7x\), so \(28 = 7x\). Divide both sides by 7: \(x = 4\).

Step4: Use segment - addition postulate for question 4

If \(DF=DE + EF\), then \(9x-39=47+(3x + 10)\). Combine like - terms: \(9x-39=3x + 57\). Subtract \(3x\) from both sides: \(9x-3x-39=3x-3x + 57\), \(6x-39 = 57\). Add 39 to both sides: \(6x=57 + 39=96\). Divide both sides by 6: \(x = 16\). Then \(EF=3x + 10=3\times16+10=48 + 10 = 58\).

Step5: Use segment - addition postulate for question 5

If \(UW=UV+VW\), then \(6x-35=19+(4x - 20)\). Combine like - terms: \(6x-35=4x - 1\). Subtract \(4x\) from both sides: \(6x-4x-35=4x-4x - 1\), \(2x-35=-1\). Add 35 to both sides: \(2x=-1 + 35 = 34\). Divide both sides by 2: \(x = 17\). Then \(UW=6x-35=6\times17-35=102-35 = 67\).

Step6: Use segment - addition postulate for question 6

If \(HJ=HI+IJ\), then \(7x-27=(3x - 5)+(x - 1)\). Combine like - terms: \(7x-27=4x-6\). Subtract \(4x\) from both sides: \(7x-4x-27=4x-4x-6\), \(3x-27=-6\). Add 27 to both sides: \(3x=-6 + 27 = 21\). Divide both sides by 3: \(x = 7\).

Step7: Use mid - point property for question 7

If \(M\) is the mid - point of \(\overline{PQ}\), then \(PQ = 2PM\). Given \(PQ = 76\) and \(PM = 5x+8\), so \(76 = 2(5x+8)\). Expand the right - hand side: \(76=10x + 16\). Subtract 16 from both sides: \(76-16=10x\), \(60 = 10x\). Divide both sides by 10: \(x = 6\).

Step8: Use mid - point property for question 8

If \(M\) is the mid - point of \(\overline{PQ}\), then \(PM = MQ\). So \(9x + 3=11x-17\). Subtract \(9x\) from both sides: \(9x-9x + 3=11x-9x-17\), \(3 = 2x-17\). Add 17 to both sides: \(3 + 17=2x\), \(20 = 2x\). Divide both sides by 2: \(x = 10\). Then \(PM=9x + 3=9\times10+3=93\), \(PQ = 2PM=186\).

Step9: Use mid - point property for question 9

If \(M\) is the mid - point of \(\overline{PQ}\), then \(PM = MQ\) and \(PQ = 2MQ\). Given \(MQ = 5x+8\) and \(PQ = 6x + 56\), then \(6x + 56=2(5x+8)\). Expand the right - hand side: \(6x + 56=10x+16\). Subtract \(6x\) from both sides: \(6x-6x + 56=10x-6x+16\), \(56 = 4x+16\). Subtract 16 from both sides: \(56-16=4x\), \(40 = 4x\). Divide both sides by 4: \(x = 10\). Then \(PM=MQ=5x + 8=5\times10+8=58\).

Step10: Use the property of a bisector for question 10

If \(\overline{DE}\) bisects \(\overline{AB}\) at \(C\), then \(AC = CB\). So \(3x-5=5x + 13\). Subtract \(3x\) from both sides: \(3x-3x-5=5x-3x + 13\), \(-5=2x + 13\). Subtract 13 from both sides: \(-5-13=2x\), \(-18=2x\). Divide both sides by 2: \(x=-9\). Then \(AC=3x-5=3\times(-9)-5=-27-5=-32\), \(CB = 5x + 13=5\times(-9)+13=-45 + 13=-32\), \(AB=2AC = 64\).

Answer:

1. \(LN = 37\)
2. \(MN = 23\)
3. \(x = 4\)
4. \(EF = 58\)
5. \(UW = 67\)
6. \(x = 7\)
7. \(x = 6\)
8. \(x = 10\), \(PQ = 186\)
9. \(PM = 58\)
10. \(AB = 64\)