Question

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.

Explanation:

Step1: Use segment - addition postulate for question 1

Since $LN = LM+MN$, substituting $LM = 22$ and $MN = 15$, we get $LN=22 + 15$.

Step2: Calculate the value of $LN$

$LN=22+15=37$

Step3: Use segment - addition postulate for question 2

Since $LN = LM+MN$, then $MN=LN - LM$. Substituting $LN = 54$ and $LM = 31$, we have $MN=54 - 31$.

Step4: Calculate the value of $MN$

$MN=54 - 31 = 23$

Step5: Use segment - addition postulate for question 3

Since $RT=RS + ST$, and $RT = 36$, $RS=6x + 1$, $ST=x + 7$, we get the equation $6x+1+x + 7=36$.

Step6: Combine like - terms

$(6x+x)+(1 + 7)=36$, which simplifies to $7x+8 = 36$.

Step7: Solve for $x$

Subtract 8 from both sides: $7x=36 - 8=28$. Then divide both sides by 7, so $x = 4$.

Step8: Use segment - addition postulate for question 4

Since $DF=DE + EF$, and $DF = 9x-39$, $DE = 47$, $EF=3x + 10$, we have $9x-39=47+3x + 10$.

Step9: Combine like - terms

$9x-39=3x + 57$. Subtract $3x$ from both sides: $9x-3x-39=3x-3x + 57$, which gives $6x-39 = 57$.

Step10: Solve for $x$

Add 39 to both sides: $6x=57 + 39=96$. Then divide both sides by 6, so $x = 16$. Then $EF=3x + 10=3\times16+10=48 + 10=58$.

Step11: Use segment - addition postulate for question 5

Since $UW=UV + VW$, and $UW = 6x-35$, $UV = 19$, $VW=4x-20$, we have $6x-35=19+4x-20$.

Step12: Combine like - terms

$6x-35=4x - 1$. Subtract $4x$ from both sides: $6x-4x-35=4x-4x - 1$, which gives $2x-35=-1$.

Step13: Solve for $x$

Add 35 to both sides: $2x=-1 + 35=34$. Then divide both sides by 2, so $x = 17$. Then $UW=6x-35=6\times17-35=102-35=67$.

Step14: Use segment - addition postulate for question 6

Since $HJ=HI + IJ$, and $HJ = 7x-27$, $HI=3x-5$, $IJ=x - 1$, we have $7x-27=(3x-5)+(x - 1)$.

Step15: Combine like - terms

$7x-27=3x+x-5 - 1$, which simplifies to $7x-27=4x-6$.

Step16: Solve for $x$

Subtract $4x$ from both sides: $7x-4x-27=4x-4x-6$, which gives $3x-27=-6$. Add 27 to both sides: $3x=-6 + 27=21$. Then divide both sides by 3, so $x = 7$.

Step17: Use segment - addition postulate for question 7

Since $BD=BC + CD$, and $BD = 7x-10$, $BC = 4x-29$, $CD = 5x-9$, we have $7x-10=(4x-29)+(5x-9)$.

Step18: Combine like - terms

$7x-10=4x+5x-29 - 9$, which simplifies to $7x-10=9x-38$.

Step19: Solve for $x$

Subtract $7x$ from both sides: $7x-7x-10=9x-7x-38$, which gives $-10=2x-38$. Add 38 to both sides: $2x=-10 + 38=28$. Then divide both sides by 2, so $x = 14$. Then $BC=4x-29=4\times14-29=56-29=27$, $CD=5x-9=5\times14-9=70 - 9=61$, $BD=7x-10=7\times14-10=98-10=88$.

Step20: Use the congruence and segment - addition postulate for question 8

Since $\overline{BD}\cong\overline{BC}$, then $BD = BC$. So $5x-26=2x + 1$.

Step21: Solve for $x$

Subtract $2x$ from both sides: $5x-2x-26=2x-2x + 1$, which gives $3x-26=1$. Add 26 to both sides: $3x=1 + 26=27$. Then divide both sides by 3, so $x = 9$. Then $BC=2x + 1=2\times9+1=19$. Since $AC = 43$, then $AB=AC - BC=43-19=24$.

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. $AB = 24$