Question

part 1 - use the segment addition postulate to answer each problem.

1. find ac.

a 13 b 5 c

1. find pq.

p 2x - 5 q x r 34

1. find ef.

d 16 e f 27

1. find su.

s 3x + 4 t 15 - x u 4x + 1

1. find hk.

h 14 i 6 j 5 k

1. find wx.

v 2x w 14 - x x x y 25

1. find mo.

l 13 m n 8 o 30

Explanation:

Step1: Recall segment - addition postulate

If a point $B$ lies on a line segment $\overline{AC}$, then $AC = AB+BC$.

Step2: Solve for $AC$

Given $AB = 13$ and $BC = 5$, by the segment - addition postulate, $AC=AB + BC=13 + 5=18$.

Step3: Solve for $EF$

Given $DF = 27$ and $DE = 16$, since $DF=DE + EF$, then $EF=DF - DE=27-16 = 11$.

Step4: Solve for $HK$

Given $HI = 14$, $IJ = 6$, and $JK = 5$, by the segment - addition postulate, $HK=HI+IJ + JK=14 + 6+5=25$.

Step5: Solve for $MO$

Given $LO = 30$, $LM = 13$, and $NO = 8$. Since $LO=LM + MO+NO$, then $MO=LO-(LM + NO)=30-(13 + 8)=30 - 21 = 9$.

Step6: Solve for $PQ$

Given $PR = 34$, $PQ=2x - 5$, and $QR = x$. Since $PR=PQ + QR$, we have $34=(2x - 5)+x$. Combining like - terms gives $34 = 3x-5$. Adding 5 to both sides: $34 + 5=3x$, so $39 = 3x$. Dividing both sides by 3, $x = 13$. Then $PQ=2x - 5=2\times13-5=26 - 5=21$.

Step7: Solve for $SU$

Given $SU = 4x + 1$, $ST = 3x + 4$, and $TU = 15 - x$. Since $SU=ST + TU$, we have $4x + 1=(3x + 4)+(15 - x)$. Simplifying the right - hand side gives $4x + 1=3x + 4+15 - x=2x+19$. Subtracting $2x$ from both sides: $4x-2x + 1=2x-2x + 19$, so $2x+1 = 19$. Subtracting 1 from both sides: $2x=18$, and $x = 9$. Then $SU=4x + 1=4\times9+1=36 + 1=37$.

Step8: Solve for $WX$

Given $VY = 25$, $VW = 2x$, $WX=14 - x$, and $XY = x$. Since $VY=VW + WX+XY$, we have $25=2x+(14 - x)+x$. Simplifying the right - hand side gives $25=2x + 14 - x+x=2x+14$. Subtracting 14 from both sides: $25 - 14=2x$, so $11 = 2x$. Dividing both sides by 2, $x=\frac{11}{2}$. Then $WX=14 - x=14-\frac{11}{2}=\frac{28 - 11}{2}=\frac{17}{2}=8.5$.

Answer:

1. $AC = 18$
2. $EF = 11$
3. $HK = 25$
4. $MO = 9$
5. $PQ = 21$
6. $SU = 37$
7. $WX = 8.5$