Question

kuta software - infinite geometry
the segment addition postulate
find the length indicated.

1. h? f

g
10

1. r? t

s
13

1. t? v

u
20
32

1. c 14? e

d
30

1. find kl

i 9 11 l
j k
26

1. find hj

g 2 12 j
h i
7

1. find ec

e 16 b
d c
49
30

1. find ik

i 12 l
j k
49
31
points a, b, and c are collinear. point b is between a and c. find the length indicated.

1. find ac if ab = 16 and bc = 12.
2. find ac if ab = 13 and bc = 9.

Explanation:

Step1: Recall segment - addition postulate

If a point lies on a line segment, the sum of the lengths of the two sub - segments equals the length of the whole segment. Let the whole segment be \(AC\) and points \(A\), \(B\), \(C\) be collinear with \(B\) between \(A\) and \(C\), then \(AC=AB + BC\). For other cases with three collinear points \(X\), \(Y\), \(Z\) where \(Y\) is between \(X\) and \(Z\), \(XZ=XY + YZ\).

Step2: Solve problem 1

Let the unknown length be \(x\). If the whole segment is \(10\) and one part is \(1\), then \(x=10 - 1=9\).

Step3: Solve problem 2

Let the unknown length be \(x\). If the whole segment is \(13\) and one part is \(1\), then \(x=13 - 1 = 12\).

Step4: Solve problem 3

Let the unknown length be \(x\). We know that the whole segment \(TV = 32\) and \(UV=20\). By the segment - addition postulate \(TV=TU + UV\), so \(x=32 - 20=12\).

Step5: Solve problem 4

Let the unknown length be \(x\). The whole segment \(CE = 30\) and \(CD = 14\). By the segment - addition postulate \(CE=CD + DE\), so \(x=30 - 14 = 16\).

Step6: Solve problem 5

Let \(KL=x\). The whole segment \(IL = 26\), \(IJ = 9\), and \(JK = 11\). Since \(IL=IJ + JK+KL\), then \(x=26-(9 + 11)=6\).

Step7: Solve problem 6

Let \(HJ=x\). The whole segment \(GJ\) has \(GH = 2\), \(IJ = 12\) and \(GI = 7\). First, find \(HI=GI - GH=7 - 2 = 5\). Then \(HJ=IJ - HI=12 - 5 = 7\).

Step8: Solve problem 7

Let \(EC=x\). The whole segment \(EB = 49\), \(DB = 30\), and \(DE = 16\). Since \(EB=ED+DB\) and \(EC=EB - BC\), and \(BC=DB - DC\), we know that \(EC=49 - 30=19\).

Step9: Solve problem 8

Let \(IK=x\). The whole segment \(IL = 49\), \(IJ = 31\) and \(JK = 12\). Since \(IL=IJ + JK+KL\), \(x=31 - 12=19\).

Step10: Solve problem 9

Given \(AB = 16\) and \(BC = 12\), by the segment - addition postulate \(AC=AB + BC\), so \(AC=16+12 = 28\).

Step11: Solve problem 10

Given \(AB = 13\) and \(BC = 9\), by the segment - addition postulate \(AC=AB + BC\), so \(AC=13 + 9=22\).

Answer:

1. 9
2. 12
3. 12
4. 16
5. 6
6. 7
7. 19
8. 19
9. 28
10. 22