Problem 5
Point B is the midpoint of AC. This means B cuts AC in half to get the smaller pieces AB and BC which are equal to one another.
AB = BC
AC = AB+BC
AC = AB+AB
AC = 2x+2x
AC = 4x
Because AC = 20 as well, this means,
4x = 20
x = 20/4
x = 5
So, AB = 2x = 2*5 = 10 and BC is this length as well.
Answers:
- x = 5
- AB = 10
- BC = 10
- AC = 20
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Problem 6
Use the segment addition postulate.
RS + ST = RT
x + 3x = 24
4x = 24
x = 24/4
x = 6
RS = x = 6
ST = 3x = 3*6 = 18
Answers:
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Problem 7
We'll use the same idea as problem 5.
DE = EF
3x+10 = 8x-5
10+5 = 8x-3x
15 = 5x
5x = 15
x = 15/5
x = 3
So
DE = 3x+10 = 3*3+10 = 9+10 = 19
EF = 8x-5 = 8*3-5 = 24-5 = 19
DF = DE+EF = 19+19 = 38
Answers:
- x = 3
- DE = 19
- EF = 19
- DF = 38
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Problem 8
This problem is similar to problem 6. We use the segment addition postulate.
GH + HI = GI
3x + 10 = 40
3x = 40-10
3x = 30
x = 30/3
x = 10
GH = 3x = 3*10 = 30
Answers:
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Problem 9
The double tickmarks indicate the segments KL and LM are equal in length. That means point L bisects KM.
KL = LM
KL + LM = KM
x+x = 25
2x = 25
x = 25/2
x = 12.5
Answer: x = 12.5