Using the segment addition theorem, the solutions are:
5. x = 17
6. x = 7
7. x = 14
BC = 27
CD = 61
BD = 88
8. AB = 26
9. LJ = 46
10. x = 3
11. FG = 15
12. QS = 34
13. BC = 26
14. EG = 19
15. QS = 68
How to Apply the Segment Addition Theorem?
The segment addition theorem would be used to solve the problems as shown below:
5. UW = UV + VW [segment addition theorem]
Substitute
6x - 35 = 19 + 4x - 20
6x - 4x = 35 + 19 - 20
2x = 34
x = 17
6. HJ = HI + IJ [segment addition theorem]
Substitute
7x - 27 = 3x - 5 + x - 1
7x - 3x - x = 27 - 5 - 1
3x = 21
x = 7
7. BD = BC + CD [segment addition theorem]
Substitute
7x - 10 = 4x - 29 + 5x - 9
7x - 4x - 5x = 10 - 29 - 9
-2x = -28
x = 14
BC = 4x - 29 = 4(14) - 29 = 27
CD = 5x - 9 = 5(14) - 9 = 61
BD = 7x - 10 = 7(14) - 10 = 88
8. BC = BD
Substitute
2x + 1 = 5x - 26
2x - 5x = -1 - 26
-3x = -27
x = 9
AB = 43 - BC
AB = 43 - 2x + 1 = 43 - 2(9) + 1 = 26
9. 7x - 10 = 9x - 11 - (x + 3)
7x - 10 = 9x - 11 - x - 3
7x - 9x + x = 10 - 11 - 3
-x = -4
x = 4
LJ = 28 + 7x - 10 = 28 + 7(4) - 10
LJ = 46
10. 8x + 11 = 12x - 1
8x - 12x = -11 - 1
-4x = -12
x = 3
11. 11x - 7 = 3x + 9
11x - 3x = 7 + 9
8x = 16
x = 2
FG = 11x - 7 = 11(2) - 7
FG = 15
12. 5x - 3 = 21 - x
5x + x = 21 + 3
6x = 24
x = 4
QS = 2(5x - 3) = 10x - 6 = 10(4) - 6
QS = 34
13. 8x - 20 = 2(3x - 1)
8x - 20 = 6x - 2
8x - 6x = 20 - 2
2x = 18
x = 9
BC = AB = 3x - 1 = 3(9) - 1
BC = 26
14. 5x - 1 = 7x - 13
5x - 7x = 1 - 13
-2x = -12
x = 6
EG = 6x - 4 - 13 = 6(6) - 4 - 13
EG = 19
15. RT - ST = RS
Substitute
8x - 43 - (4x - 1) = 2x - 4
8x - 43 - 4x + 1 = 2x - 4
4x - 42 = 2x - 4
4x - 2x = 42 - 4
2x = 38
x = 19
QS = 2(RS)
QS = 2(2x - 4) = 4x - 8 = 4(19) - 8
QS = 68
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