Segment Addition Postulate worksheet with examples, exercises, and a proof section.

Let's solve each part of this Segment Addition Postulate worksheet step by step.

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🔹 Segment Addition Postulate Definition:

> If 3 points are collinear (on the same line segment), and B is between A and C, then:
>
> AB + BC = AC

This means the total length of a segment is the sum of its parts when a point lies between two others.

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## ✔ Example 1:
> If DF = 10 and DE = 7, what is the length of EF?

We are given:
- DF = 10
- DE = 7

Since E is between D and F, we apply the Segment Addition Postulate:

$$
DE + EF = DF
$$

Substitute values:

$$
7 + EF = 10
\Rightarrow EF = 10 - 7 = 3
$$

✔ Answer: EF = 3

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## ✔ Example 2:
> If MO = 32, find the value of x.

Given:
- MN = 2x
- NO = 3x + 7
- MO = 32

Since N is between M and O:

$$
MN + NO = MO
\Rightarrow 2x + (3x + 7) = 32
$$

Simplify:

$$
5x + 7 = 32
\Rightarrow 5x = 25
\Rightarrow x = 5
$$

✔ Answer: x = 5

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## ✔ Example 3:
> If RU = 59, find TR.

Given:
- RS = 7
- ST = 3x - 4
- TU = 4x
- RU = 59

So, the whole segment RU is made up of RS + ST + TU:

$$
RS + ST + TU = RU
\Rightarrow 7 + (3x - 4) + 4x = 59
$$

Simplify:

$$
7 + 3x - 4 + 4x = 59
\Rightarrow (3x + 4x) + (7 - 4) = 59
\Rightarrow 7x + 3 = 59
\Rightarrow 7x = 56
\Rightarrow x = 8
$$

Now, we want TR, which is from T to R. That’s TS + SR.

But note: TR = TS + SR = (3x - 4) + 7

Plug in x = 8:

$$
3(8) - 4 + 7 = 24 - 4 + 7 = 27
$$

✔ Answer: TR = 27

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## ✔ Example 4:
> Solve for x. Then find the length of each segment.

We have three segments on a line:
- First: x
- Second: x + 15
- Third: 3x
- Total: 10x + 5

So,

$$
x + (x + 15) + 3x = 10x + 5
$$

Left side:

$$
x + x + 15 + 3x = 5x + 15
$$

Set equal:

$$
5x + 15 = 10x + 5
\Rightarrow 15 - 5 = 10x - 5x
\Rightarrow 10 = 5x
\Rightarrow x = 2
$$

Now find lengths:

- First segment: $ x = 2 $
- Second: $ x + 15 = 2 + 15 = 17 $
- Third: $ 3x = 3(2) = 6 $

Check total: $ 2 + 17 + 6 = 25 $

Total given: $ 10x + 5 = 10(2) + 5 = 25 $ → ✔ Matches!

✔ Answer: x = 2
Segment lengths:
- First: 2
- Second: 17
- Third: 6

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## ✔ Example 5: Complete the proof

Given: JK = MN
Prove: MN + KL = JL

We are given a diagram with points J–K–L on one line, and M–N on another, but since JK = MN, and we’re using segment addition, we assume they are collinear in their own segments.

| Statements | Reasons |
|------------|---------|
| 1. JK = MN | 1. Given |
| 2. JK + KL = JL | 2. Segment Addition Postulate (since K is between J and L) |
| 3. MN + KL = JL | 3. Substitution Property of Equality (replace JK with MN) |

✔ Completed Proof:

| Statements | Reasons |
|-----------|--------|
| 1. JK = MN | 1. Given |
| 2. JK + KL = JL | 2. Segment Addition Postulate |
| 3. MN + KL = JL | 3. Substitution (from statement 1) |

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🟩 Final Answers Summary:

1. EF = 3
2. x = 5
3. TR = 27
4. x = 2, segments: 2, 17, 6
5. Proof completed as above.

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