Geometry homework worksheet focusing on the segment addition postulate, featuring 10 problems with diagrams and algebraic expressions to solve for unknown segment lengths.

Let's solve each problem step-by-step using the Segment Addition Postulate.

---

🔷 Segment Addition Postulate (SAP)

> If point $ B $ lies on segment $ \overline{AC} $, then:
> $$
> AB + BC = AC
> $$

This means that if a point is between two endpoints of a segment, the sum of the lengths of the two smaller segments equals the total length.

We'll apply this for all problems.

---

## ✔ Problem 1: Find $ LM $

Given:
- $ KL = 11 $
- $ KM = 17 $
- Point $ L $ is between $ K $ and $ M $

a) State SAP with variables:

$$
KL + LM = KM
$$

b) Substitute:

$$
11 + LM = 17
$$

c) Solve:

$$
LM = 17 - 11 = 6
$$

✔ Answer: $ \boxed{6} $

---

## ✔ Problem 2: Find $ CD $

Given:
- $ BC = 10 $
- $ BD = 14 $
- Point $ C $ is between $ B $ and $ D $

a) SAP:

$$
BC + CD = BD
$$

b) Substitute:

$$
10 + CD = 14
$$

c) Solve:

$$
CD = 14 - 10 = 4
$$

✔ Answer: $ \boxed{4} $

---

## ✔ Problem 3: Solve for $ x $

Given:
- $ FE = -12 + 2x $
- $ ED = x - 8 $
- $ FD = 10 $
- $ E $ is between $ F $ and $ D $

a) SAP:

$$
FE + ED = FD
$$

b) Substitute:

$$
(-12 + 2x) + (x - 8) = 10
$$

c) Solve:

$$
-12 + 2x + x - 8 = 10 \\
3x - 20 = 10 \\
3x = 30 \\
x = 10
$$

✔ Answer: $ \boxed{x = 10} $

---

## ✔ Problem 4: Solve for $ x $

Given:
- $ PQ = 3 $
- $ QR = 2x + 22 $
- $ PR = x + 17 $
- $ Q $ is between $ P $ and $ R $

a) SAP:

$$
PQ + QR = PR
$$

b) Substitute:

$$
3 + (2x + 22) = x + 17
$$

c) Solve:

$$
3 + 2x + 22 = x + 17 \\
2x + 25 = x + 17 \\
2x - x = 17 - 25 \\
x = -8
$$

✔ Answer: $ \boxed{x = -8} $

---

## ✔ Problem 5: Find $ DF $

Given:
- $ DE = x + 7 $
- $ EF = 7 $
- $ DF = 4x + 2 $
- $ E $ is between $ D $ and $ F $

a) SAP:

$$
DE + EF = DF
$$

b) Substitute:

$$
(x + 7) + 7 = 4x + 2
$$

c) Solve:

$$
x + 14 = 4x + 2 \\
14 - 2 = 4x - x \\
12 = 3x \\
x = 4
$$

Now plug into $ DF = 4x + 2 $:
$$
DF = 4(4) + 2 = 16 + 2 = 18
$$

✔ Answer: $ \boxed{18} $

---

## ✔ Problem 6: Find $ CD $

Given:
- $ BC = 2 $
- $ CD = 2x - 6 $
- $ BD = x + 2 $
- $ C $ is between $ B $ and $ D $

a) SAP:

$$
BC + CD = BD
$$

b) Substitute:

$$
2 + (2x - 6) = x + 2
$$

c) Solve:

$$
2 + 2x - 6 = x + 2 \\
2x - 4 = x + 2 \\
2x - x = 2 + 4 \\
x = 6
$$

Now find $ CD = 2x - 6 $:
$$
CD = 2(6) - 6 = 12 - 6 = 6
$$

✔ Answer: $ \boxed{6} $

---

## ✔ Problem 7: Solve for $ x $

Given:
- $ PQ = 2x - 6 $
- $ QR = 1 $
- $ RS = x - 4 $
- $ PS = 18 $
- Points in order: $ P-Q-R-S $

So $ PQ + QR + RS = PS $

a) SAP:

$$
PQ + QR + RS = PS
$$

b) Substitute:

$$
(2x - 6) + 1 + (x - 4) = 18
$$

c) Solve:

$$
2x - 6 + 1 + x - 4 = 18 \\
3x - 9 = 18 \\
3x = 27 \\
x = 9
$$

✔ Answer: $ \boxed{x = 9} $

---

## ✔ Problem 8: Solve for $ x $

Given:
- $ DE = 2x + 17 $
- $ EF = 8 $
- $ FG = 2 $
- $ DG = x + 20 $
- Points: $ D-E-F-G $

So $ DE + EF + FG = DG $

a) SAP:

$$
DE + EF + FG = DG
$$

b) Substitute:

$$
(2x + 17) + 8 + 2 = x + 20
$$

c) Solve:

$$
2x + 27 = x + 20 \\
2x - x = 20 - 27 \\
x = -7
$$

✔ Answer: $ \boxed{x = -7} $

---

## ✔ Problem 9: Find $ x $ given:

Points A, B, C, D are collinear in that order.

Given:
- $ AC = 17 $
- $ BD = 2x - 6 $
- $ AD = x + 16 $
- $ BC = 6 $

We need to use the relationships.

Since points are in order: $ A-B-C-D $

So:
- $ AB + BC = AC $
- $ BC + CD = BD $
- $ AB + BC + CD = AD $

Let’s use known values.

From $ AC = 17 $ and $ BC = 6 $, we can find $ AB $:

Step 1: Use $ AB + BC = AC $

$$
AB + 6 = 17 \Rightarrow AB = 11
$$

Now, from $ AD = AB + BC + CD = x + 16 $

But we also know $ AD = AB + BC + CD = 11 + 6 + CD = 17 + CD $

So:
$$
AD = 17 + CD = x + 16 \quad \text{(1)}
$$

Also, $ BD = BC + CD = 6 + CD = 2x - 6 \quad \text{(2)} $

Let’s solve equation (2):
$$
6 + CD = 2x - 6 \\
CD = 2x - 12
$$

Now plug into equation (1):
$$
17 + CD = x + 16 \\
17 + (2x - 12) = x + 16 \\
17 + 2x - 12 = x + 16 \\
2x + 5 = x + 16 \\
2x - x = 16 - 5 \\
x = 11
$$

✔ Answer: $ \boxed{x = 11} $

---

## ✔ Problem 10: Find $ x $

Given:
- $ AD = 20 $
- $ BC = 2 $
- $ AB = x + 12 $
- $ CD = x + 12 $
- Points: $ A-B-C-D $

So:
$$
AB + BC + CD = AD
$$

a) SAP:

$$
AB + BC + CD = AD
$$

b) Substitute:

$$
(x + 12) + 2 + (x + 12) = 20
$$

c) Solve:

$$
x + 12 + 2 + x + 12 = 20 \\
2x + 26 = 20 \\
2x = -6 \\
x = -3
$$

✔ Answer: $ \boxed{x = -3} $

---

## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ \boxed{6} $ |
| 2 | $ \boxed{4} $ |
| 3 | $ \boxed{x = 10} $ |
| 4 | $ \boxed{x = -8} $ |
| 5 | $ \boxed{18} $ |
| 6 | $ \boxed{6} $ |
| 7 | $ \boxed{x = 9} $ |
| 8 | $ \boxed{x = -7} $ |
| 9 | $ \boxed{x = 11} $ |
| 10 | $ \boxed{x = -3} $ |

Let me know if you'd like these explained visually or with diagrams!