Geometry worksheet with answer key covering distance formula, number line, and perimeter problems.

It looks like you've uploaded a worksheet titled "Geometry – Distance Formula" that includes problems related to the distance formula, Pythagorean Theorem, and finding perimeters of triangles. I’ll walk through each section, solve the problems, and explain the solutions step by step.

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Section 1: Use the number line to find each measure.

These are basic distance problems on a number line.

#### 1.) $ |10 - 4| = ? $
- Distance between 10 and 4:
$$
|10 - 4| = |6| = 6
$$
✔ Answer: 6

#### 2.) $ |7 - (-1)| = ? $
- Distance between 7 and -1:
$$
|7 - (-1)| = |7 + 1| = |8| = 8
$$
✔ Answer: 8

#### 3.) $ |6 - 4| = ? $
- Distance between 6 and 4:
$$
|6 - 4| = |2| = 2
$$
✔ Answer: 2

#### 4.) $ |6 - (-8)| = ? $
- Distance between 6 and -8:
$$
|6 - (-8)| = |6 + 8| = |14| = 14
$$
✔ Answer: 14

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Section 2: Use the number line to find each measure. (continued)

#### 5.) $ |x - 2| = 9 $. What is $ x $?
We solve:
$$
|x - 2| = 9 \Rightarrow x - 2 = 9 \quad \text{or} \quad x - 2 = -9
$$
$$
x = 11 \quad \text{or} \quad x = -7
$$
✔ Answer: $ x = 11 $ or $ x = -7 $

#### 6.) $ |x - 7| = 12 $. What is $ x $?
$$
|x - 7| = 12 \Rightarrow x - 7 = 12 \quad \text{or} \quad x - 7 = -12
$$
$$
x = 19 \quad \text{or} \quad x = -5
$$
✔ Answer: $ x = 19 $ or $ x = -5 $

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Section 3: Find the length of $ RT $ for different cases. Show work.

#### 11.) Use the Pythagorean Theorem.
Given:
- Right triangle with legs $ a = 6 $, $ b = 8 $
- Find hypotenuse $ c = RT $

Use:
$$
a^2 + b^2 = c^2
\Rightarrow 6^2 + 8^2 = c^2
\Rightarrow 36 + 64 = c^2
\Rightarrow 100 = c^2
\Rightarrow c = \sqrt{100} = 10
$$
✔ Answer: $ RT = 10 $

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#### 12.) Use the Pythagorean Theorem.
Given:
- Legs: $ a = 5 $, $ b = 12 $
- Find hypotenuse $ c = RT $

$$
5^2 + 12^2 = c^2
\Rightarrow 25 + 144 = c^2
\Rightarrow 169 = c^2
\Rightarrow c = \sqrt{169} = 13
$$
✔ Answer: $ RT = 13 $

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#### 13.) Use the Pythagorean Theorem.
Given:
- Legs: $ a = 10 $, $ b = 24 $
- Find hypotenuse $ c = RT $

$$
10^2 + 24^2 = c^2
\Rightarrow 100 + 576 = c^2
\Rightarrow 676 = c^2
\Rightarrow c = \sqrt{676} = 26
$$
✔ Answer: $ RT = 26 $

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#### 14.) Use the distance formula.
Points: $ R(1, 2) $, $ T(4, 6) $

Distance formula:
$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
= \sqrt{(4 - 1)^2 + (6 - 2)^2}
= \sqrt{3^2 + 4^2}
= \sqrt{9 + 16} = \sqrt{25} = 5
$$
✔ Answer: $ RT = 5 $

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#### 15.) Use the distance formula.
Points: $ R(-2, 3) $, $ T(4, -1) $

$$
d = \sqrt{(4 - (-2))^2 + (-1 - 3)^2}
= \sqrt{(6)^2 + (-4)^2}
= \sqrt{36 + 16} = \sqrt{52}
= \sqrt{4 \cdot 13} = 2\sqrt{13}
$$
✔ Answer: $ RT = 2\sqrt{13} $

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Section 4: Use the diagram on (15).

The diagram shows a right triangle on a coordinate plane with vertices:
- $ A(-2, 3) $
- $ B(4, -1) $
- $ C(4, 3) $

We need to find the perimeter of triangle $ \triangle ABC $.

Let’s find lengths of all three sides:

#### Side $ AB $: from $ A(-2,3) $ to $ B(4,-1) $
Already computed in #15:
$$
AB = \sqrt{(4 + 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
$$

#### Side $ BC $: from $ B(4,-1) $ to $ C(4,3) $
Same x-coordinate → vertical segment:
$$
BC = |3 - (-1)| = |4| = 4
$$

#### Side $ AC $: from $ A(-2,3) $ to $ C(4,3) $
Same y-coordinate → horizontal segment:
$$
AC = |4 - (-2)| = |6| = 6
$$

Now compute perimeter:
$$
P = AB + BC + AC = 2\sqrt{13} + 4 + 6 = 10 + 2\sqrt{13}
$$

✔ Perimeter of $ \triangle ABC = 10 + 2\sqrt{13} $

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Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 6 |
| 2 | 8 |
| 3 | 2 |
| 4 | 14 |
| 5 | $ x = 11 $ or $ x = -7 $ |
| 6 | $ x = 19 $ or $ x = -5 $ |
| 11 | $ RT = 10 $ |
| 12 | $ RT = 13 $ |
| 13 | $ RT = 26 $ |
| 14 | $ RT = 5 $ |
| 15 | $ RT = 2\sqrt{13} $ |
| Perimeter of $ \triangle ABC $ | $ 10 + 2\sqrt{13} $ |

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✔ Final Boxed Answer for Perimeter:

$$
\boxed{10 + 2\sqrt{13}}
$$

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