Find the value of x in each 45°-45°-90° right triangle using the special triangle properties.

Let's solve each problem on the Special Right Triangle - 45°-45°-90° Worksheet step by step.

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🔷 Key Concept: 45°-45°-90° Triangle

In a 45°-45°-90° triangle:
- The two legs are equal.
- The hypotenuse is $ \text{leg} \times \sqrt{2} $
- So:
- If leg = $ x $, then hypotenuse = $ x\sqrt{2} $
- If hypotenuse = $ h $, then leg = $ \frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2} $

We’ll use this to solve for $ x $ in each triangle.

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🔹 Problem 1

```
x
/|
/ |
/__|
8
```

- This is a 45°-45°-90° triangle with one leg = 8.
- Since both legs are equal, the other leg is also 8.
- Hypotenuse $ x = 8\sqrt{2} $

✔ Answer: $ x = 8\sqrt{2} $

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🔹 Problem 2

```
x
/|
/ |
/__|
16
```

- Right angle at bottom right, 45° at top left → this is a 45°-45°-90° triangle.
- One leg is 16, so the other leg (x) is also 16.
- But wait — the side labeled "x" is the hypotenuse, since it’s opposite the right angle.

Wait! Let's clarify:

- The side labeled 16 is one leg.
- The side labeled x is the hypotenuse.

So:
- Leg = 16
- Hypotenuse = $ 16\sqrt{2} $

✔ Answer: $ x = 16\sqrt{2} $

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🔹 Problem 3

```
12
/|
/ |
/__|
x
```

- Right angle at bottom right, 45° at top → this is a 45°-45°-90° triangle.
- The side of length 12 is the hypotenuse.
- We need to find a leg: $ x = \frac{12}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2} $

✔ Answer: $ x = 6\sqrt{2} $

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🔹 Problem 4

```
24
/|
/ |
/__|
x x
```

- Right triangle with two legs labeled $ x $, and hypotenuse = 24.
- So: $ x\sqrt{2} = 24 $
- $ x = \frac{24}{\sqrt{2}} = \frac{24\sqrt{2}}{2} = 12\sqrt{2} $

✔ Answer: $ x = 12\sqrt{2} $

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🔹 Problem 5

```
x
/|
/ |
/__|
3√2
```

- Right angle at bottom right, 45° at top → 45°-45°-90° triangle.
- One leg = $ 3\sqrt{2} $
- Since both legs are equal, $ x = 3\sqrt{2} $
- Wait — is $ x $ the leg or the hypotenuse?

Looking at the diagram:
- The side labeled $ x $ is one leg.
- The other leg is $ 3\sqrt{2} $
- Since both legs are equal, $ x = 3\sqrt{2} $

✔ Answer: $ x = 3\sqrt{2} $

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🔹 Problem 6

```
x
/|
/ |
/__|
18
```

- Right angle at bottom right, 45° at top → 45°-45°-90° triangle.
- The side labeled 18 is one leg.
- The side labeled $ x $ is the hypotenuse.
- So: $ x = 18\sqrt{2} $

✔ Answer: $ x = 18\sqrt{2} $

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🔹 Problem 7

```
22
/|
/ |
/__|
x
```

- Right triangle with 45° angle at top right.
- One leg is $ x $, hypotenuse is 22.
- This is a 45°-45°-90° triangle → both legs are equal.
- So: $ x\sqrt{2} = 22 $
- $ x = \frac{22}{\sqrt{2}} = \frac{22\sqrt{2}}{2} = 11\sqrt{2} $

✔ Answer: $ x = 11\sqrt{2} $

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🔹 Problem 8

```
+-----+
| | 5
| |
| |
+-----+
x
```

This is a square with side 5. A diagonal is drawn, labeled $ x $. The diagonal divides the square into two 45°-45°-90° triangles.

- Each leg of the triangle is 5.
- Hypotenuse $ x = 5\sqrt{2} $

✔ Answer: $ x = 5\sqrt{2} $

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✔ Final Answers:

| Problem | Answer |
|--------|--------------------|
| 1 | $ 8\sqrt{2} $ |
| 2 | $ 16\sqrt{2} $ |
| 3 | $ 6\sqrt{2} $ |
| 4 | $ 12\sqrt{2} $ |
| 5 | $ 3\sqrt{2} $ |
| 6 | $ 18\sqrt{2} $ |
| 7 | $ 11\sqrt{2} $ |
| 8 | $ 5\sqrt{2} $ |

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