45-45-90 Triangles worksheet with 10 problems to find missing side lengths.

Let's solve each of these 45-45-90 triangle problems step by step.

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🔷 Key Concept: 45-45-90 Triangle

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

We’ll use this to find missing side lengths.

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1)

Triangle with:
- One leg: $ 2\sqrt{2} $
- Angle: 45°
- Find $ y $ and $ z $

Since it's a 45-45-90 triangle:
- Both legs are equal → $ y = 2\sqrt{2} $
- Hypotenuse $ z = (2\sqrt{2}) \cdot \sqrt{2} = 2 \cdot 2 = 4 $

✔ Answer:
$ y = 2\sqrt{2}, \quad z = 4 $

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2)

Given:
- One leg: 6
- Angle: 45°
- Find $ x $ (hypotenuse)

Hypotenuse = $ 6 \cdot \sqrt{2} = 6\sqrt{2} $

✔ Answer:
$ x = 6\sqrt{2} $

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3)

Given:
- One leg: $ 14\sqrt{2} $
- Angle: 45°
- Find $ a $ and $ b $

Legs are equal → $ a = 14\sqrt{2} $

Hypotenuse $ b = 14\sqrt{2} \cdot \sqrt{2} = 14 \cdot 2 = 28 $

✔ Answer:
$ a = 14\sqrt{2}, \quad b = 28 $

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4)

Given:
- Hypotenuse: $ 10\sqrt{2} $
- Angle: 45°
- Find $ x $ and $ y $

Each leg = $ \frac{10\sqrt{2}}{\sqrt{2}} = 10 $

So $ x = y = 10 $

✔ Answer:
$ x = 10, \quad y = 10 $

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5)

Given:
- Leg: $ 10\sqrt{3} $
- Right angle at bottom
- Find $ a $ and $ b $

Wait — this is a 45-45-90 triangle? But one leg is $ 10\sqrt{3} $. That seems inconsistent unless we double-check.

But wait: in a 45-45-90 triangle, both legs are equal. If one leg is $ 10\sqrt{3} $, then the other leg is also $ 10\sqrt{3} $, and hypotenuse is:

$$
a = 10\sqrt{3} \cdot \sqrt{2} = 10\sqrt{6}
$$

But let’s label carefully:
- $ a $ is the hypotenuse
- $ b $ is the other leg

So:
- $ b = 10\sqrt{3} $ (since legs are equal)
- $ a = 10\sqrt{3} \cdot \sqrt{2} = 10\sqrt{6} $

✔ Answer:
$ a = 10\sqrt{6}, \quad b = 10\sqrt{3} $

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6)

Given:
- One leg: $ \frac{9\sqrt{10}}{2} $
- Angle: 45°
- Find $ x $ (hypotenuse)

Hypotenuse = $ \frac{9\sqrt{10}}{2} \cdot \sqrt{2} = \frac{9\sqrt{20}}{2} $

Simplify $ \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} $

So:
$$
x = \frac{9 \cdot 2\sqrt{5}}{2} = 9\sqrt{5}
$$

✔ Answer:
$ x = 9\sqrt{5} $

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7)

Given:
- One leg: $ 3 $
- Other leg: $ m $
- Hypotenuse: $ n $
- Angle: 45°

Since it's 45-45-90:
- Legs are equal → $ m = 3 $
- Hypotenuse $ n = 3\sqrt{2} $

✔ Answer:
$ m = 3, \quad n = 3\sqrt{2} $

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8)

Given:
- One leg: $ 9\sqrt{2} $
- Hypotenuse: $ x $
- Find $ x $

Hypotenuse = $ 9\sqrt{2} \cdot \sqrt{2} = 9 \cdot 2 = 18 $

✔ Answer:
$ x = 18 $

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9)

Given:
- One leg: $ 9\sqrt{2} $
- Angle: 45°
- Find $ x $ (hypotenuse)

Same as #8:
- Hypotenuse $ x = 9\sqrt{2} \cdot \sqrt{2} = 9 \cdot 2 = 18 $

✔ Answer:
$ x = 18 $

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10)

Given:
- One leg: $ 9\sqrt{2} $
- Angle: 45°
- Find $ x $ (hypotenuse)

Again, same pattern:
- $ x = 9\sqrt{2} \cdot \sqrt{2} = 18 $

✔ Answer:
$ x = 18 $

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1) | $ y = 2\sqrt{2}, \quad z = 4 $ |
| 2) | $ x = 6\sqrt{2} $ |
| 3) | $ a = 14\sqrt{2}, \quad b = 28 $ |
| 4) | $ x = 10, \quad y = 10 $ |
| 5) | $ a = 10\sqrt{6}, \quad b = 10\sqrt{3} $ |
| 6) | $ x = 9\sqrt{5} $ |
| 7) | $ m = 3, \quad n = 3\sqrt{2} $ |
| 8) | $ x = 18 $ |
| 9) | $ x = 18 $ |
| 10) | $ x = 18 $ |

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Let me know if you'd like these explained visually or need help drawing them!