Solved Special Triangles (45-45-90 and 30-60-90 ) Worksheet ... - Free Printable

Let's solve each of these special right triangles step by step using the properties of 45-45-90 and 30-60-90 triangles.

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🔷 Key Properties:

#### 🔹 45-45-90 Triangle:
- Legs are equal.
- Hypotenuse = leg × √2
- If leg = x, then hypotenuse = x√2

#### 🔹 30-60-90 Triangle:
- Side opposite 30° = x
- Side opposite 60° = x√3
- Hypotenuse (opposite 90°) = 2x

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Now let’s go through each problem.

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🔹 Problem 1: 45-45-90 triangle

Given: Hypotenuse = $ 14\sqrt{3} $

Wait! This is a 45-45-90, so both legs should be equal. But hypotenuse = leg × √2.

So:
$$
\text{Hypotenuse} = p\sqrt{2} = 14\sqrt{3}
$$
Solve for $ p $:
$$
p = \frac{14\sqrt{3}}{\sqrt{2}} = 14 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 14 \cdot \frac{\sqrt{6}}{2} = 7\sqrt{6}
$$

So both legs $ p = q = 7\sqrt{6} $

✔ Answer:
$ p = 7\sqrt{6},\ q = 7\sqrt{6} $

> ⚠️ Note: The hypotenuse is $ 14\sqrt{3} $, which is unusual because in a 45-45-90 triangle, hypotenuse should be $ x\sqrt{2} $. So unless it's a typo, we proceed as above.

But wait — this might be a mistake in interpretation. Let's double-check.

If the triangle has two 45° angles and one right angle, then yes, it's 45-45-90. So hypotenuse = leg × √2.

So:
$$
\text{leg} = \frac{14\sqrt{3}}{\sqrt{2}} = 14 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 14 \cdot \frac{\sqrt{6}}{2} = 7\sqrt{6}
$$

Yes, correct.

✔ Answer:
$$
p = 7\sqrt{6},\ q = 7\sqrt{6}
$$

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🔹 Problem 2: 30-60-90 triangle

Given: Side opposite 30° = $ 6\sqrt{3} $, find $ z $ (hypotenuse), $ y $ (side opposite 60°)

In a 30-60-90 triangle:
- Side opposite 30° = $ x $
- Side opposite 60° = $ x\sqrt{3} $
- Hypotenuse = $ 2x $

Here, side opposite 30° is $ 6\sqrt{3} $, so:
$$
x = 6\sqrt{3}
$$

Then:
- $ y = x\sqrt{3} = 6\sqrt{3} \cdot \sqrt{3} = 6 \cdot 3 = 18 $
- $ z = 2x = 2 \cdot 6\sqrt{3} = 12\sqrt{3} $

✔ Answer:
$$
z = 12\sqrt{3},\ y = 18
$$

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🔹 Problem 3: 30-60-90 triangle

Given: Side opposite 60° = $ \sqrt{5} $, find $ u $ (opposite 30°), $ v $ (hypotenuse)

Side opposite 60° = $ x\sqrt{3} = \sqrt{5} $

So:
$$
x\sqrt{3} = \sqrt{5} \Rightarrow x = \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{15}}{3}
$$

Then:
- $ u = x = \frac{\sqrt{15}}{3} $
- $ v = 2x = \frac{2\sqrt{15}}{3} $

✔ Answer:
$$
u = \frac{\sqrt{15}}{3},\ v = \frac{2\sqrt{15}}{3}
$$

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🔹 Problem 4: 45-45-90 triangle

Given: Hypotenuse = $ \frac{6\sqrt{2}}{5} $

In 45-45-90:
$$
\text{Hypotenuse} = r\sqrt{2} = \frac{6\sqrt{2}}{5}
\Rightarrow r = \frac{6\sqrt{2}}{5} \div \sqrt{2} = \frac{6}{5}
$$

So $ r = s = \frac{6}{5} $

✔ Answer:
$$
r = \frac{6}{5},\ s = \frac{6}{5}
$$

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🔹 Problem 5: 45-45-90 triangle

Given: Hypotenuse = $ \frac{13\sqrt{2}}{2} $

Then:
$$
a = b = \frac{13\sqrt{2}}{2} \div \sqrt{2} = \frac{13}{2}
$$

✔ Answer:
$$
a = \frac{13}{2},\ b = \frac{13}{2}
$$

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🔹 Problem 6: 30-60-90 triangle

Given: Side opposite 30° = 16, find $ g $ (hypotenuse), $ h $ (side opposite 60°)

In 30-60-90:
- Side opposite 30° = $ x = 16 $
- Hypotenuse $ g = 2x = 32 $
- Side opposite 60° $ h = x\sqrt{3} = 16\sqrt{3} $

✔ Answer:
$$
g = 32,\ h = 16\sqrt{3}
$$

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🔹 Problem 7: 30-60-90 triangle

Given: Hypotenuse = 18

So:
- Hypotenuse = $ 2x = 18 \Rightarrow x = 9 $
- Side opposite 30° = $ x = 9 $
- Side opposite 60° = $ x\sqrt{3} = 9\sqrt{3} $

Labeling:
- $ j $ is opposite 60° → $ j = 9\sqrt{3} $
- $ k $ is opposite 30° → $ k = 9 $

✔ Answer:
$$
j = 9\sqrt{3},\ k = 9
$$

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🔹 Problem 8: 30-60-90 triangle

Given: Side opposite 60° = 32

So:
- Side opposite 60° = $ x\sqrt{3} = 32 \Rightarrow x = \frac{32}{\sqrt{3}} = \frac{32\sqrt{3}}{3} $

Then:
- Side opposite 30° = $ x = \frac{32\sqrt{3}}{3} $
- Hypotenuse = $ 2x = \frac{64\sqrt{3}}{3} $

Now label:
- $ k $ is opposite 30° → $ k = \frac{32\sqrt{3}}{3} $
- $ L $ is opposite 60° → $ L = 32 $

Wait! Wait — look at the triangle:

- Right angle at bottom left.
- Angle at top right = 60°
- Angle at bottom right = 30°
- Side opposite 60° is labeled $ k $? No — let’s check.

From diagram:
- $ k $ is the vertical leg, adjacent to 60° angle.
- $ L $ is the horizontal leg, adjacent to 30°.
- Hypotenuse = 32

Wait — actually, side opposite 60° is the longer leg, which is not the hypotenuse.

So:
- Hypotenuse = 32
- Angle at bottom right = 30°
- So side opposite 30° = $ \frac{1}{2} \times 32 = 16 $
- Side opposite 60° = $ 16\sqrt{3} $

Now label:
- $ k $: vertical leg → opposite 60° → $ k = 16\sqrt{3} $
- $ L $: horizontal leg → opposite 30° → $ L = 16 $

✔ Answer:
$$
k = 16\sqrt{3},\ L = 16
$$

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ p = 7\sqrt{6},\ q = 7\sqrt{6} $ |
| 2 | $ z = 12\sqrt{3},\ y = 18 $ |
| 3 | $ u = \frac{\sqrt{15}}{3},\ v = \frac{2\sqrt{15}}{3} $ |
| 4 | $ r = \frac{6}{5},\ s = \frac{6}{5} $ |
| 5 | $ a = \frac{13}{2},\ b = \frac{13}{2} $ |
| 6 | $ g = 32,\ h = 16\sqrt{3} $ |
| 7 | $ j = 9\sqrt{3},\ k = 9 $ |
| 8 | $ k = 16\sqrt{3},\ L = 16 $ |

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Let me know if you'd like these written neatly or boxed!