30 60 90 and 45 45 90 Triangle Worksheet 1 x | StudyX - Free Printable

To solve the problems involving 30-60-90 and 45-45-90 triangles, we need to use the properties of these special right triangles. Let's go through each problem step by step.

Problem 1:

Given:
- A 45-45-90 triangle with one leg \( x \) and hypotenuse \( y \).
- One leg is given as \( 2\sqrt{2} \).

Solution:
In a 45-45-90 triangle:
- The legs are equal.
- The hypotenuse is \( \sqrt{2} \) times the length of each leg.

Since one leg is \( 2\sqrt{2} \):
\[ x = 2\sqrt{2} \]

The hypotenuse \( y \) is:
\[ y = x\sqrt{2} = (2\sqrt{2})\sqrt{2} = 2 \cdot 2 = 4 \]

So, the answers are:
\[ x = 2\sqrt{2}, \quad y = 4 \]

Problem 2:

Given:
- A 30-60-90 triangle with sides \( x \), \( y \), and hypotenuse 20.
- The side opposite the 30° angle is \( x \).
- The side opposite the 60° angle is \( y \).

Solution:
In a 30-60-90 triangle:
- The side opposite the 30° angle is half the hypotenuse.
- The side opposite the 60° angle is \( \frac{\sqrt{3}}{2} \) times the hypotenuse.

The hypotenuse is 20:
\[ x = \frac{20}{2} = 10 \]
\[ y = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \]

So, the answers are:
\[ x = 10, \quad y = 10\sqrt{3} \]

Problem 3:

Given:
- A 30-60-90 triangle with sides \( x \), \( y \), and hypotenuse \( 3\sqrt{12} \).
- The side opposite the 30° angle is \( x \).
- The side opposite the 60° angle is \( y \).

Solution:
First, simplify \( 3\sqrt{12} \):
\[ 3\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3} \]

In a 30-60-90 triangle:
- The side opposite the 30° angle is half the hypotenuse.
- The side opposite the 60° angle is \( \frac{\sqrt{3}}{2} \) times the hypotenuse.

The hypotenuse is \( 6\sqrt{3} \):
\[ x = \frac{6\sqrt{3}}{2} = 3\sqrt{3} \]
\[ y = 6\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6 \cdot \frac{3}{2} = 9 \]

So, the answers are:
\[ x = 3\sqrt{3}, \quad y = 9 \]

Problem 4:

Given:
- A 30-60-90 triangle with sides \( x \), \( y \), and hypotenuse \( 11\sqrt{3} \).
- The side opposite the 30° angle is \( x \).
- The side opposite the 60° angle is \( y \).

Solution:
In a 30-60-90 triangle:
- The side opposite the 30° angle is half the hypotenuse.
- The side opposite the 60° angle is \( \frac{\sqrt{3}}{2} \) times the hypotenuse.

The hypotenuse is \( 11\sqrt{3} \):
\[ x = \frac{11\sqrt{3}}{2} \]
\[ y = 11\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 11 \cdot \frac{3}{2} = \frac{33}{2} \]

So, the answers are:
\[ x = \frac{11\sqrt{3}}{2}, \quad y = \frac{33}{2} \]

Problem 5:

Given:
- A 45-45-90 triangle with one leg \( x \) and hypotenuse 10.
- The other leg is \( y \).

Solution:
In a 45-45-90 triangle:
- The legs are equal.
- The hypotenuse is \( \sqrt{2} \) times the length of each leg.

Let \( x = y \). Then:
\[ 10 = x\sqrt{2} \]
\[ x = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \]
\[ y = 5\sqrt{2} \]

So, the answers are:
\[ x = 5\sqrt{2}, \quad y = 5\sqrt{2} \]

Problem 6:

Given:
- A 45-45-90 triangle with one leg \( x \) and hypotenuse \( 6\sqrt{2} \).
- The other leg is \( y \).

Solution:
In a 45-45-90 triangle:
- The legs are equal.
- The hypotenuse is \( \sqrt{2} \) times the length of each leg.

Let \( x = y \). Then:
\[ 6\sqrt{2} = x\sqrt{2} \]
\[ x = 6 \]
\[ y = 6 \]

So, the answers are:
\[ x = 6, \quad y = 6 \]

Problem 7:

Given:
- An isosceles right triangle (45-45-90) with one leg \( 8\sqrt{3} \).
- The other leg is \( y \).
- The hypotenuse is \( x \).

Solution:
In a 45-45-90 triangle:
- The legs are equal.
- The hypotenuse is \( \sqrt{2} \) times the length of each leg.

One leg is \( 8\sqrt{3} \):
\[ y = 8\sqrt{3} \]
\[ x = 8\sqrt{3} \cdot \sqrt{2} = 8\sqrt{6} \]

So, the answers are:
\[ x = 8\sqrt{6}, \quad y = 8\sqrt{3} \]

Problem 8:

Given:
- A 30-60-90 triangle with sides \( x \), \( y \), and hypotenuse \( 8\sqrt{5} \).
- The side opposite the 30° angle is \( x \).
- The side opposite the 60° angle is \( y \).

Solution:
In a 30-60-90 triangle:
- The side opposite the 30° angle is half the hypotenuse.
- The side opposite the 60° angle is \( \frac{\sqrt{3}}{2} \) times the hypotenuse.

The hypotenuse is \( 8\sqrt{5} \):
\[ x = \frac{8\sqrt{5}}{2} = 4\sqrt{5} \]
\[ y = 8\sqrt{5} \cdot \frac{\sqrt{3}}{2} = 4\sqrt{15} \]

So, the answers are:
\[ x = 4\sqrt{5}, \quad y = 4\sqrt{15} \]

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & x = 2\sqrt{2}, \quad y = 4 \\
2. & x = 10, \quad y = 10\sqrt{3} \\
3. & x = 3\sqrt{3}, \quad y = 9 \\
4. & x = \frac{11\sqrt{3}}{2}, \quad y = \frac{33}{2} \\
5. & x = 5\sqrt{2}, \quad y = 5\sqrt{2} \\
6. & x = 6, \quad y = 6 \\
7. & x = 8\sqrt{6}, \quad y = 8\sqrt{3} \\
8. & x = 4\sqrt{5}, \quad y = 4\sqrt{15} \\
\end{array}
}
\]