Substituting the Value into the Equation

Oct 13, 2025 by magicalceylontours.com 23 views

Introduction

In a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle, the two legs are congruent, and the hypotenuse is the side opposite the right angle. Given that the hypotenuse measures $$22 \sqrt{2}$$ units, we need to find the length of one leg of the triangle. This problem requires an understanding of the properties of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle and the use of the Pythagorean theorem.

Properties of a $$45^{\circ}-45{\circ}-90^{\circ}$$ Triangle

A $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle is a special right triangle with two acute angles measuring $$45^{\circ}$$ each and the third angle measuring $$90^{\circ}$$. The two legs of the triangle are congruent, and the hypotenuse is the side opposite the right angle. In a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle, the ratio of the length of the hypotenuse to the length of a leg is $$\sqrt{2}$$.

The Pythagorean Theorem

The Pythagorean theorem is a fundamental concept in geometry that relates the lengths of the sides of a right triangle. The theorem states that in a right triangle with legs of length $$a$$ and $$b$$ and a hypotenuse of length $$c$$, the following equation holds:

a^2 + b^2 = c^2

Applying the Pythagorean Theorem to a $$45^{\circ}-45{\circ}-90^{\circ}$$ Triangle

In a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle, the two legs are congruent, and the hypotenuse is the side opposite the right angle. Let's denote the length of one leg as $$x$$. Since the two legs are congruent, the length of the other leg is also $$x$$. The hypotenuse measures $$22 \sqrt{2}$$ units.

Using the Pythagorean theorem, we can write the following equation:

x^2 + x^2 = (22 \sqrt{2})^2

Simplifying the Equation

Simplifying the equation, we get:

2x^2 = (22 \sqrt{2})^2

Evaluating the Right-Hand Side of the Equation

Evaluating the right-hand side of the equation, we get:

(22 \sqrt{2})^2 = 22^2 \cdot (\sqrt{2})^2 = 484 \cdot 2 = 968

Substituting the Value into the Equation

Substituting the value into the equation, we get:

2x^2 = 968

Dividing Both Sides of the Equation by 2

Dividing both sides of the equation by 2, we get:

x^2 = 484

Taking the Square Root of Both Sides of the Equation

Taking the square root of both sides of the equation, we get:

x = \sqrt{484}

Evaluating the Square Root

Evaluating the square root, we get:

x = 22

Conclusion

In this problem, we were given that the hypotenuse of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle measures $$22 \sqrt{2}$$ units. We needed to find the length of one leg of the triangle. Using the Pythagorean theorem and the properties of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle, we found that the length of one leg is $$22$$ units.

Final Answer

The final answer is $$\boxed{22}$$ units.

Introduction

In our previous article, we discussed how to find the length of one leg of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle given the length of the hypotenuse. In this article, we will answer some frequently asked questions related to this topic.

Q: What is a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle?

A: A $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle is a special right triangle with two acute angles measuring $$45^{\circ}$$ each and the third angle measuring $$90^{\circ}$$. The two legs of the triangle are congruent, and the hypotenuse is the side opposite the right angle.

Q: What is the ratio of the length of the hypotenuse to the length of a leg in a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle?

A: The ratio of the length of the hypotenuse to the length of a leg in a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle is $$\sqrt{2}$$.

Q: How do I find the length of one leg of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle given the length of the hypotenuse?

A: To find the length of one leg of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle given the length of the hypotenuse, you can use the Pythagorean theorem. Let's denote the length of one leg as $$x$$. Since the two legs are congruent, the length of the other leg is also $$x$$. The hypotenuse measures $$c$$ units. Using the Pythagorean theorem, you can write the following equation:

x^2 + x^2 = c^2

Q: How do I simplify the equation?

A: Simplifying the equation, you get:

2x^2 = c^2

Q: How do I evaluate the right-hand side of the equation?

A: Evaluating the right-hand side of the equation, you get:

c^2 = (22 \sqrt{2})^2 = 22^2 \cdot (\sqrt{2})^2 = 484 \cdot 2 = 968

Q: How do I substitute the value into the equation?

A: Substituting the value into the equation, you get:

2x^2 = 968

Q: How do I divide both sides of the equation by 2?

A: Dividing both sides of the equation by 2, you get:

x^2 = 484

Q: How do I take the square root of both sides of the equation?

A: Taking the square root of both sides of the equation, you get:

x = \sqrt{484}

Q: How do I evaluate the square root?

A: Evaluating the square root, you get:

x = 22

Q: What is the final answer?

A: The final answer is $$\boxed{22}$$ units.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

Not using the Pythagorean theorem correctly
Not simplifying the equation correctly
Not evaluating the right-hand side of the equation correctly
Not substituting the value into the equation correctly
Not dividing both sides of the equation by 2 correctly
Not taking the square root of both sides of the equation correctly
Not evaluating the square root correctly

Q: How can I practice solving this problem?

A: You can practice solving this problem by:

Using different values for the length of the hypotenuse
Using different values for the length of one leg
Using different methods to solve the equation
Checking your work to make sure you are getting the correct answer

Conclusion

In this article, we answered some frequently asked questions related to finding the length of one leg of a $$45^{\circ}-45{\circ}-90^{\circ}$$ triangle given the length of the hypotenuse. We also discussed some common mistakes to avoid and provided some tips for practicing solving this problem.