Isosceles Triangle Leg Length: Hypotenuse Is 10 Inches

Let's dive into the world of geometry and tackle a classic problem involving an isosceles right triangle. Specifically, we're going to figure out how to calculate the length of one of the legs when we know the hypotenuse is 10 inches. This is a common question in mathematics, and understanding the solution involves grasping some key concepts about triangles and the Pythagorean theorem. So, grab your thinking caps, and let's get started!

Understanding Isosceles Right Triangles

Before we jump into the calculations, let's make sure we're all on the same page about what an isosceles right triangle actually is. Isosceles right triangles are special triangles that possess two defining characteristics:

• Isosceles: This means that two of the triangle's sides are of equal length. In an isosceles right triangle, these equal sides are the legs (the sides that form the right angle).
• Right Triangle: This means that one of the triangle's angles is a right angle, measuring exactly 90 degrees.

These two properties combined give us some useful information. Because the two legs are equal, the two angles opposite those legs are also equal. Since one angle is 90 degrees, and the sum of all angles in a triangle is 180 degrees, the other two angles must each be 45 degrees. So, an isosceles right triangle is also known as a 45-45-90 triangle. Understanding this 45-45-90 triangle relationship is crucial for solving this type of problem efficiently.

In our case, we're given that the hypotenuse (the side opposite the right angle) of our isosceles right triangle is 10 inches. Our mission is to determine the length of one of the legs. Remember, because it's an isosceles triangle, both legs will have the same length. This problem highlights the relationship between the sides of a right triangle, and how we can use known information to find unknown lengths. This principle is fundamental not only in geometry but also in various real-world applications, such as construction and navigation. The ability to visualize and calculate the dimensions of geometric shapes is a valuable skill, making this exploration both theoretically interesting and practically useful.

Applying the Pythagorean Theorem

The Pythagorean Theorem is the cornerstone of solving problems involving right triangles. This fundamental theorem states a crucial relationship between the sides of any right triangle. It says that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (the legs). We can express this mathematically as:

$$a^2 + b^2 = c^2$$

Where:

• a and b are the lengths of the legs of the right triangle.
• c is the length of the hypotenuse.

This theorem provides a powerful tool for finding unknown side lengths when you know the other two sides. In the case of our isosceles right triangle, we know the length of the hypotenuse (c = 10 inches), and we know that the two legs are equal in length (a = b). Let's call the length of one leg 'x'. Now we can rewrite the Pythagorean theorem equation for our specific problem:

$$x^2 + x^2 = 10^2$$

This equation sets the stage for us to solve for 'x', which will give us the length of one leg of the triangle. The beauty of the Pythagorean Theorem is its versatility and broad applicability. It allows us to connect the lengths of the sides of a right triangle in a precise and predictable way. It's a foundational concept in mathematics, and mastering its application is crucial for anyone delving into geometry and related fields. By understanding and applying the Pythagorean Theorem, we can systematically unravel geometric problems and find solutions, making it an indispensable tool in our mathematical toolkit.

Solving for the Leg Length

Now that we've set up our equation using the Pythagorean Theorem, let's proceed with solving for the length of the leg, which we've denoted as 'x'. We have the equation:

$$x^2 + x^2 = 10^2$$

First, simplify the left side of the equation by combining the like terms:

$$2x^2 = 10^2$$

Next, calculate 10 squared:

$$2x^2 = 100$$

Now, isolate $x^2$ by dividing both sides of the equation by 2:

$$x^2 = 50$$

To find 'x', we need to take the square root of both sides of the equation:

$$x = \sqrt{50}$$

We can simplify the square root of 50 by factoring out the largest perfect square, which is 25:

$$x = \sqrt{25 * 2}$$

Since the square root of 25 is 5, we can further simplify:

$$x = 5\sqrt{2}$$

However, the answer options provided in the original question are in a slightly different form. To match the options, we need to manipulate the expression. Recall that we are looking for an equivalent expression among the choices:

A. $\frac{10}{\sqrt{3}}$ B. $\frac{10}{\sqrt{2}}$ C. $10 \sqrt{2}$ D. $10 \sqrt{3}$

Notice that our current simplified answer, $5\sqrt{2}$, is not directly present. To get it into the form of one of the multiple-choice answers, we should focus on expressing 5 as a fraction with a $\sqrt{2}$ in the denominator, aiming for the form $\frac{10}{\sqrt{2}}$:

To achieve this, recognize that 5 can be written as $\frac{10}{2}$. Therefore, we have:

$$x = \frac{10}{2} * \sqrt{2}$$

Now we want to introduce $\sqrt{2}$ into the denominator. We can do this by multiplying both the numerator and the denominator by $\sqrt{2}$:

$$x = \frac{10}{2} * \sqrt{2} * \frac{\sqrt{2}}{\sqrt{2}}$$

Simplify the expression:

$$x = \frac{10\sqrt{2}}{2\sqrt{2}}$$

However, this manipulation did not directly lead us to the given answer choice B. It seems there was a misinterpretation in simplifying to match the multiple-choice answers. Let’s re-evaluate the step where we had $x = 5\sqrt{2}$.

We need to rationalize our initially derived solution $5\sqrt{2}$ or recognize an equivalent form among the provided choices directly. The correct simplification and recognition of equivalent expressions are crucial steps in mathematical problem-solving.

Let's reconsider rationalizing an alternative form that might match a provided option. Going back to our simplified expression $x = \sqrt{50}$, let’s try a different approach to see if it aligns with one of the options.

$$x = \sqrt{50} = \sqrt{2 * 25} = \sqrt{2} * \sqrt{25} = 5\sqrt{2}$$

Given the options, we need to check if we can manipulate this to match one of the multiple-choice selections. We'll focus on option B, $\frac{10}{\sqrt{2}}$, and see if it’s equivalent to $5\sqrt{2}$.

To compare these, rationalize the denominator in option B. Start with option B:

B. $\frac{10}{\sqrt{2}}$

Rationalize the denominator by multiplying both the numerator and the denominator by $\sqrt{2}$:

$$\frac{10}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2}$$

Now, simplify the fraction:

$$\frac{10\sqrt{2}}{2} = 5\sqrt{2}$$

This is exactly the simplified solution we found for the leg length! Thus, the correct answer is B. $\frac{10}{\sqrt{2}}$. This process illustrates the importance of not only solving the problem correctly but also manipulating the solution to match the format of the given answer choices.

Conclusion

In this exploration, we've successfully determined the length of a leg in an isosceles right triangle with a hypotenuse of 10 inches. We started by understanding the properties of isosceles right triangles and then applied the Pythagorean Theorem to establish a relationship between the sides. By solving the resulting equation, we found that the length of one leg is $5\sqrt{2}$ inches, which is equivalent to $\frac{10}{\sqrt{2}}$. This exercise highlights the power of geometric principles and algebraic manipulation in solving mathematical problems. It also underscores the importance of being able to simplify and express solutions in different forms to match given answer options.

To deepen your understanding of right triangles and the Pythagorean Theorem, you might find it helpful to explore additional resources. A great place to start is Khan Academy's geometry section, where you can find lessons, practice exercises, and videos on this and many other mathematical topics. Keep practicing, and you'll become a geometry pro in no time!