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Show that the leg length of a 45-45-90 right triangle with a hypotenuse of [tex]$x$[/tex] is equal to [tex]\frac{x \sqrt{2}}{2}[/tex].

Sagot :

Sure! To demonstrate that the leg length of a 45-45-90 right triangle with a hypotenuse of [tex]\(x\)[/tex] is equal to [tex]\(\frac{x \sqrt{2}}{2}\)[/tex], we will follow these steps:

1. Understand the properties of a 45-45-90 triangle:
- A 45-45-90 triangle is an isosceles right triangle.
- This means both legs have the same length.
- Let's denote the length of each leg as [tex]\(a\)[/tex].

2. Apply the Pythagorean Theorem:
- The Pythagorean Theorem states that for any right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem is given by:
[tex]\[ \text{Hypotenuse}^2 = \text{Leg}_1^2 + \text{Leg}_2^2 \][/tex]
- In our 45-45-90 triangle, both legs are equal, so we can write:
[tex]\[ x^2 = a^2 + a^2 \][/tex]
- This simplifies to:
[tex]\[ x^2 = 2a^2 \][/tex]

3. Solve for the leg length [tex]\(a\)[/tex]:
- Rearrange the equation to solve for [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = \frac{x^2}{2} \][/tex]
- Take the square root of both sides to solve for [tex]\(a\)[/tex]:
[tex]\[ a = \sqrt{\frac{x^2}{2}} = \frac{x}{\sqrt{2}} \][/tex]
- To rationalize the denominator, multiply numerator and denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{x}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{x \sqrt{2}}{2} \][/tex]

4. State the final result:
- Hence, the length of each leg [tex]\(a\)[/tex] of a 45-45-90 triangle with a hypotenuse [tex]\(x\)[/tex] is indeed:
[tex]\[ a = \frac{x \sqrt{2}}{2} \][/tex]

This step-by-step process shows that the leg length of a 45-45-90 triangle with a hypotenuse of [tex]\(x\)[/tex] is [tex]\(\frac{x \sqrt{2}}{2}\)[/tex].
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