Certainly! To find the length of one leg of an isosceles right triangle given its altitude is [tex]\(x\)[/tex] units, follow these steps:
1. Understanding the Problem:
- We are given an isosceles right triangle, which means two sides (legs) are of equal length and the angles opposite these sides are [tex]\(45^\circ\)[/tex] each. The third angle is [tex]\(90^\circ\)[/tex].
- We are given the altitude from the right angle to the hypotenuse. This altitude divides the original isosceles right triangle into two smaller 45-45-90 right triangles.
2. Properties of 45-45-90 Triangles:
- In a 45-45-90 triangle, the ratio of the lengths of the legs to the hypotenuse is [tex]\(1:1:\sqrt{2}\)[/tex].
- This means if each leg of the smaller triangle is [tex]\(a\)[/tex], the hypotenuse of the smaller triangle (which is also a leg of the original triangle) would be [tex]\(a\sqrt{2}\)[/tex].
3. Altitude Relationship:
- The altitude [tex]\(x\)[/tex] splits the original isosceles right triangle into two 45-45-90 triangles.
- Therefore, the leg of the original triangle (same as the hypotenuse of the smaller triangle) forms the base or height of the smaller 45-45-90 triangles.
- The altitude [tex]\(x\)[/tex], which is perpendicular to the hypotenuse, will be equal to one leg of the two smaller 45-45-90 triangles.
4. Finding the Leg:
- Let's denote the leg length of the original isosceles right triangle as [tex]\( \frac{x}{\sqrt{2}} \)[/tex]. Since [tex]\( x \)[/tex] corresponds to the hypotenuse of the smaller triangle, we use [tex]\( \frac{x}{\sqrt{2}} \)[/tex] to denote the one leg of smaller triangles.
5. Simplifying:
- To rationalize the denominator of [tex]\( \frac{x}{\sqrt{2}} \)[/tex], we multiply numerator and denominator by [tex]\( \sqrt{2} \)[/tex]:
[tex]\[
\frac{x}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{x \sqrt{2}}{2}
\][/tex]
Thus, the length of one leg of the original isosceles right triangle in terms of [tex]\( x \)[/tex] is:
[tex]\[
\boxed{\frac{x \sqrt{2}}{2}}
\][/tex]
Therefore, the correct answer is [tex]\( \boxed{\frac{x \sqrt{2}}{2}} \)[/tex].
