To solve this problem, we need to analyze the properties of an isosceles right triangle where the altitude is given as [tex]\( x \)[/tex] units.
### Step-by-Step Solution:
1. Identify Key Properties of the Triangle:
- An isosceles right triangle has two equal legs and a right angle between them.
- The altitude in such a triangle forms two smaller right triangles within the original triangle.
2. Understand the Relationship between Altitude and Legs:
- The altitude bisects the hypotenuse, meaning it splits the triangle into two smaller identical right triangles.
- For an isosceles right triangle, the legs are congruent.
3. Set Up the Triangle Measurements:
- Consider the whole isosceles right triangle.
- Let each leg of this larger triangle be denoted by [tex]\( L \)[/tex] (which are congruent).
4. Relate the Altitude to the Legs:
- The altitude in this case splits the larger triangle into two smaller right triangles along the altitude line.
- Each small triangle will have legs measuring [tex]\( x/\sqrt{2} \)[/tex] because when split, the two legs of the smaller triangles are half the actual legs of the larger triangle.
5. Calculate the Hypotenuse of the Smaller Triangles:
- Since the legs of these two smaller triangles comprise [tex]\( x/\sqrt{2} \)[/tex]:
[tex]\[
\text{Hypotenuse of each smaller triangle} = \sqrt{\left(\frac{x}{\sqrt{2}}\right)^2 + \left(\frac{x}{\sqrt{2}}\right)^2}
\][/tex]
- Simplifying the expression inside the square root:
[tex]\[
= \sqrt{\frac{x^2}{2} + \frac{x^2}{2}} = \sqrt{\frac{2x^2}{2}} = \sqrt{x^2} = x
\][/tex]
- This confirms that [tex]\( x \)[/tex] is indeed the length of altitude.
6. Relate it Back to Large Triangle:
- The length of one leg of the entire larger isosceles right triangle can be derived as:
[tex]\[
L = x \sqrt{2}
\][/tex]
### Conclusion:
Thus, the length of one leg of the larger isosceles right triangle, given that the altitude is [tex]\( x \)[/tex] units, is [tex]\( x \sqrt{2} \)[/tex] units.
Therefore, the correct answer is [tex]\( x \sqrt{2} \)[/tex] units.
