1. **State the problem:** We have a right triangle with one angle of 45°, the right angle at the top vertex, and the sides labeled as follows: the left slanted side is $x$, the right slanted side is $y$, and the base is $\sqrt{10}$. We need to find the values of $x$ and $y$. 2. **Recall the properties of a 45°-45°-90° triangle:** In such a triangle, the legs are congruent, and the hypotenuse is $\sqrt{2}$ times the length of each leg. 3. **Identify the sides:** Since the right angle is at the top vertex and one angle is 45°, the triangle is isosceles right triangle with the two legs equal. The base $\sqrt{10}$ is the hypotenuse. 4. **Use the formula for the hypotenuse:** $$\text{hypotenuse} = x \sqrt{2}$$ Since the hypotenuse is $\sqrt{10}$, we have: $$\sqrt{10} = x \sqrt{2}$$ 5. **Solve for $x$:** $$x = \frac{\sqrt{10}}{\sqrt{2}}$$ Show cancellation: $$x = \frac{\sqrt{10}}{\cancel{\sqrt{2}}} \times \frac{\cancel{\sqrt{2}}}{\sqrt{2}} = \frac{\sqrt{10} \times \sqrt{2}}{2} = \frac{\sqrt{20}}{2}$$ 6. **Simplify $\sqrt{20}$:** $$\sqrt{20} = \sqrt{4 \times 5} = 2 \sqrt{5}$$ 7. **Substitute back:** $$x = \frac{2 \sqrt{5}}{2} = \sqrt{5}$$ 8. **Since the triangle is isosceles right, $y = x$:** $$y = \sqrt{5}$$ **Final answer:** $$x = \sqrt{5}, \quad y = \sqrt{5}$$