To solve this problem, let's analyze the isosceles triangle shown.

Problem Breakdown

1. Triangle Properties: The triangle is isosceles, meaning two of its sides are equal in length.
2. Right Angle: There is a right angle marked in the triangle, so it’s also a right triangle.
3. Sides Given: The leg on one side of the right angle is labeled with a length of $\sqrt{10}$, and the hypotenuse is labeled as $x$.

Solution

Since this is a right triangle, we can apply the Pythagorean theorem, which states: $a^2 + b^2 = c^2$ where $c$ is the hypotenuse, and $a$ and $b$ are the legs of the triangle.

In an isosceles right triangle, the two legs are equal. This means both legs of the triangle are $\sqrt{10}$.

Let’s apply the Pythagorean theorem: $(\sqrt{10})^2 + (\sqrt{10})^2 = x^2$ Calculating each term:

1. $(\sqrt{10})^2 = 10$
2. So, $10 + 10 = x^2$ $x^2 = 20$
3. Taking the square root of both sides to solve for $x$: $x = \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$

Final Answer

The length of side $x$ in simplest radical form is: $x = 2\sqrt{5}$

Would you like additional explanation on any part of this solution?

Here are five related questions you may find helpful:

1. How would you solve for the hypotenuse if the legs were different lengths?
2. What other properties do isosceles triangles have?
3. How does the Pythagorean theorem apply to non-isosceles triangles?
4. What is the process of simplifying square roots in radical form?
5. How does identifying a triangle as isosceles help solve problems?

Tip: Remember, in a 45°-45°-90° triangle (which is an isosceles right triangle), the hypotenuse is always $\sqrt{2}$ times the length of each leg.