The problem involves a geometric figure that consists of a triangle placed on top of a square. We are given the following information:

• The two equal sides of the triangle are both $6$ units long.
• The diagonal of the square is $10$ units.
• We are tasked with finding the side length $x$ of the square, leaving the answer in its simplest radical form.

Step-by-step Solution:

1. Understanding the Square's Diagonal:
   In any square, the diagonal $d$ is related to the side length $x$ by the Pythagorean Theorem. This is because the diagonal forms a right triangle with two sides of the square, and the formula is: $d = x\sqrt{2}$ We are told the diagonal is $10$, so we can set up the equation: $10 = x\sqrt{2}$

2. Solving for $x$: To solve for $x$, divide both sides of the equation by $\sqrt{2}$: $x = \frac{10}{\sqrt{2}}$ Now, rationalize the denominator: $x = \frac{10}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

So, the side length $x$ of the square is $\boxed{5\sqrt{2}}$.

Would you like further clarification or have any additional questions?

Here are five related questions to deepen your understanding:

1. How is the Pythagorean Theorem applied to solve for the diagonal of a square?
2. Why do we need to rationalize the denominator when solving for $x$?
3. How would the problem change if the diagonal of the square were $12$ instead of $10$?
4. How can we verify that the triangle is equilateral using the given side lengths?
5. What are the properties of squares that make this problem solvable using the diagonal?

Tip:

Always remember that the diagonal of a square is related to its side length by $d = x\sqrt{2}$. This relationship is a quick way to solve problems involving diagonals in squares.