The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 18 units long. (Type an expression using \( x \) as the variable.) (b) The expression for the area of the rectangle is (Type an expression using \( x \) as the variable.)

To solve this problem, we need to use the properties of an isosceles right triangle and the relationship between the sides of a rectangle inscribed in such a triangle. Let's denote the length of the rectangle as \( x \) and the width as \( y \). Since the rectangle is inscribed in an isosceles right triangle, the length and width of the rectangle will be the legs of the triangle. In an isosceles right triangle, the legs are equal, so both \( x \) and \( y \) are equal. Therefore, \( y = x \). The hypotenuse of the triangle is given as 18 units. In an isosceles right triangle, the hypotenuse is \( \sqrt{2} \) times the length of a leg. So, we can write the following equation: \[ x\sqrt{2} = 18 \] Now, we can solve for \( x \): \[ x = \frac{18}{\sqrt{2}} \] To rationalize the denominator, we multiply the numerator and the denominator by \( \sqrt{2} \): \[ x = \frac{18}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \] \[ x = \frac{18\sqrt{2}}{2} \] \[ x = 9\sqrt{2} \] Now that we have the value of \( x \), we can find the area of the rectangle. The area \( A \) of a rectangle is given by the product of its length and width: \[ A = x \cdot y \] Since \( y = x \), we can substitute \( x \) with \( 9\sqrt{2} \): \[ A = (9\sqrt{2}) \cdot (9\sqrt{2}) \] \[ A = 81 \cdot 2 \] \[ A = 162 \] So, the expression for the area of the rectangle is \( 162 \) square units. However, you asked for the expression using \( x \) as the variable. Since we have \( x = 9\sqrt{2} \), we can express the area in terms of \( x \) as follows: \[ A = x^2 \] This is because the area of the rectangle is the square of its length (or width, since they are equal). Therefore, the expression for the area of the rectangle using \( x \) as the variable is \( x^2 \).