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.) (c) The largest area the rectangle can have is \( \square \) square units. (Type an integer or a simplified fraction.)

To find the largest area of the rectangle inscribed in the isosceles right triangle, we need to use the properties of the triangle and the rectangle. 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 we have \( x = y \). The hypotenuse of the triangle is given as 18 units. According to the Pythagorean theorem, the length of each leg of the triangle is \( \frac{18}{\sqrt{2}} \) units. Now, the area of the rectangle is given by \( A = x \cdot y \). Since \( x = y \), we can write the area as \( A = x^2 \). To find the largest area, we need to maximize \( x^2 \) under the constraint that \( x = \frac{18}{\sqrt{2}} \). The largest area the rectangle can have is when \( x \) is at its maximum value, which is \( \frac{18}{\sqrt{2}} \). So, the largest area \( A \) is: \[ A = \left(\frac{18}{\sqrt{2}}\right)^2 \] \[ A = \frac{18^2}{2} \] \[ A = \frac{324}{2} \] \[ A = 162 \] Therefore, the largest area the rectangle can have is \( 162 \) square units.