The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 18 units long. (a) Express the y-coordinate of \( P \) in terms of \( x \). (Hint: Write an equation for the line AB.) (b) Express the area of the rectangle in terms of \( x \). (c) What is the largest area the rectangle can have, and what are its dimensions?

To solve the problem, we first need to understand the configuration of the rectangle inscribed in the isosceles right triangle. ### Step 1: Understanding the Triangle Given that the hypotenuse of the isosceles right triangle is 18 units, we can find the lengths of the legs. In an isosceles right triangle, the legs are equal, and if we denote the length of each leg as \( a \), we can use the Pythagorean theorem: \[ a^2 + a^2 = 18^2 \implies 2a^2 = 324 \implies a^2 = 162 \implies a = 9\sqrt{2} \] Thus, the triangle has legs of length \( 9\sqrt{2} \). ### Step 2: Coordinates of the Triangle We can place the triangle in the coordinate system for convenience: - Vertex \( A \) at \( (0, 0) \) - Vertex \( B \) at \( (18, 0) \) - Vertex \( C \) at \( (9, 9\sqrt{2}) \) ### Step 3: Finding the Equation of Line AB The line segment \( AB \) is horizontal along the x-axis, so its equation is simply: \[ y = 0 \] ### Step 4: Finding the Equation of Line AC To find the equation of line \( AC \), we need the slope of the line from \( A \) to \( C \): \[ \text{slope of } AC = \frac{9\sqrt{2} - 0}{9 - 0} = \sqrt{2} \] Using point-slope form, the equation of line \( AC \) is: \[ y - 0 = \sqrt{2}(x - 0) \implies y = \sqrt{2}x \] ### Step 5: Finding the y-coordinate of Point P Let \( P \) be the point on line \( AC \) where the rectangle touches. If the rectangle has width \( x \) (from \( (0, 0) \) to \( (x, 0) \)), then the height \( y \) at point \( P \) can be expressed as: \[ y = \sqrt{2}x \] ### Step 6: Area of the Rectangle The area \( A \) of the rectangle can be expressed in terms of \( x \): \[ A = \text{width} \times \text{height} = x \cdot y = x \cdot \sqrt{2}x = \sqrt{2}x^2 \] ### Step 7: Maximizing the Area To find the maximum area, we can take the derivative of \( A \) with respect to \( x \) and set it to zero: \[ A = \sqrt{2}x^2 \] Taking the derivative: \[ \frac{dA}{dx} = 2\sqrt{2}x \] Setting the derivative to zero: \[ 2\sqrt{2}x = 0 \implies x = 0 \] This indicates a minimum area at \( x = 0 \). To find the maximum area, we need to consider the endpoints. The maximum value of \( x \) occurs when the rectangle reaches the height of the triangle, which is at \( x = 9 \). ### Step 8: Calculating Maximum Area Substituting \( x = 9 \) into the area formula: \[ A = \sqrt{2}(9^2) = \sqrt{2} \cdot 81 = 81\sqrt{2} \] ### Step 9: Dimensions of the Rectangle The dimensions of the rectangle at maximum area are: - Width: \( x = 9 \) - Height: \( y = \sqrt{2} \cdot 9 = 9\sqrt{2} \) ### Final Answers (a) The y-coordinate of \( P \) in terms of \( x \) is \( y = \sqrt{2}x \). (b) The area of the rectangle in terms of \( x \) is \( A = \sqrt{2}x^2 \). (c) The largest area the rectangle can have is \( 81\sqrt{2} \) square units, with dimensions \( 9 \) units (width) and \( 9\sqrt{2} \) units (height).