
Part 1
The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 14 units long. The expression for the area of the rectangle is 
​(Type an expression using x as the​ variable.

Let's define the problem step by step.

Given:
- We have a rectangle inscribed in an isosceles right triangle.
- The hypotenuse of the triangle is 14 units long.
- We need to find an expression for the area of the rectangle using \( x \) as the variable.

### Step 1: Relationship in the Isosceles Right Triangle
For an isosceles right triangle, the legs are of equal length, and the hypotenuse \( h \) is related to the legs \( l \) by the Pythagorean theorem:
\[
h = l\sqrt{2}
\]
Here, \( h = 14 \), so:
\[
l\sqrt{2} = 14
\]
Solving for \( l \):
\[
l = \frac{14}{\sqrt{2}} = 7\sqrt{2}
\]

### Step 2: Coordinates and Proportions
Now, let's consider the inscribed rectangle. If we set the base of the triangle along the x-axis, and let the height of the triangle be along the y-axis, the rectangle will be inscribed within these limits.

- Let \( x \) be the width of the rectangle, where \( 0 \leq x \leq 7\sqrt{2} \).
- The height of the rectangle is proportional to the remaining height of the triangle.

Since the triangle is similar to the smaller triangle formed by cutting off the top part by the rectangle, the height \( y \) of the rectangle can be expressed as:
\[
y = 7\sqrt{2} - \frac{x}{\sqrt{2}}
\]

### Step 3: Expression for the Area
The area \( A \) of the rectangle is the product of its width \( x \) and its height \( y \):
\[
A = x \left( 7\sqrt{2} - \frac{x}{\sqrt{2}} \right)
\]
Simplifying this expression:
\[
A = 7\sqrt{2}x - \frac{x^2}{\sqrt{2}}
\]

Thus, the expression for the area of the rectangle is:
\[
A(x) = 7\sqrt{2}x - \frac{x^2}{\sqrt{2}}
\]

Let me know if you'd like further details or have any questions.

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Here are five questions related to this topic:

1. How is the relationship between the legs and hypotenuse of an isosceles right triangle derived?
2. How does the similarity of triangles help in determining the dimensions of the inscribed rectangle?
3. Can we find the maximum area of the inscribed rectangle using calculus?
4. What are the limitations on the value of \( x \) for this rectangle to remain inside the triangle?
5. How can you generalize this approach to triangles with different hypotenuse lengths?

**Tip:** The maximum area of a rectangle inscribed in a right triangle often occurs when its width and height are carefully balanced—this balance can be found using optimization techniques!

The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 14 units long. The expression for the area of the rectangle is (Type an expression using x as the variable).

Discover how to find the expression for the area of a rectangle inscribed in an isosceles right triangle with a hypotenuse of 14 units. This problem involves using the Pythagorean theorem and proportions from similar triangles. Ideal for students in Grades 9-12.

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