To show that the rectangle of maximum area that can be inscribed in a circle of radius $a$ cm is a square with side length $a\sqrt{2}$, we will follow these steps:

1. Set up the problem using coordinates.
2. Express the area of the rectangle in terms of a single variable.
3. Maximize the area function using calculus.
4. Show that the maximum area occurs when the rectangle is a square.

Step 1: Setting up the Problem

Consider a circle centered at the origin $(0, 0)$ in a coordinate plane with radius $a$.

If we inscribe a rectangle in this circle, let the coordinates of one vertex of the rectangle in the first quadrant be $(x, y)$. Due to symmetry, the opposite vertex will be at $(-x, -y)$. The other two vertices will be at $(x, -y)$ and $(-x, y)$.

Since the rectangle is inscribed in the circle, its diagonal will be the diameter of the circle.

Step 2: Expressing the Area of the Rectangle

The equation of the circle is: $x^2 + y^2 = a^2$

The area $A$ of the rectangle can be expressed as: $A = 2x \cdot 2y = 4xy$

Now, express $y$ in terms of $x$ using the circle's equation: $y = \sqrt{a^2 - x^2}$

Substitute this into the area formula: $A(x) = 4x \sqrt{a^2 - x^2}$

Step 3: Maximizing the Area Function

To find the maximum area, differentiate $A(x)$ with respect to $x$ and set the derivative equal to zero:

$A'(x) = 4 \left(\sqrt{a^2 - x^2} + \frac{-x^2}{\sqrt{a^2 - x^2}}\right)$

Simplify $A'(x)$:

$A'(x) = 4 \left( \frac{(a^2 - x^2) - x^2}{\sqrt{a^2 - x^2}} \right)$

$A'(x) = 4 \left( \frac{a^2 - 2x^2}{\sqrt{a^2 - x^2}} \right)$

Set $A'(x) = 0$ to find the critical points:

$a^2 - 2x^2 = 0$

$2x^2 = a^2$

$x^2 = \frac{a^2}{2}$

$x = \frac{a}{\sqrt{2}}$

Step 4: Show the Maximum Area Occurs When the Rectangle is a Square

With $x = \frac{a}{\sqrt{2}}$, find $y$:

$y = \sqrt{a^2 - x^2} = \sqrt{a^2 - \left(\frac{a}{\sqrt{2}}\right)^2} = \sqrt{a^2 - \frac{a^2}{2}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$

So, $x = y = \frac{a}{\sqrt{2}}$. This means the rectangle is a square with side length $\frac{a}{\sqrt{2}}$.

The area of this square is: $A = x \times y = \left(\frac{a}{\sqrt{2}}\right) \times \left(\frac{a}{\sqrt{2}}\right) = \frac{a^2}{2}$

The maximum area of a rectangle inscribed in a circle of radius $a$ is $\frac{a^2}{2}$, which occurs when the rectangle is a square with side length $\frac{a}{\sqrt{2}}$.

Thus, we have shown that the rectangle of maximum area that can be inscribed in a circle of radius $a$ cm is indeed a square with side length $\frac{a}{\sqrt{2}}$, and its diagonal, being the diameter of the circle, is $2a$.

Further Exploration

1. How does the formula for the maximum area change if the circle is not centered at the origin?
2. Can you inscribe a rectangle of maximum perimeter in a circle? What would that look like?
3. What happens if we try to inscribe other polygons in a circle? How would the area or perimeter change?
4. Is there a general formula for the maximum area of a polygon inscribed in a circle?
5. Can you derive the formula for the area of an ellipse inscribed in a circle?

Tip: Remember that symmetry often simplifies geometric problems and helps in understanding the properties of figures.