Show that the rectangle of maximum area that can be inscribed in a circle of radius ‘𝑎’ cm is the square of 𝑎√2.

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.

Discover how to maximize the area of a rectangle inscribed in a circle of radius 'a' cm. Follow a step-by-step calculus approach to prove that the maximum area occurs when the rectangle is a square with side length 'a√2'. Suitable for Grades 11-12.

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