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 prove that the rectangle of maximum area that can be inscribed in a circle of radius \( a \) is a square with side length \( a\sqrt{2} \), we can follow these steps:

### Step 1: Set up the problem.
Consider a circle with a radius \( a \) centered at the origin \( (0, 0) \) of the coordinate plane. Let the rectangle be inscribed in this circle, and let the coordinates of the vertices of the rectangle be \( (x_1, y_1) \), \( (x_1, y_2) \), \( (x_2, y_1) \), and \( (x_2, y_2) \).

### Step 2: Use the equation of the circle.
The equation of the circle is:
\[
x^2 + y^2 = a^2
\]
If a rectangle is inscribed in the circle, its sides will be parallel to the axes. The diagonal of the rectangle will be the diameter of the circle.

### Step 3: Express the area of the rectangle.
Let the length of the rectangle be \( 2x \) and the width be \( 2y \). The area \( A \) of the rectangle is:
\[
A = 2x \times 2y = 4xy
\]
But since the rectangle is inscribed in the circle, the coordinates \( (x, y) \) must satisfy the equation of the circle:
\[
x^2 + y^2 = a^2
\]

### Step 4: Maximize the area.
We want to maximize the area \( A = 4xy \) under the constraint \( x^2 + y^2 = a^2 \). 

We can use the method of Lagrange multipliers or substitute \( y \) in terms of \( x \) using the constraint:
\[
y = \sqrt{a^2 - x^2}
\]
Substitute this into the expression for the area:
\[
A = 4x \sqrt{a^2 - x^2}
\]
Now, to find the maximum area, we take the derivative of \( A \) with respect to \( x \) and set it equal to zero:
\[
\frac{dA}{dx} = 4\left[\sqrt{a^2 - x^2} + x \cdot \frac{-x}{\sqrt{a^2 - x^2}}\right] = 0
\]
Simplifying:
\[
4\left[\sqrt{a^2 - x^2} - \frac{x^2}{\sqrt{a^2 - x^2}}\right] = 0
\]
\[
\sqrt{a^2 - x^2} = \frac{x^2}{\sqrt{a^2 - x^2}}
\]
Square both sides:
\[
a^2 - x^2 = x^2
\]
\[
2x^2 = a^2
\]
\[
x^2 = \frac{a^2}{2}
\]
\[
x = \frac{a}{\sqrt{2}}
\]

### Step 5: Determine the dimensions of the square.
Given \( x = \frac{a}{\sqrt{2}} \), we can find \( y \) using the equation \( x^2 + y^2 = a^2 \):
\[
y = \sqrt{a^2 - \left(\frac{a}{\sqrt{2}}\right)^2} = \sqrt{a^2 - \frac{a^2}{2}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}
\]
So, \( x = y = \frac{a}{\sqrt{2}} \).

### Step 6: Calculate the maximum area.
The area of the square formed is:
\[
A = 4xy = 4\left(\frac{a}{\sqrt{2}}\right)\left(\frac{a}{\sqrt{2}}\right) = 4 \times \frac{a^2}{2} = 2a^2
\]
This is the maximum area of the rectangle that can be inscribed in the circle.

### Conclusion:
The rectangle of maximum area that can be inscribed in a circle is indeed a square, and its side length is \( \frac{a}{\sqrt{2}} \). The area of this square is \( 2a^2 \), confirming that the side length of the square is \( a\sqrt{2} \).

Would you like to dive deeper into this proof or have any related questions?

Here are some related questions:
1. How would the result change if the rectangle wasn't aligned with the coordinate axes?
2. Can a rectangle of maximum area be inscribed in an ellipse instead of a circle? What would be its dimensions?
3. How does the perimeter of this inscribed square compare to the circumference of the circle?
4. What is the relationship between the diagonal of the square and the radius of the circle?
5. What if the radius of the circle is doubled? How does it affect the area of the inscribed square?

**Tip:** When optimizing areas with constraints like circles or ellipses, geometric symmetry often leads to maximized results—like squares in circles.

Learn how to prove that the rectangle of maximum area inscribed in a circle of radius 'a' is a square with side length a√2. This detailed mathematical proof involves geometric concepts, circle equations, and area maximization. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation