Q9:

Shazli took a wire of length $44$ cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take $\pi = \frac{22}{7}$)

Given Variables & Initial Setup

We are given a single continuous wire of a fixed length that is bent into two different geometric shapes (a circle and a square) in separate scenarios. The fundamental principle governing this problem is the Conservation of Perimeter, which states that the total boundary length of any shape formed by the wire must equal the total length of the wire.

• Total length of the wire, $L = 44 \text{ cm}$
• Mathematical constant, $\pi = \frac{22}{7}$

Step 1: Analyzing the Circular Shape and Finding the Radius

When the wire is bent into the shape of a circle, the entire length of the wire forms the boundary of the circle. [Per the geometric definition of circumference, the perimeter of a circle is its circumference].

Let $r$ be the radius of the circle. The formula for the circumference $C$ is:

$C = 2\pi r$

Equating the circumference to the total length of the wire:

$2\pi r = 44$

Substituting the given value of $\pi$:

$2 \left( \frac{22}{7} \right) r = 44$

$\frac{44}{7} r = 44$

Isolating $r$ by multiplying both sides by $\frac{7}{44}$:

$r = 44 \times \frac{7}{44}$

$r = 7 \text{ cm}$

Step 2: Calculating the Area of the Circle

The area $A_{circle}$ enclosed by a circle is given by the formula:

$A_{circle} = \pi r^2$

Substituting $r = 7 \text{ cm}$ and $\pi = \frac{22}{7}$:

$A_{circle} = \frac{22}{7} \times (7)^2$

$A_{circle} = \frac{22}{7} \times 49$

$A_{circle} = 22 \times 7$

$A_{circle} = 154 \text{ cm}^2$

Step 3: Analyzing the Square Shape and Finding the Side Length

When the same wire is bent into the shape of a square, the total length of the wire forms the perimeter of the square. [Per the properties of regular polygons, a square has four sides of equal length].

Let $s$ be the side length of the square. The formula for the perimeter $P$ of a square is:

$P = 4s$

Equating the perimeter to the total length of the wire:

$4s = 44$

Isolating $s$ by dividing both sides by $4$:

$s = \frac{44}{4}$

$s = 11 \text{ cm}$

Step 4: Calculating the Area of the Square

The area $A_{square}$ enclosed by a square is given by the formula:

$A_{square} = s^2$

Substituting $s = 11 \text{ cm}$:

$A_{square} = (11)^2$

$A_{square} = 121 \text{ cm}^2$

Step 5: Visualizing and Comparing the Enclosed Areas

We now compare the calculated areas to determine which geometric figure maximizes the enclosed space for a given perimeter (an application of the Isoperimetric Inequality).

• Area of the Circle: $154 \text{ cm}^2$
• Area of the Square: $121 \text{ cm}^2$

Since $154 \text{ cm}^2 > 121 \text{ cm}^2$, the circle encloses a significantly larger area than the square.

Final Solution: The radius of the circle is $7 \text{ cm}$ and its area is $154 \text{ cm}^2$. When bent into a square, the length of each side is $11 \text{ cm}$ and its area is $121 \text{ cm}^2$. Comparing the two, the circle encloses more area than the square.