A wire is bent in to the form of a square of side 27.5 cm . It is straightened and then bent into the shape of a circle What is the radius of the circle so formed ?

Options

A. `=18.5` cm
B. `=19.5` cm
C. `=17.5` cm
D. `=20.5` cm

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the perimeter of the square.
The perimeter (P) of a square can be calculated using the formula:
\[ P = 4 \times \text{side} \]
Given that the side of the square is 27.5 cm, we can substitute this value into the formula.

\[ P = 4 \times 27.5 \]
\[ P = 110 \text{ cm} \]

### Step 2: Understand that the length of the wire is equal to the perimeter of the square.
Since the wire is bent into the shape of a square, the length of the wire is equal to the perimeter we just calculated.

### Step 3: Set the length of the wire equal to the circumference of the circle.
When the wire is straightened and bent into the shape of a circle, the length of the wire will be equal to the circumference (C) of the circle. The formula for the circumference of a circle is:
\[ C = 2 \pi r \]
Where \( r \) is the radius of the circle.

### Step 4: Substitute the length of the wire into the circumference formula.
From Step 2, we know that the length of the wire is 110 cm, so we can set up the equation:
\[ 110 = 2 \pi r \]

### Step 5: Solve for the radius \( r \).
We can rearrange the equation to solve for \( r \):
\[ r = \frac{110}{2 \pi} \]

Using the approximate value of \( \pi \) as \( \frac{22}{7} \):
\[ r = \frac{110}{2 \times \frac{22}{7}} \]
\[ r = \frac{110 \times 7}{44} \]
\[ r = \frac{770}{44} \]
\[ r = \frac{35}{2} \]
\[ r = 17.5 \text{ cm} \]

### Final Answer:
The radius of the circle formed is **17.5 cm**.

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