In a municipal design, the length of a rectangular parking lot is exactly 200% of the radius of a circular roundabout. The circumference of this roundabout measures 264 meters. Assuming the rectangular parking lot covers an area of 2016 square meters, determine the percentage by which the radius of the roundabout is greater than the width of the parking lot.

Correct Answer: Option 75%

Let us solve this step-by-step to find the required percentage.

Step 1: Find the radius of the circular roundabout.
We are given that the circumference of the circular roundabout is 264 meters.
The formula for the circumference of a circle is:

$\text{Circumference}=2\πr$

Using $\π=\frac{22}{7}$, we set up the equation:

$2\×\frac{22}{7}\×r=264$

$\frac{44}{7}\×r=264$

$r=\frac{264\×7}{44}$

$r=6\×7=42\text{meters}$

So, the radius of the roundabout ($r$) is 42 meters.

Step 2: Find the length of the rectangular parking lot.
The problem states that the length ($L$) of the rectangular parking lot is exactly 200% of the radius of the roundabout.

$L=200\%\text{of}r=2\×42=84\text{meters}$

Step 3: Find the width of the rectangular parking lot.
The area of the rectangular parking lot is given as 2016 square meters.
The formula for the area of a rectangle is:

$\text{Area}=L\×W$

Substituting the known values:

$84\×W=2016$

$W=\frac{2016}{84}=24\text{meters}$

So, the width of the parking lot ($W$) is 24 meters.

Step 4: Determine the percentage by which the radius is greater than the width.
We need to calculate by what percentage the radius ($r=42\text{m}$) is greater than the width ($W=24\text{m}$):

$\text{Difference}=r-W=42-24=18\text{meters}$

Now, calculate the percentage increase with respect to the width:

$\text{Percentage Greater}=\left(\frac{r-W}{W}\right)\×100\%$

$\text{Percentage Greater}=\left(\frac{18}{24}\right)\×100\%=\frac{3}{4}\×100\%=75\%$

Hence, the radius of the roundabout is 75% greater than the width of the parking lot.