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: The correct option is 75%.

Step-by-Step Solution:

Step 1: Find the radius of the circular roundabout.
The circumference ($C$) of a circle is given by the formula:

$C=2\pi r$

We are given that the circumference of the roundabout is $264\text{meters}$. Using $\pi =\frac{22}{7}$:

$264=2\times \frac{22}{7}\times r$

$264=\frac{44}{7}\times r$

$r=\frac{264\times 7}{44}$

$r=6\times 7=42\text{meters}$

Step 2: Calculate the length of the rectangular parking lot.
The length ($L$) of the parking lot is $200\%$ of the radius ($r$):

$L=200\%\text{of}r=2\times r$

$L=2\times 42=84\text{meters}$

Step 3: Calculate the width of the rectangular parking lot.
The area ($A$) of the rectangle is given as $2016{\text{m}}^2$:

$A=L\times W$

$2016=84\times W$

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

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

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

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

$\text{Percentage greater}=\left(\frac{18}{24}\right)\times 100\%$

$\text{Percentage greater}=\frac{3}{4}\times 100\%=75\%$

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