The city of Atlantis was crafted by the God of the seas, Poseidon. It was made of alternating concentric circular rings of land (shaded) and water (not shaded) as represented in the figure. The radius of Inner Island was 2.5 stades. The water surrounding Inner Island was one stade wide (length AB). This was surrounded by two pairs of alternating rings of land and water. The first pair of land and water was two stades wide each (lengths BC and CD), and the outer pair is three stades wide each (lengths DE and EF).

The ratio of the surface area of the land to that of the water in the city of Atlantis is __________.(round off to two decimal places).

The correct option is 0.75.

To find the ratio of the surface area of the land to that of the water in the city of Atlantis, we calculate the area of each concentric circular region starting from the center and moving outwards. As shown in the provided figure, the city consists of an Inner Island at the center and alternating concentric rings of water and land, labeled with points A, B, C, D, E, and F along a radial line.

Let the center of the city be the origin. We define the radii of the concentric circles corresponding to the boundaries at points A, B, C, D, E, and F as follows:

$r_A=2.5\text{stades (radius of Inner Island)}$

The width of each successive ring is given: AB = 1 stade, BC = 2 stades, CD = 2 stades, DE = 3 stades, and EF = 3 stades. By adding these widths sequentially, we find the outer radius for each boundary:

$r_B=r_A+\text{AB}=2.5+1=3.5\text{stades}$

$r_C=r_B+\text{BC}=3.5+2=5.5\text{stades}$

$r_D=r_C+\text{CD}=5.5+2=7.5\text{stades}$

$r_E=r_D+\text{DE}=7.5+3=10.5\text{stades}$

$r_F=r_E+\text{EF}=10.5+3=13.5\text{stades}$

Now, we calculate the area of each region. The area of a circle with radius $r$ is given by $\pi r^2$, and the area of a circular ring bounded by inner radius $r_{\text{in}}$ and outer radius $r_{\text{out}}$ is given by $\pi (r_{\text{out}}^2-r_{\text{in}}^2)$.

1. Calculation of Land Areas (Shaded Regions):
- The Inner Island (from center to boundary A):

${\text{Area}}_{\text{Inner Island}}=\pi \times {2.5}^2=6.25\pi$
- The first land ring (between boundaries B and C):

${\text{Area}}_{\text{Land Ring 1}}=\pi \times ({5.5}^2-{3.5}^2)=\pi \times (30.25-12.25)=18\pi$
- The second land ring (between boundaries D and E):

${\text{Area}}_{\text{Land Ring 2}}=\pi \times ({10.5}^2-{7.5}^2)=\pi \times (110.25-56.25)=54\pi$
- Total Area of Land:

${\text{Area}}_{\text{Land}}=(6.25+18+54)\pi =78.25\pi$

2. Calculation of Water Areas (Unshaded Regions):
- The first water ring (between boundaries A and B):

${\text{Area}}_{\text{Water Ring 1}}=\pi \times ({3.5}^2-{2.5}^2)=\pi \times (12.25-6.25)=6\pi$
- The second water ring (between boundaries C and D):

${\text{Area}}_{\text{Water Ring 2}}=\pi \times ({7.5}^2-{5.5}^2)=\pi \times (56.25-30.25)=26\pi$
- The third water ring (between boundaries E and F):

${\text{Area}}_{\text{Water Ring 3}}=\pi \times ({13.5}^2-{10.5}^2)=\pi \times (182.25-110.25)=72\pi$
- Total Area of Water:

${\text{Area}}_{\text{Water}}=(6+26+72)\pi =104\pi$

3. Calculating the Ratio:
We compute the ratio of the surface area of the land to that of the water:

$\text{Ratio}=\frac{{\text{Area}}_{\text{Land}}}{{\text{Area}}_{\text{Water}}}=\frac{78.25\pi }{104\pi }=\frac{78.25}{104}\approx 0.7524$
Rounding off to two decimal places gives 0.75.