Question

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 (not to scale). The radius of Inner Island was \(2.5\) stades (a unit of length used in ancient Greece). 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).

• A. 0.45
• B. 0.60
• C. 0.75
• D. 0.90

Hint:

Add up the ring widths to get every boundary radius, then use pi(R_out^2 - R_in^2) for each ring's area before forming the ratio.

Answer 1

The Correct Option is C

Step 1: List the radii of every ring boundary.
The Inner Island is a solid circle of radius \(2.5\) stades. Moving outward, each named segment adds its own width to the running radius.
Radius up to the Inner Island (land): \(2.5\).
Radius up to B, after the water ring AB of width \(1\): \(2.5+1=3.5\).
Radius up to C, after the land ring BC of width \(2\): \(3.5+2=5.5\).
Radius up to D, after the water ring CD of width \(2\): \(5.5+2=7.5\).
Radius up to E, after the land ring DE of width \(3\): \(7.5+3=10.5\).
Radius up to F, after the water ring EF of width \(3\): \(10.5+3=13.5\).

Step 2: Sort the rings into land and water.
Land regions are the Inner Island (radius \(0\) to \(2.5\)), ring BC (radius \(3.5\) to \(5.5\)) and ring DE (radius \(7.5\) to \(10.5\)). Water regions are ring AB (radius \(2.5\) to \(3.5\)), ring CD (radius \(5.5\) to \(7.5\)) and ring EF (radius \(10.5\) to \(13.5\)).

Step 3: Work out each land area using \(\pi(R_{out}^2-R_{in}^2)\).
\[ \text{Inner Island}=\pi(2.5)^2=6.25\pi \] \[ \text{Ring BC}=\pi\left(5.5^2-3.5^2\right)=\pi(30.25-12.25)=18\pi \] \[ \text{Ring DE}=\pi\left(10.5^2-7.5^2\right)=\pi(110.25-56.25)=54\pi \]

Step 4: Add up the land area.
\[ \text{Total land}=6.25\pi+18\pi+54\pi=78.25\pi \]

Step 5: Work out and add the water areas the same way.
\[ \text{Ring AB}=\pi\left(3.5^2-2.5^2\right)=\pi(12.25-6.25)=6\pi \] \[ \text{Ring CD}=\pi\left(7.5^2-5.5^2\right)=\pi(56.25-30.25)=26\pi \] \[ \text{Ring EF}=\pi\left(13.5^2-10.5^2\right)=\pi(182.25-110.25)=72\pi \] \[ \text{Total water}=6\pi+26\pi+72\pi=104\pi \]

Step 6: Form the ratio and check the other options.
\[ \frac{\text{Land}}{\text{Water}}=\frac{78.25\pi}{104\pi}=\frac{78.25}{104}\approx0.7524 \] Rounded to two decimal places this is \(0.75\). Option (A) \(0.45\) and option (B) \(0.60\) undercount the land rings, and option (D) \(0.90\) would need more land area than the rings actually give; none of these match the ring-by-ring computation.

Step 7: Final conclusion.
\[ \boxed{0.75} \]