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).
To solve this problem, we need to find the ratio of the surface area of the land to that of the water in the city of Atlantis. Let's break down the given information and calculate step-by-step.
- Calculate the Area of Each Section:
- The radius of the Inner Island = 2.5 stades.
- The water surrounding the Inner Island is 1 stade wide, so the outer radius at B = 2.5 + 1 = 3.5 stades.
- The width of each land and water ring is as follows:
- Land BC = 2 stades (radius at C = 3.5 + 2 = 5.5 stades).
- Water CD = 2 stades (radius at D = 5.5 + 2 = 7.5 stades).
- Land DE = 3 stades (radius at E = 7.5 + 3 = 10.5 stades).
- Water EF = 3 stades (radius at F = 10.5 + 3 = 13.5 stades).
- Calculate the Surface Area of Land and Water:
- Area of Inner Island (Land) = \(\pi \times (2.5)^2\).
- Area of Water AB = \(\pi \times (3.5)^2 - \pi \times (2.5)^2\).
- Area of Land BC = \(\pi \times (5.5)^2 - \pi \times (3.5)^2\).
- Area of Water CD = \(\pi \times (7.5)^2 - \pi \times (5.5)^2\).
- Area of Land DE = \(\pi \times (10.5)^2 - \pi \times (7.5)^2\).
- Area of Water EF = \(\pi \times (13.5)^2 - \pi \times (10.5)^2\).
- Compute Total Areas:
- Total Land Area = \(\pi \times (2.5)^2 + \pi \times ((5.5)^2 - (3.5)^2) + \pi \times ((10.5)^2 - (7.5)^2)\).
- Total Water Area = \(\pi \times ((3.5)^2 - (2.5)^2) + \pi \times ((7.5)^2 - (5.5)^2) + \pi \times ((13.5)^2 - (10.5)^2)\).
- Calculate the Ratio:
- Simplifying both areas and dividing for the ratio:
- Total Land Area = \(\pi (6.25 + (30.25 - 12.25) + (110.25 - 56.25))\).
- Total Water Area = \(\pi ((12.25 - 6.25) + (56.25 - 30.25) + (182.25 - 110.25))\).
- The ratio of the Land to Water areas is \(\frac{178.5}{238}\), which rounds off to 0.75.
- Simplifying both areas and dividing for the ratio:
Hence, the ratio of the surface area of the land to that of the water in the city of Atlantis is 0.75.
Important Questions from Mensuration and Geometry
In the given figure, PQRS is a square of side 2 cm and PLMN is a rectangle. The corner L of the rectangle is on the side QR. Side MN of the rectangle passes through the corner S of the square.
What is the area (in cm²) of the rectangle PLMN?
Note: The figure shown is representative.
A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius $r$ cm as shown in the figure. The side of the dodecagon is $d$ cm. All the triangles (numbered 1 to 12) in the figure are used to form squares of side $r$ cm and each numbered triangle is used only once to form a square.
The number of squares that can be formed and the number of triangles required to form each square, respectively, are:
Note: The figure shown is representative.
- Which one of the following options has the correct sequence of objects arranged in the increasing number of mirror lines (lines of symmetry)?
- A circle with center at $(x, y) = (0.5, 0)$ and radius $= 0.5$ intersects with another circle with center at $(x, y) = (1, 1)$ and radius $= 1$ at two points. One of the points of intersection $(x, y)$ is:
- During a half-moon phase, the Earth-Moon-Sun form a right triangle. If the Moon-Earth-Sun angle at this half-moon phase is measured to be $89.85^{\circ}$, the ratio of the Earth-Sun and Earth-Moon distances is closest to
