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Difficulty: HardArea of Two-Dimensional Shapes

A landscape architect is designing two similar gardens. The smaller garden is in the shape of a regular hexagon and has an area of 24324\sqrt{3} square meters. The larger garden is also a regular hexagon, and its perimeter is 33 times the perimeter of the smaller garden. An inscribed circular fountain is to be placed in the center of the larger garden, touching all six sides. What is the area, in square meters, of the circular fountain?

  1. A
    12π12\pi
  2. B
    36π36\pi
  3. 108π108\piAnswer
  4. D
    144π144\pi

Answer

The area of the circular fountain is 108π108\pi square meters.
The side length of the smaller hexagon is calculated to be 44 meters using the area formula for a regular hexagon. Because the larger hexagon's perimeter is scaled by a factor of 33, its side length is 1212 meters. The radius of the inscribed circle is the apothem of the larger hexagon, which is 12×32=6312 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} meters. Squaring this radius and multiplying by π\pi results in an area of 108π108\pi square meters.

Step-by-Step Solution

1
Find the side length of the smaller regular hexagon.
The side length is 44 meters.
The area of a regular hexagon with side length s1s_1 is given by A=332s12A = \frac{3\sqrt{3}}{2} s_1^2. Setting this equal to 24324\sqrt{3} yields s12=16s_1^2 = 16, so s1=4s_1 = 4.
2
Determine the side length of the larger regular hexagon.
The side length is 1212 meters.
Since the two hexagons are similar, the ratio of their perimeters is equal to the linear scale factor of their sides. The perimeter of the larger hexagon is 33 times that of the smaller hexagon, so its side length is s2=3×4=12s_2 = 3 \times 4 = 12.
3
Find the radius of the inscribed circle in the larger hexagon.
The radius is 636\sqrt{3} meters.
The radius rr of the inscribed circle is the apothem of the regular hexagon. For a hexagon with side length s2=12s_2 = 12, the apothem is r=s232=1232=63r = s_2 \frac{\sqrt{3}}{2} = 12 \frac{\sqrt{3}}{2} = 6\sqrt{3}.
4
Calculate the area of the inscribed circular fountain.
The area is 108π108\pi square meters.
Using the area formula for a circle, A=πr2A = \pi r^2, we substitute r=63r = 6\sqrt{3} to get A=π(63)2=108πA = \pi (6\sqrt{3})^2 = 108\pi.

Key Concept

Area of regular polygons, similarity ratio scaling, and properties of inscribed circles.
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Topics

Area of Two-Dimensional Shapes

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