SOLUTION: A dog is tied to the corner of a house with a regular hexagonal base that measures 6 ft on each side. If the rope is 12 ft in length, what is the area in square feet of the region

Question 251485: A dog is tied to the corner of a house with a regular hexagonal base that measures 6 ft on each side. If the rope is 12 ft in length, what is the area in square feet of the region outside the house that the dog can reach ?
(A)108 pi (B) 144 pi (C) 180 pi (D) 216 pi (E) 256 pi
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i believe your answer is selection (A) = 108 * pi.

here's why.

the sum of the interior angles of a hexagon is equal to:

(6-2) * 180 = 4 * 180 = 720 degrees.

since there are 6 angles, then each angle is equal to 120 degrees.

120*6 = 600 + 120 = 720 degrees total so that checks out.

Let the hexagonal house be labeled ABCDEF with:

A = top left
B = top right
C = middle right
D = bottom right
E = bottom left
F = middle left

assume the dog is tied at corner B.

the rope is 12 feet in length which is equal to the radius of a circle that the dog can roam in up to the point where the 12 foot rope cannot be straight anymore.

That point is the extension of BA and BC.

we'll call the extension of BA point X, and the extension of BC point Y.

since interior angle ABC is equal to 120 degrees, this means that the area of the large circle with radius 12 that the dog can freely roam in is 240/360 of the circle.

since the area of the circle is pi*r^2, this means that the area of the circle is equal to pi*12^2 = pi*144.

if we take 240/360 of the area of the circle, then we get 240/360*144*pi = 96*pi.

so far the dog is able to roam 96*pi square feet.

once the dog reaches points X and Y, the dog has some additional room to roam which is equivalent to 60/360 of a circle of radius 6 feet.

since angle BCD measures 120 degrees, then angle YCD measures 60 degrees because it is supplementary to it.

similarly, since angle FAB measures 120 degrees, then angle XAF measures 60 degrees because it is supplementary to it.

At points A and C, we effectively have circles with radius 6 because if you freeze the first 6 feet of the rope at points A and C, then all that is left free to rotate is the additional 6 feet that is left.

the area of these circles is pi*r^2 = pi*6^2 = pi*36

the additional area that the dog is free to roam in is equal to 2 * 60 / 360 * 36 * pi which equals 12 * pi.

96 * pi + 12 * pi = 108 * pi which is selection (A).

a diagram of what I just did is shown in the attached link.

http://theo.x10hosting.com/problems/251485.html

The hexagon is ABCDEF
The rope is tied at point B.
On the left, the rope extends to X.
On the right, the rope extends to Y.
The area of the large circle that the dog is free to roam in is the area of the circle bounded by XMY. This is called Area LC1.
At point A, another circle is formed with a radius of 6 feet.
The area of that small circle that the dog is free to roam in that is in addition to the area already covered by the large circle is the area bounded by AXF. This is called Area LC2.
At point C, another circle is formed with a radius of 6 feet.
The area of that small circle that the dog is free to roam in that is in addition to the area already covered by the large circle is the area bounded by CYD. This is called Area LC3

The sum of areas LC1, LC2, and LC3 are the total area that the dog is free to roam in.

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