The cross section of a gazebo is a 10 foot by 10 foot square. The gazebo is located in an open, level field.
A goat is tethered to one corner of the gazebo by a forty foot rope. The goat can not enter the gazebo.
What is the area over which the goat can graze?
Solution to the Problem:
The area over which the goat can graze is approximately 4,847 square feet.
First, notice that the goat has free range over three quarters of a circle of radius 40 feet (going clockwise or counter-clockwise). So that area is given by the formula for a circle multiplied by 3/4:
The difficulty lies in that fourth section (the bottom left quadrant in the diagram below). If the goat pulls taut on the rope and walks around the gazebo, his tether will wrap around the 10 foot by 10 foot square. The gazebo is labeled ABDC in the diagram.
If the goat is tethered at point A in the diagram and he goes in a counter-clockwise direction, then his path will trace out a quarter of a circle of radius 30 feet and center B. Similarly, if the goat goes in a clockwise direction, then point C will act as the center of a circle with radius 30 feet. These two arcs intersect at E, so we are trying to find the area of section BFEGCD.
The diagram below shows the area over which the goat can graze -- it looks similar to a lima bean.
If you draw a line segment from A to D to E in the first diagram, you will see that the area we are looking for is made up of two congruent triangles (BDE and CDE) and two congruent sectors of a circle (BFE and CGE). So we will determine the areas of sector BFE and triangle BDE and multiply by two. Look at the diagram below.
Angles BAD and BDA have measures of 45 degrees. So, the measure of angle BDE = 135 degrees.
This gives us two sides and an angle in triangle BDE, so we can solve for the other parts of the triangle using the Law of Sines and the Law of Cosines:
Now use the Law of cosines to determine ED:
Now use Heron's formula to determine the area of the triangle:
We can now determine the area of the sector BEF:
Finally, we can add up the areas:
3/4 of the circle with r = 40': 3769.91 sq. ft.
Each sector with r = 30' multiplied by two: 2 (460.48) = 921 sq. ft.
Each triangle multiplied by two: 2 (78.08) = 156.16 sq. ft.
The total area = 3769.91 + 921 + 156.16 = 4847.07 square feet.
The proposer of the problem had a different way of solving the problem.
Click here for James Boyd's method.
Correctly solved by:
| 1. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 2. Veena Mg | Bangalore, Karnataka, India |
| 3. Kelly Stubblefield | Mobile, Alabama |
| 4. Dr. Hari Kishan |
D.N. College, Meerut, Uttar Pradesh, India |
| 5. Anonymous | ------------ |
| 6. Ivy Joseph | Pune, Maharashtra, India |