8 The Oregon Surveyor | Vol. 47, No. 5 Featured Article SOME GEOMETRY PROBLEMS Dr. Richard L. Elgin, PS, PE Geometry problems are fundamental to surveying. They involve all aspects of the usual surveying calculation problem such as getting the “picture” of the problem in mind or sketched (somewhat to scale), understanding what is known and what is required, then applying algebra and trigonometry to reach a solution, then considering if the answer seems reasonable. Frequently, the solution requires unit conversions and an awareness of significant digits. Practicing solving geometry problems seems mundane today, but they are fundamental and should remain an important part of high school mathematics curricula. (IMO: If high schools taught more algebra, trigonometry, and geometry and less AP calculus, college students entering surveying and engineering programs would be much better served!) So, here are some geometry problems to challenge those entering or beginning the surveying profession. Handwritten solutions are provided herein. Dr. Richard L. Elgin, PS, PE is a surveying practitioner, educator, researcher, collector, and author. He authored the books “The U.S. Public Land Survey System for Missouri,” “Riparian Boundaries for Missouri,” and “Shoulda Played the Flute” (a memoir of his year flying helicopters in Vietnam). He co-authored the Lietz/ Sokkia ephemeris and co-developed the “ASTRO” celestial observation software products. He can be reached at [email protected]. 1. For the figure, compute its area to the nearest 0.01 square foot. Consider the dimensions given to be exact. Lines EF and AG are parallel and the same length. At C, line CE is tangent to the semicircle centered at D, with radius 15.00 feet. Angle FGA is exactly 90°. X is the radius point for circle sector FH. The radius is 15.37 feet. At H, line HG is tangent to the circle centered at X. 2. For the figure, compute its area to the nearest 0.01 square foot. Consider the dimensions given to be exact. Lines GA and ED are tangent to the circle centered at F. 3. For the figure, derive an equation for its area as a function of R and θ. That is: Area = f(R,θ). At X, the line is tangent to the circle. Reduce the equation to its simplest form. [To check your solution/equation, let θ be exactly 45° and R be exactly 100 feet. If your equation does not result in the area being 43,777.09 square feet, your derived equation is wrong.] Editor’s Note: Turn page 9 upside down to see solutions.
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