# Problem 292; An Easy Solution

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John Nolton
Posts: 15

Joined: 10/11/2011
 Problem 292; An Easy Solution Flag »  Reply » The solution that Lindell put forth for problem 292 is long and more complicated than need be. To solve this problem you need to know 3 things. 1. Heron's formula 2. Pythagorean relation 3.The formula for the altitude (alt) of a general triangle #1. the area of triangle ABC is K (area)= sqrt s(s-a)(s-b)(s-c) where s= 1/2(a+b+c)      (this is Heron's formula) #3. Alt = 2K/side AC #2. If you don't know this look it up. Solve for the Area of triangle ABC; s=1/2(6+5+7) , therefore  s= 9 K= sqrt 9(9-5)(9-6)(9-7),  K= 14.696938457 Now Alt of triangle ABC= (2*14.696938457)/7 , therefore Alt= 4.199125273 If we call the foot of the Alt from B to line AC "E" then you will solve 2 triangles using the Pythagorean relation. First we will solve for distance EC in triangle BEC , EC = sqrt 5^2-Alt^2 , EC=2.714285714. Now solve for "x" (distance BD in the problem) using the Pathagorean relation again, DE= 3-EC = 0.285714286 , so x= sqrt DE^2+4.199125273^2; x= 4.208834246 Note: calculations were done on an HP20S calculator and took less than one minute for the solution to the whole problem. John Nolton Tombstone, AZ
Sunday, April 06, 2014 at 1:27:55 PM

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