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(sa)(sb)(sc) 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(95)(96)(97), 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 5^2Alt^2 , EC=2.714285714.
Now solve for "x" (distance BD in the problem) using the Pathagorean relation again,
DE= 3EC = 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 