Are these right-angle triangles?
Please can somebody help me with the following problem. I am trying to find x in terms of a,b and c (as per the attached diagram). I came up with the equation below, but this has turned out quite complicated to resolve. Can anyone think of a simpler solution?
x^2 = b^2 + (c - (ac/(x+a)))^2
May be easier by setting up an "example case"; say extending the 3-4-5 right triangle:
I'll complicate your life by changing your variables to different letters!Code:B 6 10 E 8 D 9 9 15 C 8 F 12 A
Start with right triangle ABC; a = BC = 15, b = AC = 20, c = AB = 25
Draw line ED such that similar right triangle DBE is formed; e = BE = 6, d = DE = 8, f = DB = 10
Similar right triangle ADF also is formed; DF = a-e = 9, AF = b-d = 12, AD = c-f (your x) = 15
Givens are BC = a = 15, DB = f = 10 and AF = b-d = 12
PROBLEM: what is (c-f) in terms of a, f and (b-d) ?
I'll try this out later...
That equation of yours (quite correct) can be simplified to:
x^2 = b^2 + [cx / (a + x)]^2
However, I could not find a way to end up with anything simpler than:
x^4 + 2a x^3 + (a^2 - b^2 - c^2) x^2 - 2ab^2 x - a^2 b^2 = 0
Using my example in previous post (and switching to match your variables),
we have a=10, b=12 and c=15; and x=15.
Long division by x - 15 gives:
x^3 + 35x^2 + 256x + 960; so we have:
(x - 15)(x^3 + 35x^2 + 256x + 960) = 0 ; yes, of little use!
Funny how something "appearing so simple" (similar right triangles et al...)
turns out like this; a bit like requiring Heron's area of triangle formula for
a triangle that can be divided in 2 right triangles by a height line...
Enough to make a Saint use all the French Quebec swear words