TO: Mrdavid (re your last thread)
Somehow, no more posts can be made to your thread on isosceles triangles
with same perimeter and area...
Your problem can be solved a bit easier by using 2 right triangles instead (half the isosceles):
Perimeters: a + c = d + f ; Areas: ab = de
C a B F d E
With the base ratio being 8:7, then follows that:
d = 7a/8, e = 8b/7, f = a + c - 7a/8
Now if you follow somewhat "Soroban's logic",
you'll find it all less wieldy...but still nothing "simple"!
Good way to start is using f's length:
a + c + 7a/8 = SQRT[(8b/7)^2 + (7a/8)^2]
Hope that helps; happy 2013!
Perhaps Soroban will comment...