I tried using pythagoras but I end up with equation with a lot of radicands. And to remove radicands I have to multiply each side on itself and end up with 6 defree equation.
Use right half of your triangle; change your A to a.Code:A x a D x C 10 B
Extend vertical red line to base; label as shown in my diagram.
So we have: a = AB, x = AD = BD, BC = 10 (half the given 20).
So CD = sqrt(x^2 - 100)
So a^2 = [x + sqrt(x^2 - 100)]^2 + 100
A bit of simplifying will lead to: a^2 - 2x^2 = (2x)sqrt(x^2 - 100)
Square both sides:
a^4 - 4a^2 x^2 + 4x^4 = 4x^2(x^2 - 100)
4a^2 x^2 - 400x^2 - a^4 = 0
x^2(4a^2 - 400) = a^4
above simplifies to: x = a^2 / [2sqrt(a^2 - 100]
Disagree Soroban; do you mean isosceles?
All you need is start with any right triangle ABC; a = BC, b = AC, c = AB.Code:D c d B a c C b A
Next, extend CB to CD where BD = c; let d = AD
And cases with integer solutions exist:
as example, a = 7, b = 24, c = 25;
triangle ACD: 24-32-40, thus isosceles triangle: 40-40-48
AND c = d^2 / [2sqrt(d^2 - b^2)]
With the original problem, b = 10, AD = A and AB = x:
x = A^2 / [2sqrt(A^2 - 10^2)]